Transformation Groups For Beginners PDF

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Transformation
Groups for
Beginners
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STUDENT MATHEMATICAL LIBRARY
Volume 25
Transformation
Groups for
Beginners
S.V. Duzhin
B. D. Chebotarevsky
i JAM,
AMERICAN MATHEMATICA L SOCIET Y
Editorial Boar d
Davide P . Cervon e R o b i n Forma n
Daniel L . Gorof f B r a d Osgoo d
Carl P o m e r a n c e (Chair )
T h i s wor k wa s originall y publishe d i n R u s s i a n b y
BbimBHinatf IIlKOJia , Minsk , u n d e r t h e titl e
O T OpnaMeHTO B ,zi o X[H(})(j)epeHH;HajiBHi>i x ypaBHeHH H i n 1 998 .
T h e presen t t r a n s l a t i o n wa s create d u n d e r licens e fo r t h e America n
M a t h e m a t i c a l Societ y a n d i s publishe d b y permission .
T r a n s l a t e d fro m t h e Russia n b y S . V . D u z h i n .
2000 Mathematics Subject Classification. Primar y 22-01 , 54H1 5 .
For additiona l informatio n a n d u p d a t e s o n t h i s b o o k , visi t

www.ams.org/bookpages/stml-25
Library o f Congres s Cataloging-in-Publicatio n D a t a

Duzhin, S . V . $ q (Serge i Vasil'evich) , 1 956 -
[Ot ornamento v d o difFerentsial'nyk h uravnenii . English ]
Transformation group s fo r beginner s / S . V . Duzhin , B . D . Chebotarevsky ;
[translated fro m th e Russia n b y S . V . Duzhin] .
p. cm . - (Studen t mathematica l library , ISS N 1 520-91 21 ; v. 25 )
Includes index .
Romanized record .
ISBN 0-821 8-3643- 9 (acid-fre e paper )
1. Transformatio n groups-Popula r works . 2 . Algebrai c topology-Popula r
works. I . Chebotarevskii , B . D . $ q (Bori s Dmitrievich ) II . Title . III . Series .
QA385.D8913 200 4
512 / .55-dc22 200404967 6
Copying an d reprinting . Individua l reader s o f thi s publication , an d nonprofi t

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10 9 8 7 6 5 4 3 2 1 0 9 08 07 06 05 04
Contents
Preface i x
Introduction
Chapter 1 . Algebr a o f Point s 7

§1. Checkere d plan e 7
§2. Poin t additio n 0
§3. Multiplyin
1 g point s b y number s 4
§4. Centr e o f gravit y 7
§5. Coordinate s 2 0
§6. Poin t multiplicatio n 2 4
§7. Comple x number s 3 0
Chapter 2 . Plan e Movement s 4 1

§1. Paralle l translation s 4 1
§2. Reflection s 4 4
§3. Rotation s 4 7
§4. Function s o f a comple x variabl e 5 0
§5. Compositio n o f movement s 5 5
§6. Glid e reflection s 6 1
§7. Classificatio n o f movement s 6 3
VI Contents
§8. Orientatio n 6 6
§9. Calculu s o f involution s 6 8
Chapter 3 . Transformatio n Group s 7 3

§1. A rolling triangl e 7 3
§2. Transformatio n group s 7 6
§3. Classificatio n o f finit e group s o f movement s 7 8
§4. Conjugat e transformation s 8 0
§5. Cycli c group s 8 6
§6. Generator s an d relation s 9 0
Chapter 4 . Arbitrar y Group s 9 7

§1. Th e genera l notio n o f a grou p 9 7
§2. Isomorphis m 0 6
1
1
§3. Th e Lagrang e theore m 8
Chapter1
5 . Orbit s an d Ornament s 2 7
§1. Homomorphis m 2 7
§2. Quotien t grou p 3 1
§3. Group s presente d b y generator s an d1
relation s 3 6
1
§4. Grou p action s an d orbit s 3 7
1
§5. Enumeratio n o f orbit s 4 1
§6. Invariant s 4 8
1
§7. Crystallographi c group s 5 1
Chapter 6 . Othe r Type s o1

f Transformation s 6 5
1
§1. Affin e transformation s 6 5
§2.1
Projectiv e transformation s 6 9
§3. Similitude s 7 5
§4. Inversion s 8 2
1
§5. Circula r transformation s 8 7
1
§6. Hyperboli c geometr y 9 1
Chapter 7 . Symmetrie s o f Differentia

1 l Equation s 9
Licensed to Penn St Univ, University Park. Prepared on Tue Mar 12 18:16:11 EDT 2019for download from IP 132.174.254.159. 7
Contents vn
§1. Ordinar1
y differentia l equation s 9 7
§2. Chang e o f variables 20 2
§3. Th e Bernoull i equatio n 20 3
§4. Poin t transformation s 20 7
1
§5. One-paramete r group s 2 4
§6. Symmetrie s o f differentia
1 l equation s 2 6
§7. Solvin g equation s b y symmetrie s 22 0
Answers, Hint s an d Solution s t o Exercise s 22 9
Index 24 5
Preface
The first , Russian , versio n o f thi s boo k wa s writte n i n 1 983-1 98 6 b y

B. D . Chebotarevsk y an d mysel f an d publishe d i n 1 98 8 b y "Vyshei -
shaya Shkola" (Minsk ) unde r the title "Fro m ornaments to differentia l
equations". Th e picture s wer e draw n b y Vladimi r Tsesler .
Years wen t by , an d I kep t receivin g positiv e opinion s abou t th e
book fro m bot h acquaintance s an d strangers . I n 1 99 6 I decide d t o
translate th e boo k int o English . I n th e cours e o f doin g so , I trie d t o
make th e boo k mor e consisten t an d self-contained . I delete d som e
unimportant fragment s an d adde d severa l ne w sections . Also , I cor -
rected man y mistake s ( I can onl y hope I did no t introduc e ne w ones).
The translatio n wa s finishe d b y th e yea r 2000 , an d i n tha t yea r
the Englis h tex t wa s furthe r translate d int o Japanes e an d publishe d
by Springe r Verla g Tokyo under th e title "Henkangu n Nyumon " ("In -
troduction t o Transformatio n Groups") .
The boo k i s intende d fo r colleg e an d graduat e students . It s ai m
is to introduce th e concep t o f a transformation group , usin g example s
from differen t area s o f mathematics . I n particular , th e boo k include s
an elementar y expositio n o f th e basi c idea s o f Sophu s Li e relate d
to symmetr y analysi s o f differentia l equations , whic h hav e no t ye t
appeared i n th e popula r literature .
IX
X Preface
The book contains many exercises with hints and solutions, whic h
will hel p a diligen t reade r t o maste r th e material .
The presen t version , update d i n 2002 , incorporate s som e ne w
changes, includin g th e correctio n o f errors an d misprint s kindl y indi -
cated b y the Japanes e translator s S . Yukita (Hose i University, Tokyo)
and M . Nagur a (Yokoham a Nationa l University) .
S. Duzhin
St. Petersburg
September 1 , 2002
http://dx.doi.org/10.1090/stml/025/01
Introduction
Probably, th e on e mos t famou s boo k i n th e whol e histor y o f mathe -

matics i s Euclid' s "Elements" . I n Europ e i t wa s use d a s a standar d
textbook o f geometr y i n al l school s fo r abou t 200 0 years .
One o f th e firs t theorem s i n th e "Elements " i s th e followin g
Proposition 1 .5 , o f whic h w e quote onl y th e first half .
Theorem 1 (Euclid) . In isosceles triangles the angles at the base
are equal to one another.
Proof. Ever y hig h schoo l studen t know s th e standar d moder n proo f

of thi s proposition . I t i s very short .
B & l\C
Figure 1 . A n isoscele s triangl e
STANDARD PROOF . Le t ABC b e the given isosceles triangle (Fig -

ure 1 ) . Sinc e AB = AC, ther e exist s a plan e movemen t (reflection )
that take s A t o A, B t o C an d C to B. Unde r thi s movement, /.ABC
goes int o /ACB\ therefore , thes e tw o angle s ar e equal . •
2 Introduction
It seem s tha t ther e i s nothin g interestin g abou t thi s theorem .

However, wai t a littl e an d loo k a t Euclid' s origina l proo f (Figur e 2) .
DE
Figure 2 . Euclid' s proo f
O n th e prolongation s AD an d AE
EUCLID'S ORIGINA L PROOF .
of the side s AB an d AC, choose points F an d G such that AF — AG.
Then AABG = AACF; henc e Z.ABG = ZACF. Als o ACBG =
ABCF; henc e ZCE G - ABCF. Therefor e Z.ABC = ZAB G -
ZCBG = Z.ACF - Z.BCF = ZACB. •
In mediaeval England, Propositio n 1 . 5 was known under th e nam e

of pons asinorum (asses ' bridge). I n fact , th e par t o f Figure 2 forme d
by th e point s F , B, C , G an d th e segment s tha t joi n the m reall y
resembles a bridge . Poo r student s wh o coul d no t maste r Euclid' s
proof wer e compare d t o asse s tha t coul d no t surmoun t thi s bridge .
From a moder n viewpoin t Euclid' s argumen t look s cumbersom e
and weird . Indeed , why di d h e eve r nee d thes e auxiliar y triangle s
ABG an d ACF? Wh y wa s he no t happ y jus t wit h th e triangl e ABC
itself? Th e reaso n i s tha t Eucli d jus t coul d no t us e movements i n
geometry: thi s wa s forbidde n b y hi s philosophy , statin g tha t "math -
ematical object s ar e alie n t o motion" .
This exampl e show s tha t th e us e o f movement s ca n elucidat e ge -

ometrical fact s an d greatl y facilitat e thei r proof . Bu t movement s ar e
Introduction 3
Figure 3 . Asses' s Bridg e
important no t onl y when the y ar e studied separately . I t i s very inter -

esting t o stud y th e social behaviour o f movements , i.e . th e structur e
of sets of interrelated movements (o r mor e genera l transformations) .
In thi s area , th e mos t importan t notio n i s tha t o f a transformation
group.
The theor y o f groups , a s a mathematica l theory , appeare d no t
so lon g ago , onl y i n th e nineteent h century . However , example s o f
objects tha t ar e directl y relate d t o transformatio n group s ha d bee n
created bac k i n ancien t civilizations , bot h orienta l an d occidental .
This refer s t o th e ar t o f ornament, calle d "th e oldes t aspec t o f highe r
mathematics expresse d i n a n implici t form " b y th e famou s twentiet h
century mathematicia n Herman n Weyl .
Figure 4 shows tw o example s o f ornament s foun d o n th e wall s of
the mediaeva l Alhambr a Palac e i n Spain .
Both pattern s ar e highl y symmetri c i n th e sens e tha t the y ar e
preserved b y man y plan e movements . I n fact , th e symmetr y proper -
ties o f Figur e 4 a ar e ver y clos e t o thos e o f Figur e 4b : eac h ornamen t
has a n infinit e numbe r o f translations, rotation s b y 90 ° an d 1 80° , re-
flections an d glid e reflections . However , the y ar e no t identical . Th e
difference betwee n the m i s in the wa y thes e movements ar e relate d t o
4 Introduction
ab
F i g u r e 4 . Tw o ornament s fro m Alhambr a
each othe r fo r eac h o f th e tw o patterns . Th e exac t meanin g o f thes e

words can only be explained i n terms of group theory, which says tha t
the symmetr y group s o f Figure s 4 a an d 4 b ar e no t isomorphi c (thi s
is the content s o f Exercis e 1 29 , a t th e en d o f Chape r 5) .
The proble m o f determining an d classifyin g all the possibl e type s
of wal l patter n symmetr y wa s solve d i n th e lat e nineteent h centur y
independently b y the Russia n scientis t E . S. Fedorov an d th e Germa n
scientist A . Schoenflies. I t turne d ou t tha t ther e ar e exactl y 1 7 differ -
ent type s o f plan e crystallographi c group s (se e th e tabl e a t th e en d
of Chapte r 5) .
Of course , th e significanc e o f grou p theor y goe s fa r beyon d th e
classification o f plane ornaments. I n fact, i t is one of the key notions in
the whole of mathematics, widely used in algebra, geometry , topology ,
calculus, mechanics , etc .
This boo k provide s a n elementar y introductio n int o th e theor y

of groups . W e begi n wit h som e example s fro m elementar y Euclidea n
geometry, wher e plan e movement s pla y a n importan t rol e an d th e
ideas o f grou p theor y naturall y arise . The n w e explicitl y introduc e
the notio n o f a transformatio n grou p an d th e mor e genera l notio n o f
an abstrac t group , an d discus s th e algebrai c aspect s o f grou p theor y
and it s application s i n numbe r theory . After tha t w e pas s t o grou p
Introduction 5
actions, orbits , invariants , an d som e classificatio n problems , an d fi -

nally g o as far a s the applicatio n o f continuous group s t o th e solutio n
of differentia l equations . Ou r primar y ai m i s to sho w ho w th e notio n
of group work s in differen t area s o f mathematics, thu s demonstratin g
that mathematic s i s a unifie d science .
The boo k i s intended fo r peopl e with the beginnin g o f a basic col-
lege mathematica l education , includin g th e knowledg e o f elementar y
algebra, geometr y an d calculus .
You wil l fin d man y problem s give n wit h detaile d solutions , an d
many exercises , supplie d wit h hint s an d answer s a t th e en d o f th e
book. I t goe s withou t sayin g tha t th e reade r wh o want s t o reall y
understand what' s goin g o n mus t tr y t o solv e a s man y problem s a s
possible.
Chapter 1
Algebra o f Point s
In thi s chapte r w e wil l introduc e algebrai c operations , additio n an d

multiplication, i n th e se t o f points i n th e plane . Thi s wil l allo w u s t o
apply algebr a t o geometr y an d geometr y t o algebra .
1. Checkere d plan e
Consider a plan e wit h a regula r squar e grid , i.e . tw o set s o f paralle l
equal-distanced lines , perpendicula r t o eac h other . W e wil l b e inter -
ested i n th e polygon s wit h al l vertice s a t node s o f th e grid , lik e th e
isosceles triangl e o r th e squar e show n i n Figur e 1 .
1 111 1111
HI
>H (UrHl i
Ni T >4
H
|TT
\
\
\
mm \ ML
n
\ \ \ \ \
F i g u r e 1 . Polygon s i n th e checkere d plan e
Problem 1 . Prove that a regular polygon different from a square

cannot have all its vertices at nodes of a square grid.
8 1. Algebr a o f Point s
Solution. Suppose , o n th e contrary , tha t suc h a poly -

gon A1 A2 . . . A n exists . Le t O b e it s centre . Fo r ever y
triple o f consecutiv e vertice s Ak-iAÂk+i find a poin t
Bk whic h i s th e fourt h verte x o f th e parallelogra m
Ak-\AkAk+iBk> Th e whol e constructio n o f Figur e 2
goes into itself under the reflection wit h axis OAk an d un -
der th e rotatio n throug h 360/ n degree s around th e poin t
O. Therefor e ever y poin t Bk lie s o n th e correspondin g
line OAk, an d B\B 2 . . . Bk i s a regular polygon . I f n > 6,
then thi s polygo n i s smaller tha n th e initia l one . Indeed ,
in thi s cas e th e angl e a = ^ ^ l S O 0 i s greate r tha n th e
angle (3 = f 360°; henc e the poin t Bk belong s t o th e seg -
ment OAk . I t i s a crucial observatio n tha t al l th e point s
£ 1 , £?2, . • • > B n li e agai n a t node s o f th e squar e grid .
F i g u r e 2 . Regula r polygo n
Repeating th e sam e procedur e fo r th e polygo n

B1B2 . . . B n instea d o f A1 A2 . .. A n, w e wil l arriv e a t a
third polygo n C1 C2 . .. C n whos e vertice s hav e th e fol -
lowing properties :
• the y coincid e wit h node s o f th e grid , an d
• Ck belong s t o th e segmen t OAk an d lie s closer t o O
Licensed to Penn St Univ, University Park. than Bk.
Prepared on Tue Mar 12 18:16:11 EDT 2019for download from IP 132.174.254.159.
1. Checkere d plan e 9
Since ther e ar e onl y finitel y man y intege r point s o n

the segmen t OA& , afte r severa l iteration s o f thi s proce -
dure w e will arriv e a t a contradiction .
The sam e argumen t remain s vali d als o in the cas e of
a regula r pentagon , th e onl y differenc e bein g tha t no w
the poin t B^ lie s o n th e lin e passin g throug h O an d Ak
outside o f the segmen t OAk-
Figure 3 . I s ther e suc h a n equilatera l triangle ?
If n — 3 o r 6 , th e argumen t fail s (why?) , an d w e

will giv e a differen t proo f o f ou r assertion . Not e first
of al l tha t thre e vertice s o f a regula r hexago n for m a n
equilateral triangle ; thu s i t i s sufficien t onl y t o conside r
the cas e n = 3 . Suppos e tha t a n equilatera l triangl e ha s
all it s vertice s a t node s o f th e checkere d plan e (Figur e
3). Then , b y Pythagoras' theorem , th e squar e of the sid e
of thi s triangl e mus t b e a n intege r (w e assum e tha t th e
grid is 1 by 1 ) ; hence its area S = a 2 \/3/4 is an irrationa l
number. O n the other hand , th e triangle A\ A2A 3 can b e
obtained fro m a rectangle wit h intege r side s by removin g
three righ t triangle s a s show n i n Figur e 3 ; thu s it s are a
must b e rationa l — i n fact , eithe r m o r m - f \, wher e m
is a whole number .
Exercise 1 . Suppos e tha t th e side s o f th e square s makin g th e gri d

are 1 . I s there a right triangl e with all vertices at node s such that
all its sides have integer lengths and no side is parallel to the lines
of the grid ?
2. Poin t additio n
Our solutio n o f Problem 1 was based o n the followin g nic e property of
the intege r grid : i f thre e vertice s o f a parallelogra m ar e a t nodes , s o
is th e fourth . Th e usua l mathematica l wordin g fo r thi s phenomeno n
is: the set of all nodes is closed with respect to the operation under
study. W e wil l no w giv e a n exac t definitio n o f thi s operation .
Given thre e point s i n th e plane , sa y M , N an d P , ther e ar e
three differen t way s to ad d on e more point s o that th e triangl e MNP
becomes a parallelogram. On e way is to connect P wit h th e midpoin t
K o f MN an d choose the poin t L o n the lin e PK whic h i s symmetri c
to P wit h respec t t o K (Figur e 4) .
*M + N
PM
Figure 4 . Poin t addition
Definition 1 . W e wil l cal l th e poin t L thu s constructe d the sum of

the points M and N over the pole P , an d w e will writ e L = M - f TV,
p
which shoul d b e rea d alou d a s " M plu s N ove r P " . Whe n th e pol e i s
fixed, w e may omit i t fro m th e notation an d simpl y write L = M + N.
This definitio n hold s for a n arbitrar y tripl e o f points i n the plane .

If M ,T V and P belon g t o on e straigh t line , the n th e parallelogra m
MPNL degenerate s int o a lin e segment. I f al l o f them coincide , the n
it degenerate s eve n mor e an d become s a point .
2. Poin t additio n 11
Now we can give an exact statement fo r the property of the intege r

grid tha t wa s use d i n Proble m 1 : the sum of any two nodes of the
grid over any other node is always a node.
Let u s no w forge t abou t th e gri d an d stud y th e propertie s o f
addition fo r arbitrar y points .
Exercise 2 . Give n two triangles ABC an d DEF an d a point P , de -
note by3> the set of all points M + N wher e M i s an interior point
p
of A ABC an dA T a n interior poin t o f &DEF.
a) Prov e that $ i s a polygon. Ho w many sides may it have ?
b) Prov e that it s perimeter i s the su m of the perimeters o f the two
given triangles.
Point additio n i s closel y relate d t o vecto r addition : L = M + N
p
is equivalen t t o PL — PM + PAT , and enjoy s simila r properties :
1° Th e associativ e la w
(A + B) + C = A + {B + C)
pp p p
holds fo r an y arbitrar y point s A , B, C ove r an y pol e P .

2° W e alway s hav e
P + A = A,
p
i.e., th e poin t P behave s a s a neutra l elemen t wit h respec t
to th e operatio n + .
p
3° Ove r a give n pol e P , ever y poin t A ha s a n opposite point ,
i.e., a poin t A! suc h tha t
A + Af = P.
p
In fact , on e can simply take the point A! whic h is symmetri c
to A wit h respec t t o P .
4° Th e commutativ e la w
A+B=B+A
pp
holds fo r an y thre e arbitrar y points .
The order i n which these 4 items appear i n our lis t i s not acciden -
tal — in fact, mor e fundamental rule s come first. Yo u will understan d
this bette r whe n yo u ge t t o Chapte r 4 .
Rules 2 ° — 4° ar e obvious an d d o not requir e an y proof. T o check
rule 1 ° , we first construc t th e point s M = A + B an d N = B + C (se e
pp
Figure 5) . Th e segment s AM an d CN ar e bot h equa l an d paralle l
to th e segmen t PB. Henc e th e midpoint s o f MC an d AN coincide ,
which, b y th e definitio n o f poin t addition , ensure s tha t M + C =
A + N.
F i g u r e 5 . Associativit y o f poin t additio n
Using propert y 3° , w e ca n defin e th e difference o f tw o point s

over a give n pole : B — A = B + A', wher e A! = — A i s th e poin t
pp P
opposite to A. Th e point B — Ais th e unique solution t o the equatio n
A + X = B.
p
If al l operations ar e carrie d ou t ove r the sam e pole, then additio n
and subtractio n o f point s satisfie s th e sam e rule s a s th e usua l opera -
tions o n numbers , fo r example , A — (B — C + D) = A — B + C — D.
Problem 2 . Find the sum A + B + C, where M is the intersection

MM
point of the medians in a triangle ABC.
2. Poin t additio n 13
Figure 6 . Su m o f vertice s o f a triangl e
Solution. Recal l tha t eac h media n i s divide d b y thei r

common intersection point M i n the ratio 2 : 1; therefore ,
in Figure 6, we have CM = 2MK. Th e point D = A + B
M
lies o n th e prolongatio n o f th e media n CK, an d DK —
KM = \MC. Therefore , DM = MC an d D + C = M.
Z
M
It i s interesting t o observ e tha t th e intersectio n poin t o f th e me -

dians i s the onl y point whic h satisfies A-\- B -\-C — M. T o prove this,
MM
let u s first deriv e the rule s for passin g fro m on e pole to anothe r i n th e
formulas involvin g poin t addition :
(1) A + B = A + B-Q,
QP P
(2) A-B = A-B + Q.
QP P
The firs t equalit y ca n b e rewritte n a s (A + B) + Q = A + Z? , an d

QP P
its validit y i s easil y see n fro m Figur e 7 . T o prov e th e secon d one ,
we wil l chec k tha t th e poin t A — B + Q i s a solutio n t o th e equatio n
pP
B + X = A. Indeed , usin g th e formul a w e have just proved , w e ge t
Q
B+{A-B + Q)=B+(A-B + Q)-Q = A.
QPark.
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P on TuePMar 12 18:16:11 EDTP2019for download
P P IP 132.174.254.159.
from
14 1. A l g e b r a o f P o i n t s
Figure 7 . Chang e o f bas e poin t
Note t h a t th e poin t P doe s no t appea r i n th e left-han d side s o f

(1) an d (2) ; henc e th e right-han d side s d o no t depen d o n it s choice .
This observatio n i s generalize d i n th e followin g exercise .
E x e r c i s e 3 . Investigat e th e condition s unde r whic h th e expressio n
Ai + A 2 + • • • + A k - B x - B 2 Bi
pp p p p p p
does no t depen d o n th e choic e o f th e pol e P.
Continuing th e discussio n o f Proble m 2 , suppos e t h a t a poin t

N ha s t h e sam e propert y a s t h e media n intersectio n poin t M , i.e .
A + B + C — N. W e ca n subtrac t th e pol e withou t violatin g th e
NN
equation; henc e A + B + C — N — N = N. A reade r wh o ha s don e
NN N N
Exercise 3 know s t h a t th e left-han d sid e o f thi s relatio n doe s no t
depend o n th e choic e o f N.. I n particular , substitutin g M i n plac e
of N, w e ge t A + B + C - N - N = TV ; therefor e M-N-N = N,
MM M M MM
N + N + N = N an d finally N = M. Thi s mean s t h a t th e media n
MM
intersection poin t M i s th e uniqu e poin t wit h th e propert y prove d i n
Problem 2 .
E x e r c i s e 4 . Prov e tha t A + B + C = H, wher e O i s th e centr e o f
oo
the circl e circumscribe d aroun d th e triangl e ABC an d H i s th e
intersection poin t o f it s thre e altitudes .
3. Multiplyin g point s b y number s

Over a give n pol e P , a poin t A ca n b e multiplied b y a rea l numbe r a
yielding a ne w poin t B = apA.
3. Multiplyin g point s b y number s 15
Definition 2 . Th e product of a point A by a real number a over the

pole P i s the poin t B tha t lie s on the lin e PA a t th e distanc e |a||P^4 |
from th e pol e P an d o n th e sam e sid e o f P a s A, i f a > 0 , o r o n th e
other side , i f a < 0.
PA a PA
F i g u r e 8 . Multiplicatio n o f point s b y number s
In othe r words , thi s operatio n mean s tha t yo u stretc h th e vecto r

PA, keepin g it s initia l poin t P fixed, a s i f wit h a pin : PB = aPA.
In particular ,
(1) an y poin t multiplie d b y zer o ove r P give s P , an d
(2) P multiplie d b y an y rea l numbe r give s P .
It i s eas y t o se e tha t multiplicatio n o f point s b y number s ha s
these properties :
5° 1 PA = A.
6° ap((5 PA) = {a/3) PA.
7° (a + p)pA = a PA + (3pA.
p
8° ap(A + B) = apA + CLpB.
pP
To multiply a point b y a natural numbe r n i s the same thing a s t o
add u p n equa l points : npA = A + A + ... + A(A repeate d n times) .
pp p
Using thi s fact , yo u ca n chec k tha t th e poin t \ pA i s th e (unique )
solution t o th e equatio n X + X — A.
p
Consider a linear combination over the pole P , i.e . th e su m o f
several point s wit h arbitrar y coefficient s
(3) apA + (3pB + . . . +UJ PZ = S.
pP
In general , th e resultin g poin t S depend s o n the choic e of the pol e P .
When a , / ? , . . ., u; are integer numbers , we have seen in Exercise 3 that
there ar e some occasions whe n the resul t doe s not depen d o n P . Thi s
may
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are no t integer . Fo r example , th e poin t M = | ^ 4 + \B i s alway s th e

middle poin t o f th e segmen t AB, whereve r yo u p u t th e pole .
E x e r c i s e 5 ( A generalizatio n o f Exercis e 3) . Fin d a necessar y an d
sufficient conditio n o n th e coefficient s a , (3, . . ., UJ whic h guaran -
tees tha t th e linea r combinatio n S o f (3 ) doe s no t depen d o n th e
choice o f th e poin t P .
AB c
Figure 9 . Poin t o f a segment expresse d i n terms o f endpoint s
Using poin t additio n an d multiplicatio n b y numbers , i t i s possibl e

to expres s an y poin t o f th e segmen t AB i n term s o f it s endpoints .
Indeed, suppos e t h a t th e poin t C divide s th e segmen t AB i n th e
ratio k : / (b y definition , thi s mean s t h a t / • AC = k • CB). Choos e
an arbitrar y poin t P outsid e o f th e lin e AB; w e wil l us e i t a s th e pol e
in al l subsequen t operation s o n points . Throug h th e poin t C w e dra w
two lines , paralle l t o PB an d PA, whic h mee t PA an d PB a t point s
A an d B', respectivel y (se e Figur e 9) . T h e n
PA!_ _ BC__ J_
~PA " ~BA~ k + l'
PB' AC _ k
~PB ~AB ~k +l
Put j ^ = a , ^ = /? . The n C = A! + B' = aA + /3J5 , an d th e su m
of th e tw o coefficient s a an d (3 is 1 .
T h e convers e i s als o true : i f a an d /3 ar e arbitrar y nonnegativ e
numbers suc h t h a t a + (3 = 1 , the n th e poin t C — a A + (3B belong s
to th e segmen t AB. Moreover , i f on e o f th e number s a an d (3 i n
the formul a fo r C i s negative , bu t th e su m o f t h e tw o i s stil l 1 , the n
the poin t C lie s o n th e straigh t lin e AB, bu t outsid e o f th e segmen t
4. C e n t r e o f g r a v i t y 17
AB. B y changin g Figur e 9 appropriately , yo u ca n verif y t h a t th e

relations a = ^ , / ? = ^ remai n valid , althoug h th e rati o k : I is
now negative .
Thus, th e straigh t lin e AB i s th e se t o f al l point s a A + ( 1 — a)J3,
where a i s a n arbitrar y rea l number , whil e th e segmen t AB i s it s
subset specifie d b y th e restrictio n 0 < a < 1 . Not e agai n t h a t thi s
description doe s no t depen d o n th e choic e o f th e bas e poin t (pole) .
E x e r c i s e 6 . Fin d a similar descriptio n o f the se t o f all interio r point s
of a conve x polygo n wit h vertice s Ai, A2, . . . , A n.
After doin g Exercis e 6 , yo u ca n g o bac k an d tackl e Exercis e 2
once again , usin g th e ne w technique .
E x e r c i s e 7 . A middl e lin e o f a quadrilatera l i s th e lin e joinin g th e
midpoints o f two opposite sides . An y quadrilateral ha s two middl e
lines. Prov e tha t thes e tw o lines , a s wel l a s th e segmen t joinin g
the midpoint s o f th e tw o diagonals , mee t i n on e point , an d thi s
point divide s eac h o f the m i n hal f (se e Figur e 1 0) .
Figure 1 0 . Quadrilatera l o f exercise 7
4. Centr e o f gravit y
In Proble m 2 , th e media n intersectio n poin t M o f a triangl e ABC
was describe d implicitl y a s th e (unique! ) solutio n t o th e equatio n
A + B + C = M. W e ca n no w expres s M explicitl y i n term s o f A, B
MM
and C. Indeed , multiplyin g bot h side s o f th e equatio n b y | , w e ge t

A.,(A + B + C) = M. Accordin g t o th e answe r o f Exercis e 5 , th e
6M
M M
left-hand sid e o f thi s equalit y doe s no t depen d o n th e choic e o f th e
pole; henc e w e can writ e
M=\(A + B + C).
In a simila r way , th e poin t referre d t o i n Exercis e 7 ca n b e ex -

pressed i n term s o f th e vertice s o f th e quadrilatera l a s
M = i ( A + £ + C + D) .
In general , th e arithmeti c mea n o f severa l point s i s calle d th e

centre of gravity (o r centre of mass) o f the syste m consistin g o f thes e
points: M = ^{A\ - f A2 + V A n) (ove r a n arbitrar y pole) . Thus ,
the centr e o f gravit y o f a triangl e (or , mor e exactly , o f th e se t o f it s
vertices) i s the media n intersectio n point , whil e th e centr e o f gravit y
of th e se t o f vertice s o f a quadrilatera l i s th e intersectio n o f it s tw o
middle lines .
We procee d t o som e example s wher e geometri c problem s relate d
to th e centr e o f gravit y ar e solve d usin g operation s o n points .
Problem 3 . Suppose that A, B and C are three collinear points,
while E and F are arbitrary points in the plane. Prove that the median
intersection points of the triangles AEF, BEF, CEF are collinear.
Solution. Media n intersectio n point s ar e arithmeti c
means o f th e vertices :
l
-(A + E + F) = K,
^(B + E + F) = L,
^{C + E + F) = M.
By assumption , poin t C lie s o n th e lin e AB, thu s
C = <xA + ( 1 - a)B. Henc e
aK + (l- a)L = ^(A + E + F)
= ^ - ^ (B + E + F) = \ {C + E + F) = M,
4. C e n t r e o f g r a v i t y 19
which implie s t h a t th e poin t M belong s t o th e lin e KL.
E x e r c i s e 8 . Le t A , B, C, D, E, F be th e middl e point s o f th e con -

secutive edge s o f a hexagon . Prov e tha t th e centre s o f gravit y o f
the triangle s ACE an d BDF coincide .
E x e r c i s e 9 . I n a quadrilatera l ABCD, th e poin t E i s th e midpoin t
of th e sid e AB an d K th e midpoin t o f th e sid e CD. Prov e tha t
the midpoint s o f th e fou r segment s AK, CE, BK an d ED for m
a parallelogram .
P r o b l e m 4 . Prove that the middle line of a quadrilateral (see Ex-

ercise 7) passes through the intersection point of its diagonals if and
only if this quadrilateral is a trapezoid, i. e. has two parallel sides.
Figure 1 1 . Trapezoi d
S o l u t i o n . W e choos e th e intersectio n poin t O o f th e

diagonals a s th e pol e (se e Figur e 1 1 ) . T h e n C = a A,
D = j3B fo r appropriat e number s a an d /? , an d fo r th e
middle point s K an d L w e ca n writ e K = | ( A - f B),
L=\(aA + i3B).
If A S | | C D , the n th e triangle s OB A an d ODC ar e
similar, henc e a = /? , L = OLK and th e point s if , L , O
are collinear .
Suppose, o n th e othe r hand , t h a t w e d o no t kno w
whether AB i s paralle l t o CD, bu t w e d o kno w t h a t K,
L an d O li e o n th e sam e line . Then , usin g t h e poin t
operations ove r th e pol e O , fo r a suitabl e rea l numbe r 7
we hav e L — ^K. Substitutin g th e previou s expression s

for K an d L , w e ge t aA + (3B = jA - f 7 #, o r ( a — 7 )A =
(7 - /3)B. Bu t th e point s ( a - 7)A an d ( 7 - (5)B li e
on differen t line s OA an d O B , an d i f the y coincide , thi s
must mea n t h a t the y coincid e a t O. Thus , a — 7 =
7 — / ? = 0 , a = /? , triangle s O T 4 £ an d O D C ar e similar ,
and AB \\ CD.
E x e r c i s e 1 0 . Usin g poin t additio n an d multiplicatio n b y numbers ,

find a n independent proo f of the fact tha t th e median s of a triangl e
are divide d i n proportio n 2 : 1 b y thei r intersectio n poin t (not e
that w e hav e use d thi s fac t before , i n Proble m 2) .
E x e r c i s e 1 1 . A lin e cut s 1 / 3 o f on e sid e o f a parallelogra m an d 1 / 4
of th e adjacen t sid e i n suc h a wa y tha t th e smalle r part s hav e a
common verte x (Figur e 1 2) . I n wha t rati o doe s thi s lin e divid e
the diagona l o f th e parallelogram ?
Figure 1 2 . Cuttin g th e diagona l
5. Coordinate s
In th e discussio n o f Proble m 4 , w e hav e use d th e followin g importan t
fact: if two points M, N are not collinear with the pole, then the
equality aM + @N = 7 M + 5N is possible only if a — f3 and 7 = S.
In fact , th e give n equalit y ca n b e rewritte n a s (a — ~f)M = ( 7 — S)AT,
which implie s a = 7 an d (3 = 5.
Choose a pol e P an d tw o point s M , N t h a t ar e no t collinea r wit h
P. T h e n an y poin t Z o f th e plan e ca n b e expresse d a s Z = xM + yN
for suitabl e rea l number s x an d y (Figur e 1 3) .
D e f i n i t i o n 3 . A system of affine coordinates i n th e plan e i s a n or -

dered se t o f thre e non-collinea r point s {P , M, N}. T h e firs t poin t P i s
5. Coordinate s 21
xM P aM
F i g u r e 1 3 . Affin e coordinate s
referred t o a s the pole, o r the origin, whil e the se t {M , TV} i s referre d

to a s th e basis of th e give n coordinat e system . Th e coordinate s o f a
point Z i n the coordinat e system {P , M, N} ar e the coefficients {x, y}
in th e expansio n Z — xpM + ypN.
The abov e argumen t show s tha t th e coordinate s x an d y ar e

uniquely determine d b y th e poin t Z. Thu s w e obtai n a one-to-on e
correspondence betwee n th e point s o f a plan e an d pair s o f rea l num -
bers.
If ZMNP i s a righ t angl e an d bot h PM an d PN ar e uni t seg -
ments, the n wha t w e get i s the usua l Cartesia n coordinat e system . I n
general, suc h coordinate s ar e referre d t o a s affine coordinates.
When tw o point s ar e added , thei r coordinate s ad d up :
(aM + bN) + (cM + dN) = ( a + b)M + ( c + d)N.
When a point i s multiplied b y a number, it s coordinate s ge t mul -

tiplied b y th e sam e number :
c(aM + bN) = (ca)M + (cb)N.
The correspondenc e betwee n point s o f the plan e an d pair s o f rea l

numbers ca n b e use d a s a dictionar y whic h serve s t o translat e geo -
metric proposition s int o th e languag e o f algebr a an d vic e versa . An y
geometric figur e i s th e se t o f al l point s whos e coordinate s satisf y a
certain relation . Fo r example , w e kno w tha t a poin t belong s t o th e

line MN i f an d onl y i f ha s th e expressio n xM + yN wit h x + y = 1 .
In thi s sense , x + y — 1 is the equation of the straight line MN.
Problem 5 . Find the equation of the straight line which is parallel
to MN and passes through the given point K with coordinates a, b.
Solution. Le t M' an d N f b e th e intersectio n point s o f
this lin e wit h PM an d PN, respectively . Sinc e M'N' | |
MN, w e hav e M' = tM, N' = tN fo r a n appropriat e
number t (se e Figure 1 3) . An y poin t Z o f the lin e M'N'
is equa l t o aM' + @N', wher e a + (3 = 1 , i.e. , Z =
atM + (3tN an d at + f3t = t. Thus , th e coordinate s
x = at, y = (5t of an arbitrar y poin t Z € MN satisf y th e
relation x + y = t, wher e th e valu e o f t i s ye t unknown .
To find it , not e tha t th e poin t K lie s o n th e lin e unde r
study; henc e it s coordinate s a , b satisfy th e equatio n o f
this line : a + b — t i s true. W e have found tha t t = a + 6,
and th e answe r t o th e exercis e is x + y — a + b.
Exercise 1 2 . Writ e the equatio n o f the straight lin e that contain s a

given point K(a,b) an d
(a) i s parallel to PM ,
(b) i s parallel t o P7V,
(c) passe s through P.
Problem 6 . Suppose that in a certain triangular region of the plane
the laws of optics are such that a ray of light which goes parallel to one
side of the triangle and hits the second side, after reflection assumes
the direction of the third side of the triangle. Prove that a person
standing inside this triangle and directing the beam of his flashlight
parallel to one of the sides of the triangle, is in fact shining the light
onto his own back.
Solution. Le t on e verte x o f the triangle , P , b e th e pol e

and tw o others , M an d TV , be th e tw o basi c point s o f a
coordinate syste m (Figur e 1 4) .
Suppose tha t th e perso n wit h th e flashlight stand s
at th e poin t K(a, b) and th e bea m o f hi s flashlight goe s
5. Coordinate s 23
Figure 1 4 . Zigza g insid e a triangl e
parallel t o PN an d meet s th e sid e PM a t th e poin t A.

The coordinate s o f A ar e (a , 0), because , o n on e hand ,
KA | | PN an d henc e th e firs t coordinat e o f A i s equa l
to th e firs t coordinat e o f K (se e Exercis e 1 2) , whil e o n
the other hand , poin t A lie s on PM an d henc e its secon d
coordinate i s 0 .
The nex t segmen t o f th e beam , AB, i s paralle l t o
MN. Accordin g t o Proble m 5 , th e equatio n o f th e lin e
AB i s x + y = a, becaus e a is the su m o f the coordinate s
of th e poin t A. Sinc e th e poin t B lie s o n PN, i t ha s
x = 0 ; therefor e it s secon d coordinat e mus t b e equa l t o
a.
Proceeding i n the sam e way, we successively fin d th e
coordinates o f all points wher e the bea m meet s th e side s
of th e triangle : C( l - a,a), D(l - a,0) , £ ( 0 , 1 - a) ,
F(a, 1 — a). Th e lin e FK i s parallel t o PN, whic h i s why
the bea m doe s retur n t o th e initia l poin t K — fro m th e
opposite direction .
A vigilan t reade r ma y hav e notice d a fla w i n th e previou s argu -

ment: i n fact , i t ma y happe n tha t th e bea m return s t o th e poin t
K befor e i t make s th e complet e tou r o f ABCDEF — an d hit s th e
flashlighter i n a side , no t i n th e back .
E x e r c i s e 1 3 . Describ e th e se t o f al l point s K i n th e triangl e MNP

for whic h th e trajector y o f th e flashligh t bea m consist s o f onl y
three segments , no t six .
E x e r c i s e 1 4 . A poin t K lie s insid e th e triangl e ABC. Straigh t line s
AK, BK, CK mee t th e side s BC, CA, AB a t th e point s D, E,
F , respectivel y (Figur e 1 5) . Prov e tha t KD/AD + KE/BE +
KF/CF = 1 .
A E C
Figure 1 5 . Line s i n a triangle meetin g a t on e poin t
E x e r c i s e 1 5 . Give n thre e point s D , E, F o n th e side s o f the triangl e

ABC (Figur e 1 5) , prove that th e line s AD, BE, CF pas s throug h
one poin t i f an d onl y i f AF • BD • CE = FB DC • EA (theore m
of Ceva) .
6. Poin t multiplicatio n
We hav e learne d ho w t o multipl y a poin t i n th e plan e b y a rea l num -
ber. No w recal l t h a t rea l number s ca n b e represente d a s point s lyin g
on a line . Le t u s inser t thi s lin e int o th e plan e s o t h a t it s origi n (zer o
point) coincide s wit h th e pol e P use d t o defin e th e poin t additio n an d
multiplication o f point s b y numbers . Th e uni t poin t o f th e rea l lin e
will b e E (se e Figur e 1 6a) .
Our definitio n o f poin t additio n agree s wit h t h e usua l additio n o f
real number s i n th e sens e t h a t i f t h e point s A an d B correspon d t o
numbers a an d 6 , the n th e su m A + B (ove r th e pol e P) correspond s
to th e numbe r a + b (Figur e 1 6b) .
6. Poin t multiplicatio n 25
P E
a • 0
0
A p A+B B
O " • •
0
a 0 a*b *
p A £ AB 5
—<j 0 0 «
0 o
0a 1 ab b
Figure 1 6 . Algebrai c operation s i n th e lin e
Moreover, ou r multiplicatio n o f point s b y numbers , restricte d t o

the rea l line , als o agree s wit h th e usua l produc t o f number s i n th e
sense tha t i f A <- > a, B <^> 6 , the n bot h point s apB an d bpA corre -
spond t o th e numbe r ab. I t i s natura l t o cal l thi s poin t th e product
of th e tw o point s A an d B an d denot e i t b y AB.
The nex t ste p w e wan t t o mak e i s t o exten d thi s definitio n t o
the entir e plane . W e wan t t o fin d a rul e t o assig n a ne w poin t AB
to an y pai r o f arbitrar y point s A, B i n suc h a wa y tha t thi s point
multiplication satisfie s th e usua l rule s o f multiplication :
9° Associativit y
(AB)C = A(BC).
10° Commutativit y
AB = BA.
11° Distributiv e la w wit h respec t t o poin t additio n wit h th e
same pol e
A(B + C) = AB + AC.
We also require tha t th e ne w operation agre e with th e previousl y
defined multiplicatio n o f points by numbers, i.e. , that fo r an y point Z
in th e plan e an d an y poin t A o n th e rea l lin e tha t correspond s t o th e
number a we should hav e AZ — apZ. I n particular , thi s mean s tha t
the uni t poin t E o f th e rea l lin e mus t pla y th e rol e o f th e numbe r 1

for al l point s Z o f th e plan e i n th e sens e tha t EZ = lpZ = Z.
It i s not immediatel y clea r whether i t is possible to introduce suc h
an operation fo r th e points of the plane . W e will see, however, tha t i n
fact ther e ar e man y way s to d o so, and the y com e i n three essentiall y
different types . Bu t le t u s first d o som e exercises .
Problem 7 . Let EABCDK be a regular hexagon with centre at P

(recall that E is the unit point), and suppose that A 2 = B for a
certain choice of point multiplication. Find all pairwise products of
the vertices of the given hexagon.
o
p
Figure 1 7 . Multiplicatio n o f th e vertice s o f a hexago n
Solution. Expan d al l th e vertice s ove r th e basi s E, A

taking P fo r th e pol e (Figur e 1 7) : B = A - E, C = -E,
D — —A, K = E — A. W e know the product s o f all pair s
consisting o f basi c points : E 2 = E, EA = A, A 2 = B.
Using the distributive law , we can find, fo r example , tha t
BK = (A-E){E-A) = -A 2 + 2AE-E2 = -B + 2A-
E = A. Othe r product s ca n b e foun d i n simila r fashion ,
6. Poin t multiplicatio n 2 7
yielding th e multiplicatio n table :
E A B C D K
E E A B C D K
A A B C D K E
B B C D K E A
C C D K E A B
D D K E A B C
K K E A B C D
Note tha t th e se t o f 6 vertice s o f th e hexago n turn s ou t t o b e

closed unde r th e chose n rul e o f multiplication , i.e. , th e produc t o f
any tw o vertice s i s als o a vertex .
Exercise 1 6 . I s the set of vertices of the same hexagon closed unde r

multiplication, i f (a ) A 2 = A; (b ) A 2 = P? Fil l ou t th e corre -
sponding multiplicatio n tables .
Exercise 1 7 . Fin d the multiplication table for the set of vertices of a

regular pentago n EABCD centre d a t th e pole P, if A 2 i s known
to be equal to B.
After thes e examples , w e ca n investigat e th e genera l case . Sup -

pose that w e are given a point multiplicatio n rul e that satisfie s al l th e
requirements state d above .
Besides th e tw o alread y chose n point s P = 0 an d E = 1 , pic k
an arbitrar y poin t F no t o n th e lin e PE. The n th e pai r (£" , F) i s a
basis ove r P , and , a s w e sa w i n th e discussio n o f Proble m 7 , poin t
multiplication i s completely denned , i f we only know the squar e o f F.
We have F 2 = aE+f3F fo r suitable real numbers a an d /? . Le t u s
try t o find anothe r poin t G such that th e pai r (E, G) i s also a basis in
the plane , bu t th e squar e G 2 ha s a simple r expansio n ove r thi s basis .
Let G = F - § E. The n th e line s FG an d PE ar e parallel , s o

that E an d G constitut e a basis , an d
2
=F -(3EF+^-E2
If you look closely at thi s relation, yo u will see that multiplicatio n

in th e basi s (E,G) look s simple r tha n i n th e initia l basi s (i£ , F ),
because th e squar e o f th e secon d basi c poin t i s no w jus t E wit h a
certain coefficien t — an d no t th e combinatio n o f th e tw o points , a s
before. T o furthe r simplif y th e multiplicatio n rule , w e will chang e G
once again , dependin g o n th e sig n o f thi s coefficient .
(32
(1) a + — = 0 (cf . Exercis e 1 6b) . I n thi s cas e th e produc t i s
given b y th e formula s
E2 = E, EG = G,G 2
= 0;
{aE + bG)(cE 4 - dG) = acE + (ad + bc)G.
2
1(3
(2) a + — - > 0 . Denotin g =G b y H, i n th e basi s
4 y/a + /3 2 /4
(E, H) w e will hav e th e followin g rule s o f multiplication :
E2 = E, EH = H, H 2 = E\
(aE + bH)(cE + dH) = (ac + bd)E + (ad + 6c)ff .
(Try t o fin d suc h a point H amon g th e vertice s o f the hexa -
gon i n Exercis e 1 6b. )
g2 I
(3) a + ^ - < 0 . Se t / = G . The n
4
vl a + /?2/4|
£2 = £ , E J= /, / 2
= -F ;
(aE + 61 )(c £ + d/ ) = (ac - bd)E + (a d + bc)H.
It i s easil y verifie d tha t i n eac h o f th e thre e case s ou r operatio n
satisfies al l th e law s impose d o n multiplication . Th e nex t questio n
that naturall y appear s i s whethe r thi s multiplicatio n ha s a n invers e
6. Poin t multiplicatio n 29
operation o f division, i.e. , whethe r th e equatio n AZ — B ca n alway s

be solve d fo r Z , provide d tha t A ^ 0 .
In th e firs t cas e le t u s tr y t o divid e E b y G , i.e. , find a poin t
Z — xE + yG suc h tha t GZ = E. Accordin g t o th e definition ,
G{xE-\-yG) — xG, whic h never equals E. Thus , divisio n i s in genera l
impossible.
The sam e i s true i n th e secon d case , where, a s yo u ca n check , H
is no t divisibl e b y E - f H.
We claim , however , tha t i n th e thir d cas e divisio n b y a non-zer o
point i s alway s possible . Indeed , le t M = aE + bl, N — cE + dl,
where th e coefficient s c and d do not vanis h simultaneously . W e wan t
to find th e quotien t M/N, tha t is , a poin t Z = xE + yl suc h tha t
NZ = M , o r (cE + dI)(xE + yl) = aE + 67 . Whe n expanded , thi s
equality become s equivalen t t o th e syste m o f equation s
ex — dy — a ,
dx + cy — b,
,. , , . , . ac + bd be — ad .
which ha s a uniqu e solutio n x — —z
z -r-l , y = - zr ^ , provide d tha t
c + d c + cr
c2 + d 2 ^ 0 .
The resul t o f ou r investigatio n ca n b e state d a s follows .
Theorem 2 . Multiplication of points in the plane can be introduced
in three essentially different ways, depending on the existence of an
element X with the property
(1) X 2 = 0 ,
(2) X 2 = 1 ,
(3) X 2 = - 1.
Only in case (3) is division by non-zero elements always possible.
Speaking more formally, there exist three different two-dimen-
sional algebras over the field of real numbers, and only one of them
(case 3) is an algebra with division.
Note tha t th e actua l geometri c meanin g o f multiplication , say ,

in cas e (3) , depend s o n th e mutua l positio n o f point s E an d / wit h
respect
Licensed t oUniversity
to Penn St Univ, th e origi n P.on Tue
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18:16:11 EDT 2019for t PEAB b e a squar e draw n
from IP 132.174.254.159.
on th e segmen t PE. Wher e i s th e poin t A 2 ? Thi s depend s o n th e

choice of / . I f / coincide s wit h A , the n A 2 = — E. HI coincide s wit h
B, the n
A2 = (B + E) 2 = B 2 + 2BE + E 2 =-E+ 2B + E = 2B.
Of course , othe r choice s ar e als o possible , givin g othe r answers .
Among all these possibilities w e now choose the one where / = A ,
i.e., I i s obtaine d fro m E b y a rotatio n throug h 90 ° i n th e positiv e
direction (counterclockwise) , an d w e stud y i t i n mor e detai l i n th e
next section .
7. Comple x number s
The point s o f the lin e PE ar e identified wit h real numbers . No w tha t
we hav e introduce d algebrai c operation s fo r th e point s o f th e plane ,
we ca n vie w th e se t o f al l point s a s a numbe r syste m whic h i s wide r
than rea l numbers . Thes e number s ar e calle d complex numbers. I n
the conventional notatio n fo r comple x numbers, our pol e P i s denoted
by 0, point E b y 1 , point / b y i or >/—T > a nd a+bi i s written instea d of
aE + bl. Her e are, once again, th e definition s fo r algebrai c operation s
on comple x number s i n thi s standar d notation :
(a + bi) + (c + di) = ( a + c) + (b + d)i,
(a + bi) — ( c + di) — (a — c)-\-(b — d)i,
(a + bi)(c + di) = (ac-bd) + (ad + bc)i,
a + bi ac + bd be — ad .
= 2
c + di c + d 2 + c 2 + d 2%'
To put i t shortly , th e operations ar e performed a s if on polynomi -
als in the "variable " i with the rule i 2 = — 1 applied whenever possible .
To derive the formul a fo r th e quotien t fro m thi s rule , bot h numerato r
and denominato r shoul d b e multiplie d b y th e sam e number , c — di.
The tw o basic complex numbers 1 and i ar e referred t o a s the real
unit an d th e imaginary unit, respectively .
Exercise 1 8 . Perfor m th e operations o n complex numbers:
(a) — - I ( 3 + J ) + T ,
(b) y/3=Ti,
7. Comple x number s 31
(c)
U - TV •
Let 2 = a + 6i . Th e distanc e betwee n th e point s z an d 0 i s
called th e modulus, o r absolute value, o f th e comple x numbe r z , an d
is denote d b y \z\. Sinc e a an d b ar e Cartesia n coordinate s o f th e
point z, w e hav e \z\ = \Ja 2 + 6 2. Fo r example , th e modulu s o f bot h
1 2t — t2
cos t+z sin £ and « -f ~ i is 1 for an y value of the real numbe r
1 -f11 l + tl
t.
The distanc e betwee n th e tw o point s represente d b y comple x
numbers z an d w i s \z — w\, becaus e th e fou r point s 0 , w, z, z — w
form a parallelogra m (se e Figur e 1 8) .
z— wz
Figure 1 8 . A comple x parallelogra m
Exercise 1 9 . Fin d the set of all points z in the complex plane which
satisfy:
(a) \z + 3\ = 5,
(b) \z + 4\ = \z-2i\,
(c) th e sum of the squares of the distances from z to two fixed points
is a given number .
The fac t tha t \/a 2 + b 2 i s th e distanc e betwee n tw o point s pro -
vides a mean s t o visualiz e certai n purel y algebrai c problems .
Problem 8 . Prove the inequality
\l*l + 6? + \/4 + H + • • • + V< + t> 2n

> \ / ( a i + a 2 + • • • + a n)2 + (fe i + b 2 + • • • + b n)2.
32 . Algebr a o f Point s
Solution. Pu t z\ — a\ - f b\i, . . . , z n — a n + b ni an d
consider th e broke n lin e wit h vertice s a t 0 , 21 , Zi +2:2,
. . . , z\ + Z2 H h ^n • The left-hand sid e of the inequalit y
is th e tota l lengt h o f thi s line , whil e th e right-han d sid e
is the distanc e betwee n it s endpoints .
Exercise 20 . Prov e the inequalit y

yjx\ + ( 1 - X 2)2 + yjx 22 + ( 1 - Xtf + - • • + yjx 21 0 + ( 1 - Xi) 2 > 7
for an y real numbers #i, ... , #10.
The angl e b y whic h th e hal f lin e 0 1 shoul d b e rotate d counter -
clockwise i n orde r fo r i t t o pas s throug h th e poin t z i s calle d th e
argument o f th e comple x numbe r z\ i t i s denote d b y ar g z. Her e 0
and 1 are th e point s tha t correspon d t o th e number s 0 and 1 .
Exercise 21 . Fin d the arguments of the following comple x numbers:
2, i, - 3 , -2z, 1 + i, y/3-i.
Figure 1 9 . Pola r coordinate s
A complex number i s completely define d i f one knows its modulu s

r an d argumen t <p. Indeed , a s yo u ca n se e i n Figur e 1 9 , z = x + yi,
where x — r cos (/?, y = r sin </?; thus
z = r (cos ip + i sirup).
This expressio n i s called th e trigonometric form o f the comple x num -
ber.
The correspondenc e z <- > (r , <p) betwee n comple x number s an d
pairs
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18:16:11 EDT 2019for r on
download from e thing , th e argumen t
IP 132.174.254.159.
of th e numbe r 0 i s undefined . O n th e othe r hand , th e argumen t o f

any non-zer o comple x numbe r i s onl y define d u p t o a whol e numbe r
of complet e rotations . Thus , on e i s fre e t o choos e 0 , 27r , — 27r, 47r ,
. . . a s th e argumen t o f the numbe r 1 . Nevertheless , th e pai r (r , if) i s
usually viewe d a s a pai r o f coordinate s fo r th e poin t z , calle d polar
coordinates.
These coordinate s ar e widel y use d i n practice , e.g. , i n airpor t
control centres : t o determin e th e locatio n o f an aircraft , yo u first find
the directio n an d the n measur e th e distance .
The equation s o f some figures loo k much simple r whe n writte n i n
polar coordinates .
Exercise 22 . (a ) Plot the line given in polar coordinates by the equa-
tion r = |cos3<^| . (b ) Fin d a pola r equatio n whic h describe s a
flower wit h si x petals simila r t o the on e shown i n Figur e 20 . Tr y
to rewrite i t i n Cartesia n coordinates .
F i g u r e 20 . A flower i n th e comple x plan e
Multiplication o f complex number s look s simpler whe n writte n i n

terms o f modulu s an d argument . I n fact , th e followin g tw o relation s
hold:
(4) \zw\ = \z\\wl
(5) aig(zw) = argz- f argiu .
The firs t on e i s a consequenc e o f the remarkabl e identit y

(ac - bd) 2 + {ad + be) 2 = (a 2 + b 2)(c2 + d 2 ).
To prove th e secon d one , le t
z = r (cos</? -h i sin (/?), u > = s(co s ip -f i sin -0).
Then
zu> = rs(co s </? + z sin <p)(cos ^ - f z sin 0)
= rs((cos^cos' 0 — sin(/?sin^) + z(sin<pcos ^ + c o s ^ s i n ^ ) ) ,
which simplifie s t o
zw = rs(cos((£ > + 0 ) + i sin(<p + T/J)) .
This proo f i s based o n th e well-known trigonometri c formula s fo r

the sin e an d cosin e o f th e su m o f tw o numbers . W e will giv e an -
other, mor e elegant , proo f whic h onl y relie s o n elementar y Euclidea n
geometry and , b y the way, implies the trigonometric rule s used above .
Figure 21 . Produc t o f comple x number s
Consider tw o triangle s wit h vertice s 0 , 1 , z an d 0 , w, zw (Figur e

21). Sinc e \w\ : 1 = \zw\ : \z\ = \zw — w\ : \z — 1 | , thes e tw o
triangles ar e similar , s o tha t thei r respectiv e angle s ar e equal . Th e
equality o f the tw o angle s whic h ar e marke d i n Figur e 2 1 proves tha t

aig(zw) = ar g z -f ar g w.
OK. Multiplicatio n o f complex number s mean s tha t thei r modul i
get multiplied , whil e thei r argument s ar e added . Iterate d a s appro -
priate, thi s observatio n yield s th e formul a fo r power s o f a comple x
number i n trigonometri c notation :
[r(cos(p + isinc/?)] n = r n (co s nip -f ism nip).
Exercise 23 . Prov e that , i f z i s a comple x numbe r an d a i s a rea l
number suc h that z + 1 /z = 2 cos a, the n z n + l/z n = 2 cos na.
The trigonometri c powe r formul a i s very convenien t i n problem s
such as Exercise 1 8c , which you might hav e already tried . Le t u s do it
1 \/ 3
together now . Denot e —i b y £. The n |£ | = 1 and arg £ = — 7r/6;
hence |C 1 998 | = l 1 9 9 8 - 1 and argC 1 998 = 1 99 8 • (-TT/6 ) = -33 3 • 2TT .
This implie s tha t C 1 998 = 1 .
F i g u r e 22 . Comple x root s o f 1
Note tha t a n intege r powe r o f th e numbe r £ ca n occup y onl y

one o f th e si x position s i n th e plan e — th e vertice s o f th e regula r
hexagon show n i n Figur e 22 . An y o f thes e si x comple x number s i s a
power o f £ and play s the rol e of a sixth roo t o f the numbe r 1 , becaus e
(Cfc)6 — (C 6)^ = l fc = 1 . I n general , fo r an y natura l n , ther e ar e
exactly n comple x n-t h root s o f unity , arrange d a s th e vertice s o f a
n-gon.
P r o b l e m 9 . Find the product of all diagonals and both sides that

issue from one vertex of the regular n-gon inscribed into the circle of
radius 1 .
Figure 23 . Side s and diagonal s of a regular polygo n
S o l u t i o n . P u t th e pol e (numbe r 0 ) a t th e centr e o f th e

polygon an d th e rea l unit y (numbe r 1 ) a t th e give n verte x
A\. Al l th e vertice s ar e t h e root s o f th e equatio n z n —
1 = 0 ; therefor e al l th e vertice s bu t A\ als o satisf y th e
equation z n~l+zn~2-\ f-z+ 1 = 0 obtained b y dividin g
zn — 1 by z — 1. No w compar e t h e tw o polynomial s z n~x - h
zn-2 + .. . + z + x = 0 an d ( z _ A2^z _ Asy^z_ An y
They ar e identicall y equal , becaus e the y hav e th e sam e
roots an d equa l leadin g coefficients . Hence , thei r value s
at z — A\ ar e equal :
(Ai - A 2){Al - A 3)... (Ai - An) = A^ 1 + A^ 2 + • ••+A x+ 1 .
Recalling t h a t A\ = 1 b y ou r choice , w e obtai n th e an -

swer:
\Ai - A 2\\A1 -A 3\... \Ai - A n\ - n .
E x e r c i s e 2 4 . A regular polygo n A1 A2 . . . A n i s inscribed i n the circl e

of uni t radius , an d A i s a n arbitrar y poin t o f thi s circle . Fin d
the su m o f square s o f distance s fro m A t o al l th e vertice s o f th e
polygon.
Since division i s an operation invers e to multiplication, i t satisfie s

the formula s invers e t o (4) :
I I '
il!
\w\
z
arg-w arg z — argiu.
F i g u r e 24 . Angl e expresse d throug h comple x number s
The latte r equalit y i s interestin g fro m th e poin t o f vie w o f ele -

mentary geometry : i t allow s u s to expres s th e magnitud e o f a n angl e
in term s o f it s verte x an d tw o point s belongin g t o it s sides :
ip = ar g
(Figure 24) . Her e are two examples where this observation i s applied :
in th e first one , w e solv e a geometri c proble m usin g th e algebr a o f
complex numbers , an d i n th e secon d one , conversely , w e solv e a n
algebraic proble m b y a geometri c method .
Problem 1 0 . Three squares are placed side by side as shown in Fig-
ure 25. Prove that the sum of ZKAH, Z.KDH and ZKFH is a right
angle.
Solution. Evidently , ZKFH = 7r/4 , S O we hav e t o
prove tha t ZKAH + ZKDH = 7r/4 , too . Assumin g
that A = 0 , D = 1 and B = i , w e hav e F = 2 , K = 3 ,
H = 3 + i. Therefore , Z.DAH = ar g § E^ = arg( 3 + i) ,
B C G H
Figure 25 . Su m o f thre e angle s
Z.FDH = a r g f f g = arg( 2 + z) , whenc e Z.DAH +

Z.FDH = arg( 3 + i) (2 + i) = arg( 5 + 5i) = TT/4 , whic h i s
just wha t wa s required .
Problem 1 1 . Prove that if z\, Z2, z%, z± are different complex num-
bers with equal absolute values, then
Z\ ~ Zs # Z\ — Z 4
Z2 ~ Z 3 ' Z 2- Z 4
is a real number.
Solution. Th e fou r give n point s li e o n th e sam e circl e
centred a t 0 . Point s z\ an d z 2 spli t thi s circl e int o tw o
arcs. Th e othe r tw o point s z$, z± ca n belon g eithe r t o
the sam e arc , o r t o differen t arcs . I n th e firs t cas e th e
angles z\z^z 2 an d Z1 Z4Z 2 a re equal , because they subten d
the sam e arc . Therefore , ar g = ar g an d
Z2 - 2 3 z 2~ Z±
Z\ — £ 3 Z\ — £4
arg : =0 , i.e . th e numbe r i n questio n
z2 — Z3 z 2 — Z4
is rea l an d positive . I n th e secon d cas e th e tw o angle s
Z1Z3Z4 and z 2z4zi hav e the same orientation an d togethe r
make 1 80° . Therefore , th e number i n question is real an d
negative.
The assertio n o f Proble m 1 1 evidentl y generalize s t o an y se t o f

four comple x number s tha t belon g t o a n arbitrar y circl e o r straigh t
line i n th e plane . Th e convers e i s also true: i f the give n expressio n i s

real, the n th e fou r number s mus t belon g eithe r t o th e sam e circl e o r
to th e sam e straigh t line .
Exercise 25 . Le t ci , C2, ..., c n b e the vertices of a convex polygon.
Prove that al l complex roots of the equatio n
Z — C\ Z — C2 Z — C n
are interior point s of this polygon .
Chapter 2
Plane Movement s
Plane movement s ar e transformations o f the plane that d o not chang e

the lengths of segments and, a s a consequence, preserve all parameter s
of geometri c figures, suc h a s areas , angles , etc .
We begi n thi s chapte r wit h th e discussio n o f som e well-know n
problems o f elementar y geometr y tha t allo w a shor t solutio n usin g
plane movements. Al l these problems share the same underlying idea :
change th e positio n o f certai n part s o f th e give n geometri c configu -
ration i n suc h a wa y tha t th e hidde n relation s betwee n th e element s
become transparent .
We the n procee d t o a detaile d discussio n o f th e compositio n o f
movements, whic h wil l provid e experimenta l materia l fo r th e intro -
duction o f transformatio n group s i n th e nex t chapter .
1. Paralle l translation s
Definition 4 . A parallel translation (o r simpl y a translation) i s a
transformation o f th e plan e tha t send s ever y poin t A int o th e poin t
A' suc h tha t A A' i s equa l t o a give n constan t vecto r v . Thi s trans -
formation i s denoted b y T v .
Problem 1 2 . Two villages A and B are located across the river from
each other. The sides of the river are rectilinear and parallel to each
other. Where should one build the bridge MN so that the distance
42 2. Plan e Movement s
AMNB be as small as possible? The bridge must be perpendicular to

the sides of the river.
Figure 1 . Bridg e ove r a rive r
Solution. I f ther e wer e n o river , th e shortes t pat h join -

ing A an d B woul d b e a straigh t line . Le t u s tr y t o ge t
rid o f th e rive r b y movin g on e o f it s side s toward s th e
other perpendicularl y unti l bot h side s coincid e (Figur e
1). Le t B' b e th e ne w positio n o f th e poin t B. Th e
lengths o f AMB'B an d AMNB ar e equal . Th e positio n
of point B' doe s not depen d o n the choice of the place for
the bridge . Henc e we only have to minimize th e distanc e
AMB', whic h i s can b e don e simpl y b y makin g AMB' a
straight line .
Exercise 26 . Construc t th e shortes t pat h tha t connect s tw o point s

A an d B separate d b y two rivers (Figur e 2) . Bot h bridge s mus t
be perpendicular t o the sides of the rivers.
Problem 1 3 . Inscribe a given vector in a given circle (i.e., construct

a chord of a given circle which is equal and parallel to a given seg-
ment).
Solution. Le t AB b e th e give n vecto r an d C th e give n

circle wit h centr e O an d radiu s r (Figur e 3) . W e hav e
to mov e AB, keepin g i t paralle l t o itself , toward s C s o
that i t get s inscribe d int o th e circle . I n fact , i t i s muc h
1. P a r a l l e l t r a n s l a t i o n s 43
Figure 2 . Tw o bridges over tw o rivers
Figure 3 . Inscribin g a vector int o a circle
easier t o perfor m th e revers e operation : mov e th e circl e

in th e opposit e directio n s o t h a t i n th e ne w positio n i t
will pas s throug h b o t h endpoint s o f th e vector , A an d
B. T o d o so , w e construc t th e triangl e ABD suc h t h a t
AD = BD = r. Th e poin t D i s th e centr e o f th e move d
circle. No w i f w e translat e th e point s A an d B b y th e
vector D O , w e wil l obtaine d th e segmen t inscribe d int o
the initia l circle .
Here ar e tw o mor e problem s whic h ca n b e solve d usin g paralle l

translation.
E x e r c i s e 27 . Inscrib e a give n vecto r i n a give n triangle , i.e. , fin d a
segment whos e endpoint s li e on th e side s o f the give n triangle an d
which i s equa l an d paralle l t o a give n segment .
Exercise 28 . Construc t a trapezoid i f the lengths of its parallel sides

and diagonal s ar e known.
2. Reflection s
Definition 5 . Le t I be a line in the plane. Th e reflection with respect
to I i s a transformatio n o f th e plan e tha t send s ever y poin t A int o
the poin t A' suc h tha t I is the perpendicula r bisecto r o f the segmen t
A A''. Thi s transformatio n i s denote d b y Si. an d i s als o calle d axial
symmetry with axis l.
Problem 1 4 . Two points A and B are on one side of the straight
line I. Find the point M G I such that the length of the broken line
AMB is minimal. If you prefer 'real life' problems, you may imagine
a person with an empty bucket at point A, a fire at point B and a
straightline river I.
Figure 4 . Shortes t pat h
Solution. I f bot h point s A an d B wer e situate d o n dif -

ferent side s of the lin e / , the solutio n woul d b e a straigh t
line AB. Le t u s try t o reduce our problem t o this case by
reflecting th e give n poin t B i n th e lin e I (se e Figur e 4) .
If B' i s the imag e o f B, the n th e line s AMB an d AMB'
have equal length s for an y arbitrar y positio n o f the poin t
Mel. T o minimiz e thi s distance , w e dra w th e straigh t
2. Reflection s 45
line AB' an d set M t o be the intersection poin t o f this

line wit h / . Not e tha t i n this cas e th e angles forme d by
either of the tw o lines AM an d BM wit h I are the same,
which agree s wit h th e well-known la w of optics.
Exercise 29 . Insid e an angle XOY', tw o points, A and £? , are given.

Among al l broken line s AMNB wher e M e XO, N e YO, find
the line of minimal length 1 .
F i g u r e 5 . Tw o river s o f Exercis e 2 9
We proceed wit h on e mor e proble m relate d t o shortest paths .

Problem 1 5 . Into a given acute triangle inscribe a triangle of min-
imal perimeter.
Solution. Le t UVW b e an arbitrary triangl e inscribe d

in the give n triangle ABC. Le t K an d L be the symmet -
ric images of the point U with respect t o the lines AB an d
Figure 5 refer s t o a Russia n fol k tal e wher e a rave n h a s t o brin g tw o kind s o f

water, t h e 'dead ' wate r a n d t he 'live ' water , t o reviv e t h e prince .
Figure 6 . Inscribe d triangl e o f minima l perimete r
BC (se e Figure 6) . Th e path s UVWU an d KVWL hav e

equal lengths . T o minimiz e thi s lengt h amon g al l trian -
gles UVW wit h a fixe d verte x J7 , w e hav e t o choos e V
and W s o that KVWL become s a straight line , i.e. t o set
V = M an d W = N. No w amon g al l triangle s AUMN
that correspon d t o differen t position s o f the poin t U, we
will choose th e on e wit h th e minima l perimeter , an d i t
will giv e th e solutio n t o th e problem . W e hav e t o fin d
the positio n o f U fo r whic h th e segmen t KL i s shortest .
Note that ABKL i s an isosceles triangle wit h BK —
BU — BL. It s angl e a t verte x B doe s not depen d o n th e
position o f th e poin t U: ZKBL = 2ZABC. Therefore ,
to minimiz e th e lengt h o f th e sid e KL w e hav e t o mak e
sure tha t th e sid e BK i s a s smal l a s possible . Sinc e
BK = BU, thi s minimu m i s attaine d whe n U i s th e
base poin t o f th e altitud e draw n i n th e triangl e ABC
from th e verte x B: BU J L AC.
Because o f th e symmetr y betwee n th e thre e point s
[/, V an d W , w e conclude tha t V an d W i n th e minima l
triangle UVW ar e als o basepoint s o f th e correspondin g
altitudes o f th e triangl e ABC.
Exercise 30 . Construc t a triangle, if one of its vertices and the three

lines that contai n it s bisectors ar e given.
3. Rotation s 47
E x e r c i s e 3 1 . A ra y o f ligh t enter s a n angl e o f 45 ° forme d b y tw o

mirrors. Prov e tha t afte r severa l reflection s th e ra y wil l exi t th e
angle movin g alon g a lin e paralle l t o it s initia l trajectory . Ar e
there othe r value s o f th e angl e wit h th e sam e property ?
3. Rotation s
D e f i n i t i o n 6 . Le t O b e a poin t i n th e plan e an d tp a rea l number ,
understood a s a n angle . Th e rotation around O through angle <p i s a
transformation o f the plan e t h a t send s ever y poin t A int o th e poin t A!
such t h a t \OA\ = \OA!\ an d Z.AOA' = <p, wher e th e angl e i s counte d
with sign , th e counterclockwis e directio n bein g considere d a s positive .
This transformatio n i s denote d b y RQ.
Look a t Figur e 7 . I t i s eviden t t h a t th e su m o f al l vertice s o f a

regular polygo n wit h a n eve n numbe r o f vertice s ove r it s centr e P i s
equal t o P (se e p . 1 0 fo r th e definitio n o f poin t addition) . Indeed , th e
set o f vertice s split s int o pair s o f mutuall y opposit e points . I t i s no t
so eas y t o prov e th e sam e propert y fo r a polygo n wit h a n od d numbe r
of vertices . I f yo u t r y t o directl y comput e th e coordinate s o f al l th e
vectors, yo u wil l hav e t o dea l wit h rathe r complicate d trigonometri c
expressions. However , th e proble m look s difficul t onl y a s lon g a s th e
plane doe s no t move .
P r o b l e m 1 6 . Prove that the sum of vertices of a regular polygon over

its centre P coincides with P.
Figure 7 . Su m of vertices of a polygo n

Solution. Le t n b e th e numbe r o f vertices. Unde r a ro-

tation throug h 360/ n degree s aroun d P th e give n poly -
gon goe s int o itself . Therefore , th e su m o f vertice s re -
mains unchanged . Bu t i n th e plan e ther e i s onl y on e
point tha t goe s int o itsel f unde r a rotation: i t i s the cen -
tre o f th e rotation .
Exercise 32 . A point M lie s inside a convex polygon. Perpendicular s

are drawn from M t o all sides of the polygon, and on each of these
half-lines, a point A% i s taken whos e distance fro m M equal s th e
length o f the correspondin g side . Prov e that th e sum o f all these
points over M i s zero.
Problem 1 7 . Construct an equilateral triangle, if the distances of its
vertices from a given point D are a, b and c.
Figure 8 . Constructin g a n equilatera l triangl e
Solution. Everyon e know s ho w t o construc t a triangl e

when give n th e length s o f it s sides . Unfortunately , th e
three segment s a, 6 , c in Figure 8 do not for m a triangle .
Let u s rotat e th e plan e b y 60 ° aroun d th e poin t C. Th e
point B goe s into A an d D goe s into D''. A rotation pre -
serves distances ; therefor e th e length s o f the side s o f th e
3. Rotation s 49
triangle A ADD' ar e a, b and c . W e wil l construc t thi s

triangle first , the n fin d th e poin t C (ACDD f i s equilat -
eral), an d finall y fin d th e poin t B.
Exercise 33 . Construc t a n equilateral triangl e whos e vertices li e on

three given parallel lines, one on each.
Problem 1 8 . Inside a given triangle, find the point the sum of whose
distances from the vertices is minimal
Figure 9 . Minimiz e th e su m o f distance s
Solution. Le t K b e a n arbitrar y poin t insid e th e trian -

gle ABC. Rotat e the point s C an d K aroun d A counter -
clockwise through 60 ° an d denot e thei r ne w position s b y
C an d K' (se e Figure 9). Th e sum of the three distance s
in question , AK + BK + CK, equal s th e lengt h o f th e
broken lin e C'K'KB. I t i s minima l i f K an d K f li e o n
the straigh t lin e BC. Thus , th e optima l positio n fo r K
is th e poin t Ko o n BC suc h tha t th e angl e AK§C' i s
60° or , i n othe r words , ZAK 0B = 1 20° . B y symmetry ,
we also hav e Z.BK 0C = ZCK 0A = 1 20° .
Note tha t ou r analysis , a s wel l a s th e answer , hold s
only for triangles whose angles are smaller tha n 1 20° . W e
leave it t o th e reade r t o gues s the answe r i n th e opposit e
case.
Exercise 34 . M i s an arbitrary poin t insid e a square ABCD. Dra w

four line s which pass through A, B, C an d D an d ar e perpendic-
ular t o M , CM, DM an d AM, respectively . Prov e tha t thes e
four line s pass through a common point .
The rotatio n throug h 1 80 ° i s als o referre d t o a s half turn, o r

central symmetry. Speakin g abou t centra l symmetries , w e will ofte n
leave 1 80 ° ou t o f the notation , writin g RA instea d o f R]^° . Her e ar e
two problem s wher e thi s kin d o f movement s i s used .
Exercise 35 . Throug h th e intersectio n poin t o f tw o circles , dra w a

line on which these circles cut equa l chords.
Exercise 36 . Ther e i s a roun d tabl e an d a n unlimite d numbe r o f

equal round coins. Tw o players take turns at placin g the coins on
the table i n such a way that the y d o not touc h each other. Wha t
is the winnin g strategy fo r th e first player ?
4. Function s o f a comple x variabl e

We retur n onc e agai n t o Proble m 1 6 (se e pag e 47) . Apar t fro m th e
geometric solution give n above , this problem als o has a n algebrai c so-
lution. T o explai n it , w e introduc e a complex structure i n th e plane .
More precisely , w e choose a one-to-one correspondenc e betwee n com -
plex number s an d point s i n th e plan e i n suc h a wa y tha t 0 corre -
sponds t o th e centr e o f th e polygo n an d 1 corresponds t o on e o f it s
vertices. I f £ i s th e verte x adjacen t t o 1 (i n th e counterclockwis e
direction), the n th e remainin g vertice s ar e £ 2 , . . . , C n _ 1 . W e ar e
interested i n # = 1 -f £ + £ 2 + • • • + C n_1 - Sinc e £ n = 1 , w e hav e
x( = £ + C 2 + ' ' • + C n - 1 + 1 — x i whic h implie s tha t x = 0 , becaus e
Note tha t thi s algebrai c proo f i s essentiall y th e sam e a s th e geo -

metric proo f give n above . Mor e exactly, i t i s nothing bu t th e transla -
tion o f th e geometri c argumen t int o algebrai c language . Indeed , th e
new proo f i s base d o n th e fac t tha t th e onl y numbe r whic h satisfie s
the equatio n £ x = x i s x = 0 . Bu t wha t happen s wit h a comple x
number whe n i t i s multiplie d b y £ ? Accordin g t o th e genera l rule ,
its modulu s remain s th e same , becaus e |£ | = 1 , an d it s argumen t
4. Function s o f a comple x variabl e 51
Figure 1 0 . Regula r polygo n wit h comple x vertice s
increases b y 360 ° /n. I n geometri c terms , thi s mean s tha t th e corre -

sponding poin t rotate s throug h 360/ n degree s aroun d th e centr e o f
the polygon .
In general , i f point s ar e viewe d a s comple x numbers , the n trans -
formations o f th e plane , an d i n particular , plan e movements , shoul d
be understoo d a s functions of a complex variable w = f(z), wher e z
denotes a n arbitrar y poin t an d w it s image . Fo r example , a rotatio n
around 0 i s represente d b y th e functio n w = az, wher e \a\ = 1 (w e
have i n this cas e a — co s ip + i sin<p, where ip i s the angl e o f rotation) .
It i s likewise eviden t tha t th e formul a fo r a paralle l translatio n i s
(6) w = z -f a ,
where a i s a certai n comple x number .

Now le t u s deriv e th e formul a fo r th e rotatio n o f th e comple x
plane aroun d a n arbitrar y poin t p. Figur e 1 1 shows tha t th e rotatio n
of th e poin t z aroun d p throug h a n angl e ip can b e spli t int o thre e
steps:
(1) translatio n z «—> • z — p;

(2) rotatio n aroun d th e origi n z — pi— • a(z — p);
(3) invers e translatio n a(z — p) >- > a(z — p) + p.
a(z-p)+p
Figure 1 1 . Plan e movemen t i n comple x coordinate s
The rotatio n aroun d p throug h th e angl e <p i s thu s describe d b y th e

function
(7) w — OLZ + ( 1 — a)p,
where a — co s ip + i sin (p.
Parallel translations an d rotation s ar e referred t o as proper move-
ments. Thi s expressio n i s accounted fo r b y the fac t tha t on e doe s no t
have t o leav e th e plan e i n orde r t o physicall y effectuat e on e o f thes e
transformations, wherea s a reflection i n a line requires a rotation (flip -
ping) o f th e plan e i n th e surroundin g three-space .
Theorem 3 . The set of proper movements of the plane coincides with

the set of all transformations described by the functions of a complex
variable
(8) w = az -bra ,
where a and m are complex numbers and |a | = 1 .
Proof. Formula s (6 ) an d (7 ) impl y that an y proper movemen t o f th e

plane i s describe d b y a linea r function s o f type (8) .
We will prove that th e converse also holds, i.e. tha t ever y functio n
(8) define s a prope r movement . Indeed , i f a = 1 , the n (8 ) become s
4. Function s o f a comple x variabl e 53
(6) an d w e deal wit h a paralle l translation . I f a ^ 1 , then (8 ) ca n b e

rewritten a s
/m \ m
w = az + m = a[z — + ,
\I — a1 J —a
which i s the expressio n o f the rotatio n aroun d p = m/(l — a) throug h
the angl e (p suc h tha t co s ip + i sin cp = a. D
c
}z
<^z
Figure 1 2 . Comple x conjugatio n
To find a simila r descriptio n fo r th e imprope r movements , fo r ex -

ample reflections , apar t fro m additio n an d multiplicatio n o f comple x
numbers, w e nee d on e mor e operation : complex conjugation. Th e
conjugate o f the numbe r z = x -f iy i s defined a s z = x — iy. Geomet -
rically, conjugatio n correspond s t o reflectio n i n th e rea l axi s (Figur e
12). Recal l tha t w e hav e alread y use d conjugatio n t o deriv e th e for -
mula fo r th e quotien t o f tw o comple x number s (se e p . 30) .
Exercise 37 . Prov e th e followin g formula s fo r reflectio n i n th e lin e
y = kx + b:
(9) w = z + 2bi, i f k = 0,
2
nn\ l-k + 2ki, 6 b
(10) w =
1 + fe2 (*+fc)- p l f
^ °
2 2 2
(note that ( 1 — k + 2hi)/(lH - k ) = a , wher e a — cos ip -f i sin cp
and 9 9 i s the angl e between the give n line and th e sc-axis).
In th e followin g exampl e w e use th e algebr a o f comple x number s
to solv e a geometri c problem .
Problem 1 9 . A pirate is hunting for a hidden treasure. According
to a letter he has got, he has to go to the Treasure Island, find two
trees A and B, a rock C (Figure 1 4) and dig for the buried treasure
at the point K which is the middle point of the segment DE, where D
is obtained by rotating C around A clockwise through 90° ; and E is
obtained by rotating C around B counterclockwise through 90°. When
the pirate arrived at this place, he found that the trees A and B are
there, but the rock C disappeared. Is it still possible to recover the
position of point K ?
Figure 1 3 . A pirat e
Solution. Le t u s introduc e a comple x structur e i n th e

plane, i.e. , associat e th e point s wit h comple x numbers ,
in suc h a wa y tha t A correspond s t o 0 , whil e B an d C
correspond t o number s b and c (se e Figur e 1 4) .
Then, b y formul a (7) , point s D an d E ar e repre -
sented b y th e number s — ic an d i(c — b) -f b; therefore ,
1—2
point K i s b. A s this expressio n doe s no t involv e c ,
we see that th e positio n o f th e hidin g plac e doe s no t de -
pend o n the choic e of the poin t C. W e also see that K i s
the verte x o f the isoscele s right triangl e wit h hypotenus e
AB, an d a s suc h ca n b e foun d b y ou r treasur e hunter .
5. C o m p o s i t i o n o f m o v e m e n t s 55
Figure 1 4 . Wher e i s the treasure ?
E x e r c i s e 3 8 . Tw o side s of a triangle ar e rotate d throug h 90 ° aroun d

their commo n verte x i n opposit e directions . Prov e tha t th e lin e
joining th e ne w endpoint s i s perpendicula r t o th e media n o f th e
triangle.
5. Compositio n o f movement s
Given tw o movement s o f th e plane , / an d g , on e ca n construc t a thir d
movement g o /, th e composition, o r th e product o f th e give n two , b y
performing firs t / , the n g.
D e f i n i t i o n 7 . T h e composition f o g o f tw o movement s / an d g i s
defined b y th e relatio n
(f o g)(x) = f(g(x))
for an y poin t x.
The transformatio n fog thu s define d i s reall y a movement , be -

cause i t evidentl y preserve s th e distance s betwee n th e points . I n thi s
section, w e wil l stud y th e compositio n o f specia l type s o f movements :
translations, reflection s an d rotations .
P r o b l e m 2 0 . Find the composition of two reflections.
S o l u t i o n . Le t 5 / denot e th e reflectio n i n t h e lin e I. Sup -

pose t h a t tw o lines , I an d m , ar e given , an d w e hav e t o
find th e compositio n S m 2019for
Licensed to Penn St Univ, University Park. Prepared on Tue Mar 12 18:16:11 EDT
o Si. Le t A' b e th e imag e o f a n
download from IP 132.174.254.159.
arbitrary poin t A unde r th e movemen t 5/ , an d A" th e

image o f A' unde r S m.
A"
'1 m
i
u
*A'
I
A
(a)
F i g u r e 1 5 . Produc t o f tw o reflection s
We first conside r th e case when the two lines / and m

are parallel to each other (Figur e 1 4a) . The n al l the thre e
points A, A', A" li e o n on e line , perpendicula r t o / an d
m, an d the distance between the points A an d A" i s twice
the distanc e betwee n th e line s I an d ra, independen t o f
the positio n o f A. Therefore , th e compositio n o f the tw o
reflections Si an d S m ha s th e sam e effec t a s translatio n
by th e vecto r 2u , wher e u i s the vecto r perpendicula r t o
/ and ra, o f length equa l to th e distanc e betwee n th e tw o
lines an d directe d fro m I t o m :
(ii) Sm ° Si = T 2 u .
Now suppose that th e lines / and m mee t a t a certai n

point C (Figur e 1 4b) . I f cp i s th e angl e betwee n I an d
m, then , a s yo u ca n se e fro m th e figure, AACA!' = 2(p.
Note als o tha t al l th e thre e point s A , A' an d A" ar e a t
the sam e distanc e fro m C. I t follow s tha t
(12) Sm ° Si
where R^ denote s rotatio n aroun d C throug h th e angl e

2(p (clockwise i f (p < 0 and counterclockwis e i f (p > 0).
5. Compositio n o f movement s 57
The reade r ma y wis h t o conside r othe r location s fo r th e poin t A

in th e plane , differen t fro m tha t o f Figur e 1 5b , an d mak e sur e tha t
formulas (1 1 ) an d (1 2 ) ar e alway s true . Bea r i n min d tha t th e angl e
<p should b e measure d fro m lin e / to line m, i.e. , fo r example , <p = 7r/ 4
means that m ca n be obtained fro m I by a positive (counterclockwise )
rotation throug h 45° .
Formula (1 2 ) implies , b y th e way , tha t th e compositio n o f tw o
movements i n genera l depend s o n th e orde r i n whic h the y ar e taken :
thus, Si o Sm i s the movemen t invers e t o 5 m o Si.
Exercise 39 . Le t Z , m an d n b e thre e line s meetin g a t on e point .
Find th e movement (S n ° Sm ° Si) =
We suggest that th e reader first experimen t b y applying the given
composition t o a n arbitrar y poin t o f th e plane , an d the n prov e
the resul t usin g the formula s w e have established .
Formulas (1 1 ) an d (1 2) , rea d fro m righ t t o left , sho w ho w t o
decompose a translatio n o r a rotatio n int o a produc t o f reflections .
This decompositio n i s no t unique , an d th e freedo m w e hav e i n th e
choice of the axe s of reflection ma y prov e quit e usefu l fo r th e solutio n
of a specifi c problem .
Problem 21 * Find the composition of two rotations.
Solution. I f the centre s o f both rotation s coincide , the n
the answe r i s obvious:
(13) RôR^^R^.
F i g u r e 1 6 . Produc t o f tw o rotation s
Now conside r tw o rotation s R^ an d Rg wit h differ -

ent centres . T o fin d thei r composition , w e wil l repre -
sent eac h rotatio n a s th e produc t o f tw o reflection s an d
then us e th e formula s tha t w e alread y know . W e hav e

RA ~ Srn ° Si, wher e th e line s I an d m for m a n angl e
ip/2 a t th e poin t A, an d Rg — Sp o 5n , wher e th e line s
n an d p for m a n angl e ip/2 a t th e poin t B (se e Figur e
16a). T h e n i ^ o i ^ = SmoSioS poSn. Thi s expressio n
simplifies t o 5 m o Sn whe n th e tw o lines / and p coincide ,
because i n thi s cas e Si o Sp = i d i s the identity transfor -
mation, i.e. , th e transformatio n whic h take s ever y poin t
into itself .
After thi s analysis , w e star t ane w fro m Figur e 1 6b .
We denote b y c the lin e joining A an d B; then , rotatin g c
around A throug h th e angl e p/2 an d aroun d B throug h
—^/2, w e obtai n th e line s b an d a . I f th e line s b an d
c hav e a commo n point , w e denot e i t b y C , an d i n thi s
case w e can writ e
R%oR*=ShoScoScoSa = S hoSa = #£ +V\

or, settin g a — ip/2, (3 = -0/2, 7 = 7 r — a — /?,
(14) ^ 0 ^ /= ^ ,
where C i s the thir d verte x o f th e triangl e wit h tw o ver -

tices A an d £ ? and angle s a t thes e vertices equal to a an d
/3; 7 i s the angl e o f this triangl e a t C.
After bot h side s of (1 4 ) ar e multiplie d b y R^ o n th e
right, i t take s a mor e symmetri c form :
2
(15) R foR2jPoR2J = id.
The convers e i s als o true : i f th e thre e point s A, B,
C an d thre e angle s a , /? , 7 betwee n 0 an d 1 80 ° satisf y
equation (1 5) , then a , / ? and 7 ar e equa l t o th e angle s of
the triangl e ABC.
Equality (1 5 ) ca n b e checke d directly . Sinc e 2 a +
2)8 + 2 7 = 360° , th e compositio n R 2/ o i ?^ o R 2J i s
a paralle l translation . T o prov e tha t i t i s th e identity ,
we just nee d t o chec k tha t i t ha s on e fixed point . Bu t
Figure 1 7 show s tha t th e poin t A remain s fixed unde r
the successiv e mapping s Rj , Rg , R 2^.
5. Compositio n o f movement s 59
Figure 1 7 . Compositio n o f thre e rotation s
Should th e line s a an d b b e paralle l (thi s happen s

when (p 4- ip is a multipl e o f 27r) , the n
2 a 2
(16) R A oR BP = T 2u,
where u i s defined i n Figur e 1 6c .

Using complex numbers , on e can deriv e a n algebrai c
formula fo r th e compositio n o f tw o rotations . W e tak e
a comple x numbe r z an d appl y successivel y firs t th e ro -
tation Rg, the n th e rotatio n Rf A. Accordin g t o formul a
(7), w e ca n writ e
R%(z) = q(z-b) + b,
R^(w) — p[w — a) + a,
where p — co s ip + isinip, q ~ co s ip + i sin ip. No w w e
substitute Rg(z) instea d o f w an d tr y t o rewrit e th e
result i n a simila r form :
(B%oR*)(z) = p( q(z-b)+b-a) + a
,a — pa+pb — pqb x a — pa + pb — pqb
= M\ z i ) +i •
v
1 — pq1 ' — PQ
Note tha t pq = cos(<p + i/j) + i sin((p + I/J). Therefore , th e
result obtaine d mean s tha t
R*12oR%=B*+*,
Licensed to Penn St Univ, University Park. Prepared on Tue Mar A 18:16:11 EDT 2019for download from IP 132.174.254.159.
where th e point C corresponds t o the complex numbe r

a-pa+pb-pqb
(17) c =.
1-pq
We se e that geometri c an d algebraic argument s lea d t o two dif-

ferent formula s fo r the composition o f rotations. W e can benefit fro m
this fac t b y deriving th e following corollary :
/ / two vertices A and B of a triangle ABC correspond to complex
numbers a and b, and the angles at these vertices are ip/2 and ip/2,
then the third vertex, as a complex number, is determined by formula
(17).
We pas s t o example s wher e th e composition o f movements and
the formula s w e have foun d ar e used.
Problem 22 . Three equilateral triangles are built on the sides of an

arbitrary triangle ABC (Figure 1 8). Prove that their centres M, N,
P form an equilateral triangle. 2
F i g u r e 1 8 . Proble m o f Napoleo n
Solution. Triangle s AMB, BNC an d CPA ar e isosce-

les with obtus e angle s of 120°. Conside r th e composition
of thre e rotation s F = R^°° o R$Q° o R™°. Formula s
T h i s proble m i s known a s the problem of Napoleon, althoug h t h e famous Frenc h

general i s not it s author .
6. G l i d e r e f l e c t i o n s 61
(14) an d (1 5 ) sho w t h a t F i s eithe r a rotatio n o r a par -

allel translation . Sinc e th e su m o f th e thre e angle s o f
rotation i s 360° , F mus t b e a paralle l translation . Le t
us trac e ho w th e poin t A i s move d b y F. I t i s clea r t h a t
B
RM°°(A) = , RT°( B) = C, R¥°°(C) = A, an d thu s
F{A) — A. I t follow s t h a t F i s a translatio n b y zer o
vector, i.e. ,
RôRôR™0 =id.
Comparing thi s t o (1 5) , w e conclud e t h a t M i s th e thir d

vertex o f the triangl e havin g tw o vertice s a t T V an d P an d
angles 60 ° an d 60 ° a t thes e vertices , i.e . a n equilatera l
triangle.
E x e r c i s e 4 0 . Fin d a solution o f the previou s proble m base d o n com -

putations wit h comple x numbers .
E x e r c i s e 4 1 . O n th e side s o f a n arbitrar y quadrangl e fou r square s
are built . Prov e tha t thei r centre s for m a quadrangl e whos e diag -
onals ar e mutuall y perpendicula r an d hav e equa l length .
E x e r c i s e 4 2 . Fin d th e compositio n o f
1. tw o centra l symmetries ,
2. a centra l symmetr y an d a reflection .
E x e r c i s e 4 3 . Construc t a pentagon , give n th e midpoint s o f al l it s
sides.
6. Glid e reflection s
We hav e studie d thre e type s o f plan e movements : translations , rota -
tions an d reflections . However , thes e thre e type s d o no t cove r al l plan e
movements. Fo r example , i n Exercis e 42 , th e produc t o f a reflectio n
and a centra l symmetr y doe s no t belon g t o an y o f thes e types .
D e f i n i t i o n 8 . A glide reflection with axis I and vector v i s a move -

ment t h a t consist s i n a reflectio n wit h respec t t o a lin e I an d a trans -
lation b y t h e vecto r v, whic h i s assume d t o b e paralle l t o th e lin e I
(see Figur e 1 9) .
Figure 1 9 . Glid e reflectio n
Denoting th e glid e reflectio n b y Uf, w e ca n writ e th e definitio n

as Uf —T voSi = Si oT v. Th e movement s Si an d T v commute , i.e. ,
the tw o product s take n i n differen t orde r ar e indee d equal , becaus e
the figure AA\A'A<i i s alway s a rectangle .
Glide reflections , lik e al l othe r type s o f plan e movements , ca n b e
successfully use d fo r solvin g geometri c problems .
Problem 23 . Construct a line parallel to the side AC of a given
triangle ABC and intersecting its sides AB and BC at points D and
E such that AD = BE.
Solution. Th e solutio n relie s on the followin g tw o prop-
erties o f glide reflections , whic h immediatel y follo w fro m
Figure 1 9 :
(1) th e midpoint o f a segment joining an arbitrary poin t
with it s image under a glide reflection alway s lies on
the axis ;
(2) th e axi s o f th e glid e reflectio n i s preserved .
There is a glide reflection U which takes the half-lin e
AB int o the half-lin e BC. It s axi s i s the lin e NK, wher e
N i s the midpoin t o f the segmen t AB whil e K belong s t o
BC an d BK = NB. B y th e premises , AD = BE; henc e
U(D) = E an d th e midpoin t o f DE mus t belon g t o th e
line NK. But , sinc e DE | | AC, th e midpoin t o f DE lie s
on th e media n BM. Therefor e th e thre e segment s DE,
BM an d NK hav e a point i n common , an d th e require d
7. Classificatio n o f movement s 63
construction ca n b e effectuate d i n th e followin g order .

First w e fin d th e point sT V and K a s mentione d above .
Then we draw the median BM. Finally , w e draw the lin e
parallel to AC throug h th e intersection poin t otBM an d
NK. Thi s i s the desire d line .
Exercise 44 . A point an d three straight line s are given. Dra w a line

I passing throug h th e give n poin t i n suc h a wa y tha t it s imag e
under th e thre e reflection s wit h respec t t o th e thre e give n line s
(in a prescribed order ) i s parallel to /.
Exercise 45 . Usin g complex numbers , fin d a n algebrai c formul a fo r
glide reflection .
7. Classificatio n o f movement s
In th e previou s section , w e hav e gotte n acquainte d wit h a ne w kin d
of plan e movement . S o far , w e hav e encountere d fou r type s o f plan e
movements: translations , rotations , reflection s an d glid e reflections .
A natura l questio n arises : are there any plane movements that do
not belong to any of these four types? Th e answe r i s give n b y th e
following theorem .
Theorem 4 . Any plane movement is either a translation, a rotation,

a reflection, or a glide reflection.
Proof. Firs t o f all , w e not e tha t a plan e movemen t i s completel y

defined b y th e image s o f thre e non-collinea r point s .A , B, C. I n fact ,
if A\ B' ', C ar e th e image s o f thes e points , the n fo r an y poin t D
there exist s exactl y on e point D' whos e distances fro m A! , B' , C ar e
equal t o th e distance s o f D fro m A , B, C.
The secon d usefu l observatio n i s that fo r an y tw o differen t point s
M an d M' ther e is a reflection tha t carrie s M ove r to M f. I n fact, thi s
reflection i s uniquely defined : it s axi s i s the perpendicula r bisecto r o f
the segmen t MM'.
Using thes e tw o observations , w e ar e goin g t o decompos e an y
plane movemen t a s th e produc t o f severa l reflections . Not e tha t w e
have alread y use d thi s tric k earlier : se e the discussio n o f Problem 21 .
Figure 20 . Decompositio n o f a plan e movemen t int o reflection s
Let / b e a n arbitrar y movemen t o f the plane . Choos e thre e non -

collinear point s A, B an d C . Denot e f(A) = A', f(B) = B', f(C) =
C". Suppos e tha t A! i s differen t fro m A, denot e b y S\ th e reflectio n
that take s A t o A!, an d se t B x = S t(B), d = S t(C) (se e Figur e 20) .
If B\ i s differen t fro m B' , the n w e denot e b y 5 m th e reflectio n tha t
takes B\ t o B' whil e preservin g A! ', and se t C 2 = SVn(Ci) . Finally , i f
C2 7 ^ C", w e fin d a thir d reflection , SVi , which take s C2 int o C" . W e
thus se e tha t i n th e wors t case , whe n al l th e step s o f thi s procedur e
are necessary , / ca n b e represente d a s th e compositio n S n o 5m o Si.
If som e step s tur n ou t t o b e unnecessary , w e ca n represen t / a s on e
reflection o r a compositio n o f tw o reflections .
Now we will prove tha t th e produc t o f no mor e tha n thre e reflec -
tions i s a movemen t belongin g t o on e of the fou r type s tha t w e know.
Indeed, on e reflectio n i s a reflection , an d that' s it . Tw o reflection s
make eithe r a rotation o r a translation. Th e onl y nontrivia l cas e is to
analyze th e produc t o f thre e reflection s S n o Sm o Si.
Three lines in a plane can be arranged in one of the four essentiall y
different pattern s depicte d i n Figur e 21 . W e wil l sho w tha t i n case s
(a) an d (b) , th e produc t i s a reflection , an d i n case s (c ) an d (d) , a
glide reflection .
In cas e (a) , the compositio n 5 m o Si o f the las t tw o reflections i s a
rotation throug h a n angl e equa l t o twic e th e angl e betwee n th e line s
7. Classificatio n o f movement s 6 5
H=X 3L b e d
Figure 21 . Thre e line s i n th e plan e
m an d / . W e ca n choose anothe r lin e V passin g throug h th e sam e

point, s o that 5 m o Si = S n o S^. The n
Sn ° Sm O Si = S n O S n O 5/ / = S// .
In cas e (b) , a simila r argumen t holds .
Figure 22 . Adjustin g tw o rotation s
Now conside r cas e (c) . I n th e initia l produc t o f thre e reflections ,

we will make two changes. Firs t w e replace the produc t S m ° Si b y a n
equal product Sm' °Si', wher e the line m' i s chosen to be perpendicula r
to n (Figur e 22) . W e hav e 5 n o Sm ° Si = S n o 5 m / o 5^. Nex t w e
replace th e produc t S n o Smi b y 5 n / o 5m //, where n' i s perpendicula r
to V. W e obtai n
f= S oS o Si
Licensed to Penn St Univ, UniversitynPark. Prepared
m = S OS
on Tue Mar 12 n '° 5/ / = 5
m 2019for download from
18:16:11 EDT / o Sm" ° S'i/ .
n IP 132.174.254.159.
Note tha t th e line s I' an d m" ar e parallel ; therefor e th e compositio n

S m " o Si' i s a paralle l translatio n i n th e directio n o f lin e n' ', an d th e
whole movemen t i s a glid e reflection .
Finally, cas e (d ) i s reduce d t o cas e (c) , becaus e th e compositio n
Sn ° S m ° Si remain s th e sam e i f tw o o f th e thre e line s ( n an d m o r
m an d I) ge t rotate d b y th e sam e angle . •
Plane movement s hav e a simpl e descriptio n i n term s o f comple x

functions. Theore m 3 , proved above , say s tha t translation s an d rota -
tions correspon d t o function s az + m wit h \a\ = 1 .
Theorem 5 . The set of reflections and glide reflections of the plane
coincides with the set of all transformations described by complex for-
mulas
(18) w = az + m ,
where a and m are complex numbers and \a\ = 1 .
Proof. Th e fac t tha t reflection s an d glid e reflection s ar e indee d de -

scribed b y suc h formula s follow s directl y fro m th e resul t o f Exercise s
37 an d 45 .
To prov e th e secon d hal f o f th e theorem , not e tha t th e compo -
sition o f transformatio n (1 8 ) wit h th e standar d reflectio n z \— > z i s
given b y th e formul a z y—> az + m , which , b y Theore m 3 , i s eithe r a
translation o r a rotation . •
8. Orientatio n
We have learned tha t ther e ar e four type s o f plane movements : trans -
lations, rotations , reflections , an d glid e reflections. Movement s o f th e
first tw o type s ca n b e represente d a s th e produc t o f a n eve n numbe r
(two) o f reflections ; the y ar e referre d t o a s proper movements . Th e
remaining tw o type s ar e product s o f a n od d numbe r (on e o r three )
of reflections ; the y ar e referre d t o a s improper movements , becaus e
one ha s t o exi t th e plan e i n orde r t o physicall y implemen t suc h a
movement.
The distinctio n betwee n th e tw o kind s o f plan e movement s ca n
be bes t understoo d usin g th e notio n o f orientation.
8. O r i e n t a t i o n 67
We sa y t h a t th e ordere d tripl e o f non-collinea r point s A, B, C

is positively oriented, i f thi s orderin g agree s wit h a counterclockwis e
walk aroun d th e triangl e ABC, or , i n othe r words , i f i n th e sequenc e
AB, BC, CA ever y nex t vecto r i s a t u r n t o th e lef t wit h respec t t o
the previou s one . I f th e orde r i s clockwise , th e tripl e i s sai d t o b e
negatively oriented.
Figure 23 . Thre e puck s
E x e r c i s e 4 6 . Thre e puck s for m a triangl e i n th e plane . A hockey -

player choose s a puc k an d send s i t alon g a straigh t lin e s o tha t i t
passes betwee n th e tw o remainin g pucks . I s i t possibl e tha t afte r
25 shot s eac h o f th e thre e puck s return s t o it s initia l position ?
It i s remarkabl e t h a t an y movemen t / o f th e plan e eithe r pre -

serves o r reverse s th e orientatio n o f al l triples : t h e orientatio n o f
f(A), f(B), f(C) eithe r coincide s wit h t h a t o f A, B, C fo r al l triples ,
or differ s fro m i t fo r al l triples . Mor e specifically , i t i s eas y t o se e
t h a t prope r movement s (translation s an d rotations ) preserv e t h e ori -
entation, whil e imprope r movement s (reflection s an d glid e reflections )
reverse it .
As a consequenc e o f thi s observation , w e obtai n t h e followin g
fact: th e compositio n o f a n od d numbe r o f reflection s ca n neve r b e
an identit y transformation .
The notio n o f orientatio n ha s a simpl e interpretatio n i n t e r m s o f
complex numbers .
Exercise 47 . Prov e that th e triple (21 ,22 , z$) is positively oriented if

and only if the argument of the complex number (2 3 —21)/ (22 —21)
is between 0 and 1 80° .
9. Calculu s o f involution s
Definition 9 . A transformation / i s called a n involution, i f it i s no t
the identity , bu t it s squar e i s th e identity : / 7 ^ id, f 2 = f o / = id .
This i s th e sam e a s t o sa y tha t / i s invers e t o itself : / = / _ 1 , tha t
is, f(A) = B i f an d onl y i f f(B) = A.
There ar e tw o type s o f involutiv e movement s o f th e plane :

• RA — hal f tur n aroun d poin t A (se e Sectio n 3) .
• Si — reflection i n a lin e I (se e Sectio n 2) ,
We se e tha t involutiv e movement s correspon d t o geometri c element s
of tw o kinds : point s an d lines . Thi s correspondenc e i s i n fac t one -
to-one, becaus e differen t point s an d differen t line s produc e differen t
involutions. Therefore , th e passag e fro m geometri c object s t o involu -
tions preserve s al l information, an d ever y fac t abou t point s an d line s
can b e reformulate d i n term s o f th e correspondin g involutions .
Problem 24 . Find the property of a pair of reflections Si, S m which

is equivalent to the fact that the lines I and m are mutually perpen-
dicular.
Solution. Th e compositio n 5 m o Si i s a translatio n i f

ra | | Z , or a rotatio n throug h 2(p if ra an d I intersec t a t
an angl e cp. Unless ra = Z , this compositio n ca n neve r b e
the identity . It s squar e (S m o Si)2 i s either a translatio n
(in the first case ) o r a rotation throug h 4<p (i n the secon d
case). Henc e th e line s ra an d I ar e perpendicula r i f an d
only i f
2
(19) ( S mo50 = id ,
i.e., th e produc t S m ° Si i s a n involution . Not e tha t thi s
involution i s a hal f tur n aroun d th e intersectio n poin t o f
the tw o give n lines .
9. Calculu s o f involution s 69
If w e multipl y (1 9 ) b y S m o n th e lef t an d b y Si o n
the right , the n i t become s
(20) S mo5z = Sio5 m,
i.e., th e tw o involutions S m an d Si commute. Thi s i s th e

required conditio n fo r th e tw o line s t o b e perpendicular .
This i s a n appropriat e momen t t o discus s th e notion s o f commu -

tativity an d associativity . Multiplicatio n o f movement s i s i n genera l
non-commutative. A s we have just seen , two different reflection s com -
mute i f and onl y if the corresponding line s are perpendicular. Bu t th e
composition o f movements, like that o f any arbitrary transformations ,
always ha s th e propert y o f associativity.
Let u s b e give n fou r set s an d thre e mapping s betwee n them , ar -
ranged accordin g t o th e schem e
V -^W -^X ^Y.

Then on e ca n for m th e followin g compositions : g o / : V — > X ,
hog: W - > y , ho(gof) : V - > Y, (hog) of : V -+ Y. Associativit y
means tha t th e tw o doubl e composition s h o (g o f) an d (h o g) o /
coincide.
To find a formal proof of this almost eviden t property , i t is enough
to understand th e meaning o f composition. Thus , th e mapping gofis
defined b y th e equatio n (g o f)(v) = g{f{v)) fo r a n arbitrar y elemen t
v G V. I n th e followin g chai n o f equation s thi s definitio n i s use d
several times :
(ho (go /))(« ) = h((g o /)(«)) = h(g(f(v)))
{
' =(hog)(f(v)) = ((hog)of)(v).
Since th e value s o f h o (g o /) an d (h o g) o / o n an y elemen t ar e th e
same, thes e tw o mapping s coincide .
The followin g analog y migh t b e usefu l t o bette r understan d th e
meaning o f associativity . Imagin e tha t / , g an d h ar e th e action s o f
putting o n you r socks , boot s an d overshoes , respectively . The n th e
composition (h o g) o f mean s tha t on e firs t put s o n th e socks , the n
puts th e boot s insid e th e overshoes , an d put s thi s objec t o n th e fee t
in socks . Th e othe r compositio n h o (g o /) mean s tha t on e firs t put s
the sock s insid e of the boots , put s o n thi s combinatio n an d the n put s
the overshoe s o n top . Evidently , th e resul t i n bot h case s i s the same !
F i g u r e 24 . Associativit y
The sam e analog y show s tha t th e compositio n o f operation s i n

question i s no t commutative : t o pu t o n th e socks , the n th e boot s i s
not th e sam e thin g a s t o pu t o n th e boots , the n th e socks ! However ,
commuting operation s d o exist , fo r example , puttin g a soc k o n on e
foot an d puttin g a soc k o n anothe r foot .
Using thi s analogy , i t i s eas y t o understan d th e formul a fo r th e
operation invers e to a composition of several operations. Fo r example,
if you pu t o n th e socks , the n th e boots , the n th e overshoes , the n th e
inverse operation mean s that yo u take off first th e overshoes , then th e
boots, the n th e socks :
(hogof)-1=r1og"1oh-1.
Now w e return t o th e calculu s o f involution s i n th e plane .

Problem 25 . Express in terms of involutions the property of four
points A, B, C, D forming a parallelogram.
Solution. Accordin g t o Exercis e 42 , th e compositio n
RA O RB i s translatio n b y th e vecto r 2 BA, whil e th e
composition RJJ o RQ i s translation b y th e vecto r 2 CD.
The figure ABCD (wit h thi s orde r o f vertices!) i s a par -
allelogram i f an d onl y i f BA = CD, whic h i s equivalen t
to th e followin g conditio n fo r th e fou r involutions :
RA O RB = RD °RC-
9. C a l c u l u s o f i n v o l u t i o n s 7 1
Multiplying b o t h side s o f thi s relatio n b y appropriat e in -

volutions, w e ca n rewrit e i t i n tw o mor e equivalen t forms :
RA ° RB ° Re ° RD — id an d
(22) R AoRBoRc = R D.
The las t equatio n ma y b e viewe d a s a formul a t h a t ex -

presses th e fourt h verte x o f a parallelogra m i n term s o f
the thre e give n ones .
E x e r c i s e 4 8 . Expres s th e followin g geometri c fact s a s algebrai c rela -

tions betwee n th e correspondin g involutions : (a ) poin t A belong s
to th e lin e I; (b) poin t A i s th e midpoin t o f th e segmen t BC.
E x e r c i s e 4 9 . Fin d th e geometri c meanin g o f th e followin g relations :
(a) R AoSi = Sio R B; (b ) (S n oSmO Si) 2 = id .
You ca n se e t h a t th e algebr a o f involution s ofte n provide s a shor t

and convenien t wa y t o writ e dow n fact s abou t point s an d line s i n th e
plane. Her e i s a mor e complicate d exampl e wher e thi s techniqu e i s
essential.
P r o b l e m 2 6 . Let M, N, P, Q be the centres of the four squares built

on the sides of a quadrangle ABCD (Figure 25). What conditions
should be imposed on ABCD in order that MNPQ be a square?
F i g u r e 2 5 . Square s o n th e side s o f a quadrilatera l
Solution. W e kno w tha t th e diagonal s o f MNPQ ar e

always equa l an d mutuall y perpendicula r (se e Exercis e
41). Therefore , MNPQ i s a squar e i f an d onl y i f i t i s a
parallelogram. Usin g th e resul t o f Proble m 25 , w e ca n
write thi s a s th e followin g conditio n o n th e fou r involu -
tions:
(23) R MoRNoRPoRQ = id.
By formul a (1 4 ) w e hav e
RM = RA ° R>B •>
RN — RB ° Re 5
Rp = Re ° RD 5
RQ = RD°RAI
where d = 90° , an d w e recal l tha t i f th e angl e o f ro -
tation i s no t specified , i t i s assume d t o b e 1 80° . Upo n
substitution int o (23) , this give s
d d
(24) R AoRBoRcoRDoR A = id ,
or, afte r multiplicatio n b y R Ad o n bot h sides ,
RB ° Re ° RD — RA •
According t o formul a (22) , thi s mean s tha t ABCD i s a
parallelogram. Hence , th e necessar y an d sufficien t con -
dition fo r MNPQ t o for m a squar e i s tha t th e initia l
quadrangle ABCD b e a parallelogram .
Chapter 3
Transformation Group s
The notio n o f a group unifie s tw o different ideas : a geometric on e an d

an algebrai c one .
On the geometri c side , the notio n o f a transformation group give s
a mathematica l expressio n o f th e genera l principl e o f symmetry : th e
more transformation s preserv e a give n object , th e mor e symmetri c i t
is.
On the algebraic side, the notion of an abstract group contains th e
common feature s o f operations tha t mos t ofte n appea r i n mathemat -
ics. Example s o f suc h operation s — additio n an d multiplicatio n o f
numbers an d points , additio n o f vectors , compositio n o f movement s
— were considere d i n th e previou s chapters .
1. A rollin g triangl e
We begin with an introductory proble m where a transformation grou p
comes u p i n a natura l way .
Problem 27 . An equilateral triangle ABC lies on the plane. One
can roll it over the plane by turning it through 1 80 ° around any of its
sides. Show that if after a certain number of such steps the triangle
returns to the initial place, then each of its three vertices will return
to its initial position.
Solution. Le t a , b an d c b e th e line s containin g th e

sides o f th e give n triangl e i n it s initia l position . After
any numbe r o f turnover s th e side s o f th e triangl e wil l li e
74 3. Transformatio n Group s
Figure 1 . Ornamen t o f th e rollin g triangl e
on th e line s o f th e triangula r networ k show n i n Figur e

la.
The allowe d transformation s ar e composition s o f re-
flections i n thes e lines . Le t G b e th e se t o f al l suc h
transformations. Thi s se t contains , fo r example , rota -
tions throug h 1 20 ° aroun d th e vertices , an d glid e reflec -
tions whos e axe s coincid e wit h th e middl e line s o f ou r
triangle.
The proble m wil l b e solve d i f we prove tha t th e onl y
plane movement tha t belong s to G and leaves the triangl e
1. A rollin g t r i a n g l e 75
in it s plac e i s the identit y transformation . Apar t fro m th e

identity, ther e ar e five movement s t h a t tak e th e triangl e
into itself : tw o nontrivia l rotation s aroun d it s centr e an d
three reflection s i n it s altitudes . W e hav e t o sho w t h a t
none o f thes e belong s t o th e se t G.
The reade r ha s probabl y encountere d problem s t h a t
are solve d b y constructin g a n appropriat e exampl e (o r
counterexample). W e wil l us e th e sam e tric k her e — bu t
the exampl e w e ar e goin g t o construc t i s unusual : i t i s a n
ornament possessin g th e followin g tw o properties . First ,
it i s symmetri c wit h respec t t o an y o f th e line s show n
in Figur e la . Second , i t i s not symmetri c wit h respec t
to th e altitude s o f th e triangl e ABC ', an d it s centr e i s
not th e centr e o f rotationa l symmetr y fo r th e ornament .
The firs t propert y implie s t h a t th e ornamen t i s preserve d
by an y movemen t t h a t belong s t o th e se t G , whil e th e
second mean s t h a t G contain s non e o f th e five nontrivia l
movements o f th e triangle . Thu s w e prov e th e require d
result.
It remain s t o construc t a n ornamen t wit h al l th e
specified properties . A n appropriat e exampl e i s provide d
by a n ancien t Chines e ornamen t ( a grating ) show n i n
Figure l b . O f course , thi s exampl e i s no t unique . T h e
general recip e t o buil d suc h a n exampl e ca n b e state d a s
follows. Choos e a completel y asymmetri c p a t t e r n insid e
the triangl e ABC ', i.e . a figure whic h i s no t preserve d b y
any non-identit y movemen t o f th e triangle . T h e n tak e
the unio n o f al l figures t h a t ar e obtaine d fro m thi s pat -
tern b y successiv e reflection s i n th e side s o f th e triangle .
One ma y imagin e t h a t th e p a t t e r n insid e th e triangl e i s
dyed wit h paint , leavin g a colou r trac e o n th e plan e whe n
the triangl e roll s over .
E x e r c i s e 5 0 . Introduc e a coordinat e syste m i n th e plan e (se e defini -

tion 3 o n pag e 20 ) suc h tha t th e poin t A ha s coordinate s (0,0) ,
point B ha s coordinates (6 , 0), while point C ha s coordinates (0 , 6)
76 3. T r a n s f o r m a t i o n G r o u p s
(triangle ABC i s still suppose d t o b e equilateral!) . Tak e th e pat -

tern consistin g o f one point K(3, 1 ). Dra w th e ornamen t resultin g
from thi s pattern , an d describ e th e coordinate s o f al l it s points .
E x e r c i s e 5 1 . Wil l th e assertio n o f Proble m 2 7 still hol d i f th e equi -
lateral triangl e i s replace d b y a triangl e wit h angle s (a ) 45° , 45° ,
90°? (b ) 30° , 60° , 90° ? (c ) 30° , 30° , 1 20° ?
2. Transformatio n group s
D e f i n i t i o n 1 0 . A transformation group i s a se t G o f transformation s
of a certai n se t whic h ha s th e followin g tw o properties :
(1) I f tw o transformation s / an d g belon g t o G , s o doe s thei r

composition fog.
(2) Togethe r wit h ever y transformatio n / , th e se t G als o con -
tains th e invers e transformatio n f~ l.
These tw o propertie s mea n t h a t th e element s o f th e se t G ar e

interrelated an d for m a whol e whic h i s close d unde r compositio n an d
taking th e inverse . Th e notio n o f a se t close d wit h respec t t o a cer -
tain operatio n ha s appeare d severa l time s i n thi s book , startin g fro m
Problem 1 .
E x a m p l e 1 . T h e se t o f transformation s G considere d i n th e
discussion o f Proble m 2 7 i s a transformatio n group . Propert y (1 )
is crucia l fo r th e solutio n o f Proble m 27 . I t hold s b y construction .
Property (2 ) i s als o vali d becaus e th e invers e t o a serie s o f reflection s
is th e serie s o f th e sam e reflection s performe d i n th e invers e order .
E x a m p l e 2 . Th e se t o f al l transformation s o f a give n se t M
forms a transformatio n group , denote d b y T r ( M ) .
E x e r c i s e 5 2 . Prov e tha t ever y transformatio n grou p contain s th e
identity transformation .
Apart fro m th e grou p G , i n th e discussio n o f Proble m 2 7 w e hav e

dealt wit h tw o mor e groups : th e grou p M o f al l movement s o f th e
plane an d t h e symmetr y grou p D3 o f symmetrie s o f a n equilatera l
triangle.
T h e group s G an d D 3 ar e containe d i n M; thi s fac t i s usuall y
expressed b y sayin g t h a t the y ar e subgroups o f M. I n general , give n
2. Transformatio n group s 77
an arbitrar y plan e figur e <£ , on e ca n conside r th e se t o f al l plan e

movements tha t tak e $ int o itself . Thi s se t i s denote d b y Sym( )
and calle d th e symmetry group o r group of movements o f th e figur e
$. Ever y elemen t o f thi s grou p i s calle d a symmetry o f th e figur e $ .
Thus, D% i s the symmetry grou p of an equilateral triangle: Sym ( A) =
D 3 . Th e grou p G in Problem 2 7 is also a symmetry grou p o f a certai n
figure, namely , o f th e ornamen t depicte d i n Figur e lb . T o verif y
this fact , w e onl y hav e t o chec k tha t an y movemen t preservin g th e
ornament i s a compositio n o f severa l reflection s i n th e line s o f th e
triangular grid .
Exercise 53 . Mak e sure this is indeed true .
The notio n o f a symmetr y grou p i s a sourc e o f numerou s inter -
esting example s o f transformatio n groups . Le t u s conside r som e o f
them.
Problem 28 . Figure 2 shows ancient Japanese family insignias (ka-

mon). Find the symmetry group of each kamon. Which symmetry
groups are the same and which are different?
a b e d
Figure 2 . Firs t se t o f kamon
Solution. Figur e $ 1 is axially symmetric with respect t o

the four axe s at angle s of 45° from eac h other; i t als o does
not chang e unde r rotation s throug h 90° , 1 80 ° an d 270° .
The grou p Sym($i) , lik e any transformatio n group , als o
contains th e identit y transformation ; therefore , th e to -
tal numbe r o f it s element s i s 8. Th e grou p Sym($2 ) als o
contains eigh t elements , but i t i s different fro m Sym($i) ,
because al l it s element s ar e rotations . Figur e $ 3 ha s th e
same symmetry a s i: four reflections an d four rotations ,

including th e identity . Finally , figur e $ 4 ha s n o nontriv -
ial transformations , an d it s symmetr y grou p consist s o f
only on e element , th e identit y transformation .
Exercise 54 . Fin d th e symmetr y group s o f Figure s 3 , a—d . Com -
• E3O 0
pare the m wit h eac h other, an d als o with th e group s o f Problem
28.
a b e d
F i g u r e 3 . Secon d se t o f kamon
3. Classificatio n o f finite group s o f movement s

The symmetr y group s o f al l th e kamo n displaye d i n Figure s 2 and 3
consist eithe r o f severa l rotation s o r o f several rotation s an d a n equa l
number o f reflection s (includin g th e identity) . Thi s observatio n ca n
be generalize d a s a theorem . T o stat e it , w e nee d som e terminology .
A transformation grou p i s said t o b e finite i f it consist s o f a finite
number o f elements. Th e symmetry group s of all the kamon in Figures
2 and 3 are finite. A group which which consists o f an infinit e numbe r
of element s i s calle d a n infinite group. Th e grou p J\A of al l plan e
movements an d th e grou p G (Proble m 27 ) ar e infinite .
The order o f a grou p i s th e numbe r o f element s i t contains . A
finite grou p i s a grou p o f finite order .
The grou p tha t consist s o f rotation s aroun d a commo n centr e
through multiple s o f 360 ° jn i s called th e cyclic group of order n an d
denoted b y C n. Th e grou p tha t contain s th e sam e rotation s an d n
reflections i n th e line s passing throug h th e sam e centr e an d suc h tha t
3. Classificatio n o f finit e group s o f movement s 7 9
the angl e betwee n an y tw o neighbourin g line s i s 1 80 ° /n i s calle d th e

dihedral group of order 2n an d denote d b y D n.
For example , th e symmetr y group s o f th e eigh t figure s tha t w e
considered (Figure s 2 and 3 ) ar e D4, Cg , D4, Ci , C3 , D3, D\, C2 .
Theorem 6 . Any finite group of plane movements is either a C n or

a D n.
Proof. T o prove the theorem, w e first note that a finite group canno t
contain paralle l translations , because , i f i t contain s a translatio n b y
vector a , i t mus t als o contai n a n infinit e numbe r o f translation s b y
multiple vector s na .
If the grou p contain s a glide reflection, i t als o contains it s square ,
which i s a paralle l translation , an d therefor e canno t b e finite. W e
conclude tha t an y finite grou p o f plan e movement s consist s entirel y
of rotation s an d reflections .
All rotation s belongin g t o th e grou p mus t hav e a commo n cen -
tre, becaus e th e followin g exercis e show s that a group containin g tw o
rotations wit h differen t centre s als o contain s a paralle l translation .
Exercise 55 . Prov e tha t i f A an d B ar e tw o different point s o f th e

plane and th e angle s cp an d ij) ar e not multiple s of 360°, then th e
product RQ O R^ O R^ O R^ i s a nontrivial translation .
Denote al l rotation s tha t belon g t o a give n finite grou p b y i?^ ,

i?^, . . . , JR^ , wher e th e angle s <p , . . ., uo ar e chosen to be positiv e an d
not exceedin g 360° . Suppos e tha t ip is th e smalles t o f thes e angles .
Then al l the remaining angles must b e multiples of?/;. Indeed , suppos e
that if i s no t divisibl e b y ip. The n i t ca n b e writte n a s tp — kip -f £ ,
where k i s a n intege r an d 0 < £ < ip. Th e rotatio n throug h £ mus t
also belong to the group under study , an d we arrive at a contradiction .
Note tha t ip must b e equa l t o 360°/ h fo r som e intege r n —other-
wise a certain powe r o f the rotatio n R^ woul d represen t a rotation b y
an angl e smalle r tha n ip. W e hav e thu s prove d tha t al l th e rotation s
present i n an y finite grou p o f plan e movement s ar e R A , wher e k =
0 , 1 , . . . , n — 1 and tp = 360 ° jn. I f th e grou p contain s n o reflections ,
then i t i s th e grou p C n.Tue Mar 12 18:16:11 EDT 2019for download from IP 132.174.254.159.
Licensed to Penn St Univ, University Park. Prepared on
Now suppose tha t th e grou p contain s n rotation s an d a t leas t on e

reflection. W e ar e goin g t o prov e tha t th e numbe r o f reflection s i n
the grou p i s exactly n .
Indeed, i f i?i , R2, • • •, Rn ar e n differen t rotation s an d S a re -
flection, the n th e n composition s S o R1, S o R2, . . . , S o Rn represen t
n differen t reflection s tha t belon g t o th e group . Thus , th e numbe r o f
reflections i s n o smalle r tha n n. Similarly , th e numbe r o f rotation s
in th e grou p i s no smalle r tha n th e numbe r o f reflections , because , i f
5i, 52 , . . . , Sm ar e m differen t reflections , the n S\o S\, S\o S2, . . . ,
S\ ° S m ar e m differen t rotation s (includin g th e identity) .
We hav e thu s prove d tha t an y finit e grou p o f plan e movement s
either is a C n o r it consists of n rotations with a common centre and a n
equal numbe r o f reflections. I f n = 1 , what w e get i s the grou p D\ o f
order 2 , which contains one reflection an d the identity transformation .
If n > 2 , we have to sho w that th e axe s o f all reflections pas s throug h
the centr e o f rotations . First , observ e tha t a finit e grou p ma y no t
contain tw o reflections whos e axe s ar e parallel , becaus e thei r produc t
would produc e a translation b y a non-zero vector . Thu s an y tw o axe s
must hav e a common point . Th e produc t o f the two reflections whos e
axes intersec t a t a poin t P makin g th e angl e <p belong s t o th e grou p
and i s a rotatio n aroun d P throug h th e angl e 2ip. Hence th e poin t P
coincides wit h th e commo n centr e A o f al l rotations , whil e th e angl e
(p is a multipl e o f 1 80° /n. Th e grou p unde r stud y i s thu s D n. Thi s
completes th e proof . •
Exercise 56 . Ca n a plane figure have

(1) exactl y tw o symmetry axes ?
(2) exactl y tw o centres of symmetry ?
Exercise 57 . Whic h is the most symmetrical (i.e. , having the biggest
symmetry group ) bounde d plan e figure?
4. Conjugat e transformation s
In th e discussio n o f Proble m 2 8 abov e w e sai d tha t th e symmetr y
groups o f figure s $ 1 an d $ 3 ar e th e same : Sym($i ) = Sym(3) .
What i s th e precis e meanin g o f thi s equality ? I n a mor e genera l
setting: wha t i s th e precis e meanin g o f th e classificatio n theore m w e
4. Conjugat e transformation s 81
proved i n th e previou s section ? Le t u s thin k a littl e abou t thes e

questions.
Equality o f two symmetry group s lik e Sym(i) an d Sym($3 ) ha s
the verba l meaning , i.e. , th e tw o set s o f plan e movement s coincid e
only i f the tw o figure ar e place d i n the plan e i n suc h a way tha t thei r
centres and symmetry axe s coincide. Otherwis e the sets G = Sym(3>i )
and H = Sym($3 ) woul d be different, althoug h closel y related t o each
other. W e will no w elucidat e thi s relation .
Figure 4 . Conjugat e symmetr y group s
Denote th e centr e o f rotation s an d th e symmetr y axe s fo r th e

group G by A an d ai , . . . , a<±. Denot e th e sam e object s fo r th e grou p
H b y B an d bi, . . . , 64 , respectively (se e Figure 4) . Le t / b e a plan e
movement tha t carrie s B int o A an d eac h bi into a^. Th e transforma -
tions o f th e grou p G ca n b e obtaine d fro m th e transformation s tha t
belong t o th e grou p H i n th e followin g way . Consider , fo r example ,
the reflectio n i n th e lin e 63 . Firs t mov e th e figur e $ 1 wit h th e hel p
of th e movemen t / - 1 , the n appl y reflectio n i n 6 3 to it , the n mov e i t
back wit h th e hel p o f / . I t i s eas y t o se e tha t afte r al l thes e action s
the figure ha s undergon e a reflectio n i n th e lin e a^.
In general , fo r an y elemen t ft o f th e grou p H th e compositio n
f oho f~ l i s a n elemen t o f G. Mor e precisely , i f ft i s a reflectio n i n
bi, the n / o / j o / " 1 i s a reflectio n i n a* ; i f ft i s a rotatio n aroun d B,
then / o ft o /- 1i s a rotatio n aroun d A throug h th e sam e angle .
Definition 1 1 . Tw o transformation s ft G H an d g = / o ft o f~l e
G ar e sai d t o b e conjugate. Th e transitio n fro m ft t o g i s calle d
conjugation by f.
We have encountere d conjugat e transformation s before , whe n we

derived th e complex numbe r formul a fo r a rotation wit h a n arbitrar y
centre, an d in Exercise 3 7 (Chapter 2) .
Two group s o f movement s ar e sai d t o b e conjugate, i f th e lis t
of element s o f on e grou p become s th e lis t o f element s o f th e othe r
upon conjugatio n b y a certai n (on e and the same ) movement . Th e
symmetry group s o f tw o copies o f on e and th e sam e figure , place d
arbitrarily i n th e plane , ar e alway s conjugate . Th e conjugatio n i s
effectuated b y a movemen t tha t carrie s on e o f th e copie s int o th e
other.
In general, conjugatio n shoul d be understood a s looking at an ob-
ject fro m a different viewpoint . Th e conjugating movemen t i s the one
which relate s th e two viewpoints (o r systems o f reference, i n physical
terminology).
The most importan t propert y o f conjugate subgroup s is that the y
have th e sam e interna l structure . Le t u s explai n th e exac t mathe -
matical meanin g o f this phrase . Le t G and H b e two groups o f plane
movements tha t ar e conjugated b y a movement / . I f g = f oho / - 1 ?
then w e will sa y tha t g an d h correspond t o eac h other , an d writ e
g «-» h. Thi s correspondenc e i s one-to-one, becaus e h can be uniquely
expressed i n terms of g as f~ l o g o f.
Then th e following tw o facts hold :
(1) I f gi <- » h\ an d g ft, then g" 1 <- • h~ l.
Both fact s ar e verified i n a straightforward way :

(1) ^ l O ^ = ( / o / l 1 o / - 1 ) o ( / o / l 2 o / - 1 ) = /o(ft 1 oft2)orl.
l
(2) ( f o h o f - ^ = foh- of-\
Thus, bot h group operations (compositio n and taking the inverse)

in on e group correspon d t o thei r respectiv e counterpart s i n anothe r
group unde r th e correspondenc e unde r study . Thi s phenomeno n i s
called isomorphism, an d we will study it in detail later i n this chapter .
Now we will derive formulas fo r conjugation i n the group of plane
movements.
Problem 29 . Find the movement which is conjugate to the rotation

R^ by means of the reflection Si.
Solution. B y virtue o f the general remark tha t w e made
above, t o ge t th e conjugat e movemen t on e ha s t o loo k
at th e give n rotatio n 'fro m unde r th e plane' , displacin g
one's viewpoin t b y mean s o f the reflectio n Si. I t i s fairl y
evident tha t th e result i s the rotation through —a around
the poin t A! whic h i s symmetri c t o A wit h respec t t o / .
We will perform a rigorous chec k of this result , i.e . prov e
that SioRôSf 1 =RA?-
Figure 5 . Conjugatio n o f a rotatio n b y a reflectio n
Indeed, le t M b e a n arbitrar y poin t i n the plan e (se e

Figure 5) . Le t M i b e it s image under th e movemen t S^ 1
(which, in fact, i s the same thing as Si). Suppos e that M i
goes int o M 2 unde r R% and M 2 goe s int o M 3 unde r Si.
Then AMA'M 3 = AM 1 AM2, whenc e M 3 = R~A?{M).
Exercise 58 . Compil e a complet e tabl e o f al l conjugation s i n th e

group o f plan e movements , i.e. , find / o g o f~l fo r ever y g =
Ta, R%,Sh Ut an d ever y / = T b, RfiBiSm, U*.
To acquir e freedo m i n manipulation s wit h plan e movements , w e
suggest tha t th e reade r mak e an d practic e usin g a simpl e too l tha t
we cal l a 'dihedra l instrument' . I t consist s o f a regula r polygo n cu t

from cardboar d an d o f th e sam e polygo n draw n o n a piec e o f paper .
The vertice s o f each polygo n shoul d b e consecutivel y numbere d b y 1 ,
2, . . . .
3
4 2
5 1
Figure 6 . Instrumen t fo r studyin g th e dihedra l grou p
With th e hel p o f thi s instrument , on e ca n stud y th e dihedra l

group D n. T o find th e produc t o f two element s o f D n, on e mus t first
place th e cardboar d polygo n o n th e pape r on e i n it s initia l position ,
so tha t th e vertice s wit h th e sam e number s coincide , the n perfor m
the give n movement s on e afte r anothe r and , comparin g th e number s
of vertices, tr y t o figure ou t wha t i s the compose d movement . I n th e
same wa y on e ca n als o comput e th e tabl e o f conjugations .
Exercise 59 . Usin g appropriate dihedral instruments, fill in the mul-
tiplication an d conjugatio n table s fo r th e group s D3 and D4.
Looking a t th e tabl e o f conjugate movement s (th e answe r t o Ex -
ercise 58) , on e ca n se e tha t a translatio n T a doe s no t chang e whe n
conjugated b y anothe r translatio n T\>:
ThoTaoT~1=Ta.
This equatio n i s equivalen t t o T\> o Ta = T a o Tt>, whic h mean s tha t
any tw o translation s commute; i n othe r words , the y ar e interchange -
able wit h respec t t o composition . (W e recal l tha t w e hav e alread y
mentioned commutativit y i n th e discussio n o f Proble m 24. )
A transformatio n grou p wher e an y tw o element s commut e i s
called
Licensed to Penn Stcommutative.
Univ, University Park. Prepared on Tue Mar 12 18:16:11 EDT 2019for download from IP 132.174.254.159.
Problem 30 . Find all finite commutative groups of plane move-

ments.
Solution. W e kno w th e lis t o f al l finit e group s o f plan e

movements: i t i s mad e u p o f C n an d D n fo r al l n =
1,2,3,....
Every cycli c grou p C n i s commutative . Thi s follow s
from th e fac t tha t th e grou p o f al l rotations wit h a com -
mon centr e i s commutative, becaus e th e produc t o f rota -
tions through angle s a an d (3 is a rotation throug h a + /?,
no matte r whic h orde r th e compositio n i s taken in .
The grou p Di i s commutative , becaus e i t consist s
of onl y tw o elements , on e o f whic h i s th e identity . I t i s
easy t o se e tha t th e grou p D2 i s commutative , too . I n
fact, i t consist s o f tw o reflection s S± an d S2 whos e axe s
are mutuall y perpendicular , a half-tur n rotatio n R an d
an identit y transformation . Accordin g t o th e rule s w e
have derived earlier , th e compositio n 5 i o 5 2 i s a rotatio n
through 1 80 ° i n one direction , whil e S2 ° S± is a rotatio n
through 1 80 ° i n the opposit e direction . Hence , S\o S2 =
S2 o Si. Therefore , Si o R = Si o S2 o Si = R o Si .
Similarly, S2 ° R = R ° £2.
Now conside r th e grou p D n fo r n > 3 . Choos e tw o
reflections tha t belon g t o thi s grou p an d whos e axe s ar e
adjacent, i.e . mak e a n angl e o f 1 80°/n . Thei r product s
are rotations throug h th e angle 360°/n, i n the positive or
negative direction , dependin g o n th e orde r i n whic h th e
product i s taken . Sinc e n > 3 , thes e tw o product s ar e
different. Therefore , th e grou p D n i s non-commutative .
The complet e lis t o f finite commutativ e group s o f
plane movement s thu s include s Di , J9 2 an d al l group s
The propert y o f commutativit y o f a grou p ca n b e easil y rea d of f

its multiplicatio n table : a grou p i s commutativ e i f an d onl y i f it s
multiplication tabl e i s symmetri c wit h respec t t o th e mai n diagonal .
Readers wh o di d Exercis e 5 9 have visual evidenc e o f the fac t tha t th e
groups Ds an d D4 are not commutative . However , eve n if the table is

not symmetri c a s a whole , i t alway s contain s pair s o f equa l element s
that occup y symmetri c positions . Suc h pair s correspon d t o th e pair s
of commutin g element s o f th e group .
Exercise 60 . Indicat e all pairs of commuting elements in the groups
D3 and D 4 .
5. Cycli c group s
There i s anothe r wa y t o explai n wh y th e cycli c grou p O n i s commu -
tative. Le t R denot e th e rotatio n throug h 360° /n. The n al l th e ele -
ments of the group can be represented a s powers of R, i.e . R 2 — RoR,
R3 — R o R o R, . .., R n — id, an d i t i s clea r tha t w e alway s hav e
Rk 0R 1 = R k+l = R loRk.
A transformation / whos e power s exhaus t th e se t o f al l element s
of a group i s called a generator, o r a generating element o f the group .
To bette r understan d th e meanin g o f thi s notion , le t u s imagin e tha t
we hav e n o group , bu t onl y on e transformatio n / o f a certai n se t
M. Th e questio n i s whether ther e exist s a transformatio n grou p tha t
contains thi s transformatio n / . Th e answe r t o this questio n i s always
positive. Th e smalles t grou p containin g a given transformatio n / ca n
be constructe d i n th e followin g way .
If a group contains an element / , then , accordin g to the first defin -
ing property o f a group, i t mus t contai n al l it s powers / 2 , / 3 , etc . B y
the secon d property , i t als o mus t contai n th e invers e transformatio n
Z" 1 an d therefor e al l it s power s ( / _ 1 ) 2 , ( / _ 1 ) 3 , etc .
Exercise 61 . Prov e that (f~ x)k = (Z*)" 1 -
The transformation (/~ 1 ) / c , wher e k is a natural number , i s called
a negative power of / an d is also denoted b y f~~ k. Th e zeroth power of
any transformatio n i s by definitio n th e identit y transformation . No w
observe tha t th e se t o f al l intege r power s o f a give n transformation ,
. . . , / ~ 2 , / - 1 , / ° , Z 1 , f 2, • • -> always form s a group , becaus e o f th e
identities f k o f l = f k+l an d (f k)~1 = f~ k, whic h hol d no t onl y
for natural , bu t als o fo r al l intege r value s o f k an d I. Thi s se t o f al l
powers o f f i s calle d the group generated by f.
Two possibilitie s ma y arise .
5. Cycli c group s 87
(1) Al l th e power s f k ar e different . I n thi s cas e th e grou p gen -

erated b y / i s infinit e an d i s calle d a n infinite cyclic group.
(2) Amon g th e power s o f / ther e ar e som e tha t coincide . The n

there i s a positiv e powe r o f / whic h i s equal t o th e identit y
transformation. Indeed , i f f k an d f l i s an y pai r o f coincid -
ing power s wit h k > / , the n f k~l = id . Denot e b y n th e
smallest positiv e exponen t satisfyin g f n = id . Th e num -
ber n i s calle d th e order o f th e transformatio n / . I n thi s
case the grou p generate d b y / consist s o f exactly n differen t
transformations / , / 2 , . . . , f n. (Al l o f thes e ar e differen t
indeed, becaus e i f w e had f k = f l wit h 0 < I < k < n, the n
we would ge t f k~l = id , contrar y t o th e choic e o f n.)
In thi s cas e th e grou p generate d b y / i s a finite cyclic
group of order n. I n particular , thi s notio n include s th e
cyclic group s o f rotation s C n considere d above .
When th e transformatio n / generate s a n infinit e group , w e ca n

also sa y tha t / ha s infinite order. Th e orde r o f th e identica l trans -
formation i s 1 b y definition , an d th e grou p i t generate s consist s o f
only on e elemen t an d i s calle d th e trivial group. A n involutiv e trans -
formation generate s a grou p o f orde r 2 , consistin g o f itsel f an d th e
identity.
Problem 31 . Which elements of the group C\2 are generators of this

group? What are the subgroups generated by other elements?
Solution. C1 2 consist s o f 1 2 rotation s throug h angle s

which ar e multiple s o f 30° . Al l thes e rotation s ar e pow -
ers of the rotatio n throug h 30° , which i s therefore a gen -
erator o f th e group . Th e invers e rotatio n (b y 330° ) i s
obviously a generator , too . T o facilitate th e stud y o f th e
other elements , w e us e Figur e 7 , where ever y elemen t o f
the grou p i s represente d b y a verte x o f th e regula r 1 2 -
gon: AQ corresponds t o th e identity , Ai i s th e rotatio n
through 30° , etc. Conside r th e elements of the grou p on e
by one and, fo r every element, mar k al l the vertices of the
F i g u r e 7 . Cycli c grou p C\2 an d it s subgrou p
polygon tha t correspon d t o th e power s o f thi s element .

We wil l se e that :
• Ther e ar e tw o mor e rotation s — throug h 1 5 0 an d
210 degree s — that generat e th e whol e group .
• Rotation s throug h 6 0 and 30 0 degree s hav e orde r 6
and generat e th e grou p CQ visualized a s a hexago n
AoA2A4A6AsA10 i n Figur e 7 .
• Rotation s throug h 9 0 and 27 0 degree s generat e th e
group o f orde r 4 (th e squar e AOASAQAQ).
• Rotation s throug h 1 2 0 and 24 0 degrees generate th e
group o f orde r 3 (th e triangl e AQA±A%).
• Th e rotatio n throug h 1 80° , whic h i s a n involution ,
generates th e grou p C% depicte d a s the lin e segmen t
A0A6.
This resul t ca n b e summarize d i n a tabl e wher e th e
upper lin e i s fo r th e value s o f /c , whil e th e lowe r lin e
shows the orde r o f the rotatio n R k (R bein g th e rotatio n
through 30°) :
0 1 2 3 4 5 6 7 8 9 10 11
1 12 6 4 3 12 2 12 3 4 6 12
5. C y c l i c g r o u p s 89
E x e r c i s e 6 2 . Usin g the notion of the greatest commo n divisor (GCD )

of two numbers, fin d a general formula fo r th e orde r o f the elemen t
fk i n a cycli c grou p o f orde r n generate d b y / .
A transformatio n belongin g t o a finit e grou p i s a generato r o f thi s

group i f an d onl y i f it s orde r i s equa l t o th e orde r o f th e group . T h e
number o f generatin g element s i n th e cycli c grou p C n i s denote d b y
</?(n), an d th e functio n (p is calle d th e Euler function. Fo r example ,
the tabl e abov e show s t h a t </?(1 2 ) = 4 .
E x e r c i s e 6 3 . (a ) Compil e the tabl e o f values of the functio n (p(n) fo r
n — 2 , 3 , . . ., 1 5 . (b ) Fin d a genera l formul a fo r <p(n) i n term s o f
the prim e decompositio n o f th e numbe r n.
Now le t u s se e wha t ar e th e order s o f differen t plan e movement s

according t o thei r type . I t i s clea r t h a t non-identica l translation s an d
glide reflection s hav e a n infinit e order , becaus e unde r th e repeate d
action o f suc h a transformatio n an y poin t occupie s infinitel y man y
new position s (Figur e 8 , a an d b) .
/ / /
AW d
F i g u r e 8 . Orde r o f translation , reflectio n an d glid e reflectio n
E x e r c i s e 6 4 . I s i t possibl e fo r a plan e figur e t o remai n unmove d un -

der a non-trivia l translatio n o r a glid e symmetry ?
Any reflectio n ha s orde r 2 (Figur e 8c) .

Rotations ma y hav e differen t orders . I f th e angl e o f rotatio n i s

measured b y a rationa l numbe r o f degree s 360 ° • m/n wher e m/n i s
an irreducibl e fraction , the n th e rotatio n ha s a finit e orde r n (Figur e
8d). Fo r a n irrationa l numbe r o f degree s th e orde r i s infinite .
Exercise 65 . Verif y th e previous assertion for the rotations of Prob-
lem 31 . Prove the general fact .
6. Generator s an d relation s
Cyclic groups , i.e. , group s generate d b y on e element , constitut e th e
simplest clas s of groups. No w we will consider th e group s that canno t
be generate d b y on e element .
To begi n with , le t u s prov e tha t th e grou p D n, wher e n > 2 , i s
not cyclic . I n fact , w e have alread y see n that i f n > 2 , then D n i s no t
commutative an d henc e i s no t cyclic . I n th e cas e n — 2 , not e tha t
every non-identity elemen t o f the grou p D2 is of order 2 , and thu s th e
group doe s no t contai n an y elemen t o f orde r 4 .
A natura l questio n arises : wha t i s th e smalles t se t o f element s
of th e grou p D n whic h generate s th e whol e group , i.e. , allow s u s t o
express an y elemen t o f th e grou p usin g multiplication s an d takin g
the inverse ? I t turn s ou t tha t tw o element s ar e enough , an d {R,S},
where R i s the rotatio n throug h 360 ° /n an d S a n arbitrar y reflection ,
is a n exampl e o f suc h a set . I n fact , ever y rotatio n belongin g t o D n
can b e represente d a s R k, an d ever y reflectio n a s R k o S.
The firs t o f thes e tw o assertion s i s evident . T o prov e th e other ,
note that al l the movements R k oS fo r k = 0 , 1 , . . . , n— 1 are different .
Indeed, a n equalit y R k o S = rf o S , whe n multiplie d b y S o n th e
right, woul d impl y R k = R l, whic h i s a contradiction . No w observ e
that al l thes e movement s ar e improper , i.e . reflections , no t rotations .
Since th e tota l numbe r o f reflections i n the grou p D n i s n, w e deduc e
that eac h o f the m mus t appea r i n th e lis t 5 , R o 5, . . . , R n~1 o 5.
We hav e thu s prove d tha t th e pai r {R, S} i s a se t o f generator s
of th e grou p
D = {id , i J ,. . . , R k~\ S , R o 5 , . . ., R k~l o S).

n Park. Prepared on Tue Mar 12 18:16:11 EDT 2019for download from IP 132.174.254.159.
Licensed to Penn St Univ, University
6. Generator s an d relation s 91
n ?
Figure 9 . Th e grou p D n
Let u s describ e th e multiplicatio n rul e o f th e grou p D n i n term s o f

these expression s throug h R an d S.
• Th e compositio n o f R k an d R l i s R k+l. Ifk + l>n, thi s ca n
be replace d b y R k+l~n t o coincid e literall y wit h a n elemen t
of th e abov e list .
• I n th e sam e way , R h o (Rl o S) = R k+l o S, or , i f k + I > n ,
it i s better t o writ e R k o (i?z o 5) = J Rfc+^"n o S.
• Le t u s find th e compositio n (R k o S) o Rl. Recal l (Proble m
29) tha t SoR k oS = R- k. Multiplyin g thi s equalit y b y S
on th e right , w e get S o Rk = R~ k o S. Therefore ,
(Rk o S) o Rl = R k o (S o iJz) = i^ fc o (RTl oS) = R k~l o 5.
• A simila r argumen t show s tha t (R k o 5) o (i^z o S) = i? fc~z.
In term s o f th e generator s R an d 5 , th e multiplicatio n tabl e fo r

the grou p D n ca n b e represente d a s follows :
Rl RloS
Rk Rk+i Rk+l oS
RkoS Rk~l oS Rk-i
where, whe n necessary , th e exponen t o f R ca n b e increase d o r de -

creased b y n , usin g th e fac t tha t R n = id .
For specifi c value s o f n , thi s shor t tabl e ca n b e expande d t o it s
full form . Fo r example , i f n — 3 , we get th e followin g ful l table :
id R R2 Sa sb Sc
id id R R 2
Sa sb Sc
R R R 2
id sb Sc Sa
R2 R2 id R Sc Sa sb
Sa sa Sc sb id R2
R
sb sb Sa Sc R id R2
Sc sc sb Sa R2 R id
An importan t observatio n i s that th e complet e multiplicatio n ta -
ble o f th e grou p D n follow s fro m jus t thre e relation s betwee n th e
generating element s R an d S:
2 2
(25) 5 = id ; (Soj?) = id ; i T = id .
All othe r relation s betwee n R an d S ca n b e formall y deduce d fro m
these thre e usin g th e definitio n o f a group . A s a n example , le t u s
check thi s fo r th e relatio n S o Rk o S = R~ k tha t w e have use d whe n
working o n th e multiplicatio n table .
The secon d relatio n i n (25 ) ca n b e expande d a s
SoRoSoR = id
or, equivalently , a s
1
SoRoS = R- .
2
Taking int o accoun t tha t S — id , th e n-t h powe r o f the las t equalit y
gives
SoRkoS = R~ k.
It turn s ou t tha t relation s (25 ) constitut e a complete se t o f defin-
ing relations fo r th e grou p D n i n th e followin g sense : i f a grou p i s
generated b y tw o element s S an d R whic h satisf y th e relation s (25 )
and no other relation except for those that are formal consequences
6. Generator s an d relation s 93
of these three, the n th e orde r o f thi s grou p i s 2 n an d it s structur e i s

the sam e a s th e structur e o f th e grou p D n.
We wil l no w tr y t o giv e a genera l definitio n o f generator s an d
defining relation s o f a transformation group . Le t S = { s i , . . . , s n } b e
a finite subse t o f a group G. B y a monomial over S w e understand a
product o f the for m M = sf 1 ^ 2 • • • s^™, where ii , ... , i m ar e number s
between 1 and n an d th e exponent s fci, ... , k n ar e arbitrar y integers .
A relation betwee n si , ... , s n i s a monomial r ove r S whic h i s equal t o
id i n th e grou p G. Relation s ca n als o be writte n i n th e for m n = r2 ,
which i s equivalent t o ryr^ 1 = id .
Definition 1 2 . Le t G b e a transformatio n group , S C G a subse t

and R a certai n se t o f relation s betwee n th e element s o f S
We say that th e set S i s a set of generators an d R a set of defining
relations fo r th e grou p G , i f
1. an y elemen t o f G ca n b e represente d a s a monomia l ove r 5 ,
and
2. an y relatio n betwee n the elements of S i s a formal consequenc e
of th e relation s belongin g t o th e se t R.
By a forma l consequenc e o f relation s r\ = id , ... , r m = id , wher e

every r^ is a monomial over the set 5 , we mean a new monomial r = i d
which ca n b e deduce d fro m th e give n se t usin g th e grou p operation s
(multiplication an d takin g th e inverse ) an d thei r properties , suc h a s
associativity, th e simplificatio n rule s s ksl = s fc+*, s° — id, an d th e
fact tha t th e equalitie s ab = i d an d ba = i d ar e equivalent . Fo r
example, th e relation s ab 2 = i d an d ba 2 = id impl y tha t b — a~ 2 an d
hence a 3 = id .
Let u s not e tha t i n a certai n sense , th e natur e o f th e element s
S i , . . . , s n i s here irrelevant: late r (pag e 1 3 6 in Chapter 5 ) we will give
a definitio n o f a n abstrac t grou p wit h a prescribe d se t o f generator s
and relations .
We pas s t o som e exercise s wher e th e notio n o f definin g relation s
is crucial .
Problem 32 . Suppose that A and B are two transformations that

satisfy the relations
3 5
(26) A = id ; B = id ; AB = B 4A
and do not satisfy any relations that do not follow from these three by
group axioms. Prove that the group generated by A and B is a cyclic
group of order 3.
Solution. T o simplif y th e formulas , bot h i n th e state -

ment an d th e discussio n o f Problem 3 2 we omit th e sym -
bol of composition (smal l circle) an d correspondingl y us e
the wor d 'multiplication ' instea d o f 'composition' .
An arbitrar y elemen t o f the grou p G generated b y A
and B ca n b e writte n a s
Bk*AhBk2Al2 ...B km lm
A ,
where, i n virtu e o f (26) , w e ca n assum e tha t 0 < ki < 4

and 0 < U < 2 . Le t u s transfor m thi s 'word ' usin g th e
following rule : each time that an A appears next to a
B on the left, replace AB by B 4A. Applyin g thi s rul e
sufficiently man y times , w e wil l soone r o r late r pus h al l
i?'s t o th e lef t an d arriv e a t a wor d o f th e typ e B kAl,
where, a s before , w e ar e i n a positio n t o assum e tha t
0 < ki < 4 an d 0 < l % < 2.
Since th e intege r k ma y tak e 5 differen t value s an d
the intege r I three differen t values , w e see tha t th e tota l
number o f products B k A 1 i s no greate r tha n 1 5 . I t turn s
out, however , tha t no t al l of these element s ar e different .
Indeed, w e ca n deduc e th e followin g chai n o f equalitie s
from th e definin g relation s (26) :
B = A 3B = A 2B4A = AB 1 6A2 = B 6AA3 = B\
where w e hav e use d th e abov e describe d rul e an d th e

observation tha t the letter A, when going through a B
from left to right, multiplies its exponent by 4- Therefore ,
B3 — id , which , togethe r wit h th e know n identit y B 5 =
id, implie s tha t B = id .
6. G e n e r a t o r s a n d r e l a t i o n s 95
T h e grou p G i s thu s i n fac t generate d b y onl y on e

element A satisfyin g A 3 — id . I t remain s t o not e t h a t
this generato r A cannot b e trivial , becaus e t h e relatio n
A — id i s not a consequenc e o f the relation s (26) . Indeed ,
if w e tak e a 1 2 0 degre e rotatio n fo r A an d t h e identit y
transformation fo r 5 , the n al l th e thre e relation s (26 )
are true , whil e th e relatio n A = i d i s false .
E x e r c i s e 6 6 . Le t A an d B b e tw o nontrivial movement s o f the plan e

such tha t ABA 2 = i d an d B 2A = id . Wha t i s th e orde r o f th e
group generate d b y A an d J5 ? Wha t kin d o f movement s ar e A
and B?
Similar argument s ca n b e use d i n th e followin g exercise , whic h

at firs t glanc e look s totall y unrelate d t o th e theor y o f transformatio n
groups.
E x e r c i s e 6 7 . Th e languag e o f the trib e Aiu e ha s onl y 4 letters: A , I ,
U an d E . Th e lette r E i s special. Whe n use d b y itself , i t mean s a
certain word , bu t whe n adde d t o an y wor d i n th e beginning , th e
middle o r th e end , i t doe s no t chang e it s meaning . Furthermore ,
each of the letter s A , U , I, pronounced seve n times i n a row, make s
a synony m o f th e wor d E . Th e followin g wor d fragment s ar e als o
considered a s synonyms : UUU I an d IU , AA I an d IA , UUU A an d
AU. Th e tota l numbe r o f peopl e i n th e trib e i s 400 . I s i t possibl e
that al l o f the m hav e differen t names ?
E x e r c i s e 6 8 . Fin d al l pair s o f generator s i n th e grou p D n. Fo r eac h
pair, indicat e th e definin g relations .
The notio n o f definin g relation s ca n b e use d fo r group s t h a t hav e

any numbe r o f generators . Fo r example , th e cycli c grou p o f orde r n
is th e grou p wit h on e generato r R an d on e definin g relatio n R n — id.
An infinit e cycli c grou p i s th e grou p wit h on e generato r T whic h i s
not subjec t t o an y relations . W e shoul d lik e t o emphasiz e onc e agai n
the meanin g o f th e las t phrase . W h a t w e mea n her e i s t h a t ther e ar e
no non-trivial relation s fo r T , i.e. , n o relation s t h a t involv e T an d i d
and d o no t follo w fro m th e genera l propertie s o f groups . Her e i s a n
example o f a trivia l relation : T~ 1 T4T~3 = id .
We shal l no w giv e a n exampl e o f th e grou p wit h thre e generators .
Consider th e grou p o f plan e movement s whic h wa s denote d b y G
in Proble m 2 7 an d whic h w e wil l no w denot e b y p3ral , it s officia l

crystallographic symbo l (se e Chapte r 4 fo r a detaile d discussio n o f
crystallographic groups) . B y definition , th e grou p pSml i s generate d
by a n infinit e numbe r o f reflections , namely , th e reflection s i n al l th e
lines show n i n Figur e la .
It turn s out , however , tha t thi s grou p i s als o generate d b y onl y
three reflection s 5 a , Sb, S c, wher e a, b an d c ar e th e side s o f a n
equilateral triangle which forms the unit o f the lattice shown in Figur e
la. T o prov e thi s fact , w e hav e t o sho w tha t th e reflectio n S x i n
any lin e x tha t belong s t o th e triangula r lattic e unde r stud y ca n b e
expressed i n term s o f S a, 5& , 5C .
Consider, fo r example , th e reflectio n Si (notatio n o f Figur e la) .
Since th e lin e I is symmetri c t o a wit h respec t t o c , w e se e tha t th e
movement Si ca n b e obtaine d fro m S a usin g conjugatio n b y 5 C , i.e. ,
Si = S c o Sa ° S c. Also , sinc e Si(b) = m, w e have 5 m = 5 / o 5b o Si =
Sc o Sa o 5C o Sb o Sc o Sa o 5 C.
A similar argumen t allow s us to express any reflection S x throug h
5 a , Sb, S c, becaus e any line x belongin g to the lattice ca n be obtaine d
from th e line s a, 6 , c by an appropriat e serie s of reflections 5 a , 5& , 5C .
We no w observ e tha t th e thre e generator s satisf y th e followin g
relations:
(27) sl = S 2b=S2c = id ,
3 3
(28) (S a o Sbf = (S b o Sc) = (S c o Sa) = id .
Exercise 69 . Le t F\, F2, F 3 be three plane movements that generat e
an infinit e grou p an d satisf y th e relation s
F? = Fl - Fi = (Fi o F 2f - (F 2 o F 3 ) 3 - (F 3 o F1) 3 = id .
Show that Fi , F 2, F 3 are reflections i n the sides of an equilatera l
triangle.
This exercise shows that relation s (27 ) and (28) , supplied wit h th e
additional requiremen t o f infiniteness , determin e th e grou p uniquel y
as th e grou p o f plan e movements . Actually , i t ca n b e prove d tha t
these relation s ar e th e defining relations fo r th e grou p p3ml i n th e
sense explaine d above .
Chapter 4
Arbitrary Group s
In this chapte r w e will introduce th e genera l notio n o f a group, whic h

includes transformatio n group s a s a particula r case . W e wil l discus s
the basi c propertie s o f group s i n thi s genera l settin g an d conside r
some application s o f group s i n arithmetic .
1. T h e genera l notio n o f a grou p

In our study o f transformation groups , i t often di d not matte r tha t w e
dealt wit h movements o r transformations. Wha t mattere d wa s th e
properties o f the group G expressed a s the following fou r requirement s
imposed o n th e composition define d fo r an y pai r o f element s o f G:
(1) th e compositio n o f tw o movement s belongin g t o G als o be -
longs t o G ;
(2) th e compositio n o f movement s i s associative ;
(3) th e grou p contain s th e identit y transformation , whic h i s
characterized b y the propert y tha t it s compositio n wit h an y
movement, / i s equal t o / ;
(4) togethe r wit h ever y movement , th e grou p contain s th e in -
verse movement .
We arriv e a t th e genera l notio n o f a group , i f w e conside r a n ar -
bitrary set , supplie d wit h a n operatio n whic h associate s a n elemen t
98 4. Arbitrar y Group s
of thi s se t wit h an y pai r o f it s element s an d whic h enjoy s a simila r

set o f properties . Fo r example , th e se t o f al l rea l number s wit h th e
operation o f additio n (R , +) possesse s al l the liste d propertie s an d i n
this sense constitutes a group. W e should lik e to draw the reader's at -
tention t o th e fac t tha t alread y i n Chapte r 1 (Problems 1 , 7, etc.) w e
have use d thes e propertie s fo r element s o f variou s nature s (numbers ,
points, vectors ) an d differen t operation s (addition , multiplication) .
The analog y betwee n Proble m 3 2 and Exercis e 6 7 is also noteworthy .
All thes e fact s testif y tha t th e notio n o f a grou p ough t t o b e state d
in a genera l setting . S o here w e go.
Definition 1 3 . A group i s a se t G wit h th e followin g properties :
(1) ther e i s a rul e ( a (binary) operation), accordin g t o whic h

for an y ordere d pai r (a , b) of elements o f the se t G a certai n
element a * b G G i s defined ;
(2) th e operatio n * i s associative , i.e . fo r an y thre e element s
a,b,c G G the followin g equalit y holds : (a*b)*c — a*(b*c);
(3) i n G , ther e i s a neutral element , i.e. , a n elemen t e such tha t
a* e = e* a — a for an y a G G;
(4) fo r ever y element a G G there is a symmetric elemen t a' G G
which satisfie s a * a' = a' * a = e.
The fou r propertie s 1 - 4 ar e als o calle d th e group axioms. Not e

that an y transformatio n grou p i s a grou p i n thi s genera l sense , al -
though th e definitio n o f a transformatio n grou p (p . 76 ) consist s o f
only tw o out o f the fou r grou p axioms : numbe r 1 and numbe r 4 . Th e
reason i s tha t axio m 2 alway s hold s fo r th e compositio n o f transfor -
mations (p . 69) , an d axio m 3 follow s fro m axiom s 1 and 4 , becaus e
the neutra l elemen t wit h respec t t o th e compositio n o f transforma -
tions i s the identit y transformation , an d th e questio n i s only whethe r
it belong s t o th e give n se t G.
Exercise 70 . Prov e that fo r a finite set G consisting o f transforma -
tions axiom 4 follows fro m axiom s 1 -3 .
Instead o f the symbol s * (asterisk), ' (prime) an d th e term s 'neu -

tral' an d 'symmetric' , whos e meaning i s explained i n the definitio n o f
1. Th e genera l notio n o f a grou p 99
a group , othe r symbol s an d word s ar e used i n various specifi c circum -

stances:
• I n cas e o f transformation group s th e grou p operatio n (com -
position) i s denoted by a small circle (o), the neutral elemen t
is denoted b y i d (identit y transformation) , an d th e rol e o f a
symmetric elemen t / ' i s playe d b y th e invers e transforma -
tion f- 1 .
• Fo r th e grou p o f number s (integer , rational , rea l o r com -
plex) wit h th e operatio n o f additio n th e neutra l elemen t i s
0 (zero) , an d th e elemen t symmetri c t o a give n numbe r a
is th e opposit e numbe r —a. (Whe n w e considere d comple x
numbers a s point s i n th e plane , w e calle d th e neutra l ele -
ment th e pole an d denote d i t b y P.)
• Fo r numeric group s with the operation o f multiplication (fo r
example, th e se t o f all positive rea l numbers) , th e symbo l of
the operatio n i s usuall y omitted , i.e . on e write s ab instea d
of a*6, th e neutra l elemen t i s 1 , and th e symmetri c elemen t
is the invers e numbe r a - 1 .
Groups tha t consis t o f numbers, vectors , etc. , with th e operatio n
of addition , ar e referre d t o a s additive groups. I f th e grou p operatio n
is multiplication , the n th e grou p i s called multiplicative.
The system o f notation adopte d fo r the multiplication o f number s
is th e mos t convenient , s o i t i s ofte n use d fo r arbitrar y groups . W e
shall als o us e i t b y default . Th e onl y thin g tha t ha s t o b e kep t i n
mind i s tha t grou p multiplication , unlik e numeri c multiplication , i s
not i n genera l commutative ; henc e ab and ba need no t b e th e sam e
element o f the group .
Problem 33 . For each of the following sets with binary operations,
determine whether it is a group:
(1) all even integers with the operation of addition;
(2) all odd integers with the operation of addition;
(3) all real numbers with the operation of subtraction;
(4) all natural numbers with the operation of addition;
(5) all non-negative integers with the operation of addition;
(6) all real numbers with the operation x*y = x + y — 1 .

Solution. I n exampl e 1 all th e grou p axiom s ar e obvi -
ously satisfied .
In exampl e 2 th e firs t axio m fails : th e su m o f tw o
odd number s i s not a n od d number .
In th e thir d exampl e th e firs t axio m holds , bu t th e
second fails , becaus e subtractio n i s not associative :
(6-5)-3^6-(5-3).
In the next exampl e the first tw o axioms ar e fulfilled ,
but th e operatio n doe s no t hav e a neutra l element : th e
only numbe r tha t coul d pla y th e rol e o f th e neutra l el -
ement wit h respec t t o additio n i s zero , bu t i t doe s no t
belong t o th e give n set .
Example 5 differs fro m th e previou s on e only i n tha t
the number 0 is added t o the set , s o that no w the neutra l
element exists . Bu t axio m 4 is not valid, becaus e positiv e
numbers d o no t hav e inverse s i n thi s set .
The las t exampl e require s mor e attention , becaus e
the operatio n i s unusual . Le t u s chec k al l grou p axiom s
one b y one . I t i s clear tha t fo r a n arbitrar y pai r (x , y) o f
real number s x + y — 1 is also a rea l number , s o that th e
first axio m holds . T o verify th e associativity , w e have t o
calculate th e tw o expression s (x * y) * z an d x * (y * z),
using th e definitio n o f * . W e hav e
(x * y) * z = (x + y — l)*z = x + y — 1 + z — l=x+y+z— 2 ,
x * (y * z) — x * (y + z — l) = x + y + z — 1 — \ — x-\-y-\-z — 2.
We pass t o th e thir d axiom . Th e neutra l elemen t e must
satisfy th e identit y x + e — 1 = x fo r an y x. Thi s i s
true i f an d onl y i f e = 1 . Finally , sinc e th e operatio n *
is commutative , t o determin e th e numbe r symmetri c t o
the give n numbe r x wit h respec t t o * , w e hav e onl y on e
equation instea d o f two : x * x' — 1, i.e . x -f - x' — 1 = 1 ,
whence x' — 2 — x. Al l the four axiom s ar e thus satisfied ,
and th e give n se t (R , *) i s a group .
1. T h e g e n e r a l n o t i o n o f a g r o u p 101
E x e r c i s e 7 1 . Chec k whethe r th e followin g set s wit h binar y opera -

tions ar e groups . I n cas e o f a negativ e answer , indicat e whic h o f
the fou r axiom s fails .
(1) Th e se t o f al l irrationa l number s wit h th e operatio n o f
addition.
(2) Th e se t o f al l rea l number s x > 2 wit h th e operatio n
x * y = xy — x — y + 2.
(3) The set of all binary rational numbers (i.e . fraction s whos e
denominator i s a powe r o f 2 ) wit h th e operatio n o f addition .
(4) Th e se t o f al l non-zer o binar y rationa l number s wit h th e
operation o f multiplication .
(5) Ca n yo u find a se t o f rea l number s whic h form s a grou p
with respec t t o th e operatio n x *y = (x + y)/(l — xy)?
We wil l indicat e severa l simpl e bu t importan t corollarie s o f grou p

axioms.
(1) T h e neutra l elemen t i n a grou p i s unique , i.e . ther e i s onl y

one elemen t e t h a t satisfie s th e requirement s o f t h e grou p
axiom 3 . Indeed , suppos e t h a t w e hav e tw o element s e\ an d
e 2 suc h t h a t th e followin g relatio n hold s fo r ever y a G G:
ae\ — e\a = ae 2 — e 2 a = a.
Setting successivel y a = e\ an d a = e 2 , w e deriv e t h a t e\ —

e i e 2 = e 2.
(2) An y equatio n ax — b i s uniquel y solvabl e i n a group . Thi s
means t h a t fo r an y a,b e G ther e i s a uniqu e elemen t x G G
such t h a t ax — b. Indeed , usin g grou p axiom s 2 an d 4 , w e
can multipl y th e give n equatio n b y a~ x o n t h e lef t an d ge t
x — a~ lb.
E x e r c i s e 7 2 . Fin d a solutio n o f th e equatio n xa — b an d
prove tha t i t i s unique .
(3) T h e previou s assertion s impl y t h a t :
• T h e invers e elemen t fo r a given a G G, define d b y axio m
4, i s unique .
• I n ever y ro w an d ever y colum n o f th e multiplicatio n ta -
ble of a group eac h elemen t o f the grou p appear s exactl y
once. On e o f thes e fact s follow s fro m th e uniqu e solv -

ability o f equation s ax = 6 , the othe r fro m th e uniqu e
solvability o f equation s xa = b.
(4) Grou p element s impl y tha t th e elemen t ( a _ 1 ) n i s inverse t o
an. Therefore , a s i n th e cas e o f transformatio n groups , w e
can defin e zerot h an d negativ e power s o f th e give n elemen t
by settin g a 0 = e , a~ n — (a~ 1 )n fo r n > 0 . Then , fo r
arbitrary intege r value s o f k an d / we will hav e
k l
(29) a a =ak+l.
(5) Th e las t relatio n implie s that th e se t o f all integer power s of
an elemen t a form s a group . Suc h a grou p i s calle d cyclic,
and th e elemen t a is its generator. Th e order o f the elemen t
a is defined a s the smallest positiv e intege r n suc h that a n =
e. I f a = e , the order i s 1 by definition; i f a n i s different fro m
e for every n > 0, we say that th e order of a is infinite. I n the
latter cas e th e subgrou p generate d b y a i s a n infinite cyclic
group. Not e tha t th e orde r (th e numbe r o f elements ) o f th e
group generate d b y a is equal t o th e orde r o f the elemen t a.
(6) Th e axio m o f associativit y mean s tha t th e produc t o f thre e
elements o f a group , whic h involve s tw o multiplications ,
does no t depen d o n th e orde r i n whic h thes e multiplica -
tions ar e computed . Usin g induction , on e ca n prov e tha t
this propert y i s also true fo r an y numbe r o f multiplications :
any bracketin g o f th e produc t aia 2 . . . a n give s on e an d th e
same result . Fo r example , (ai(a 2 as))a4 = ((aia 2 )as)a4 =
(aia 2 )(a 3 a 4 ) = ai(a 2 (a 3 a 4 )) = ai((a 2 a 3 )a 4 ).
So far, w e have deal t wit h group s consistin g o f either transforma -
tions (transformatio n groups ) o r number s (numeri c groups) . No w we
will give a n exampl e o f a group whos e element s hav e quit e a differen t
nature.
Problem 34 . A 3-switc h is an electric circuit with three inputs and

three outputs connected by wires in such a way that every input corre-
sponds to a certain output. The total number of 3-switches is six, and
they are all displayed in Figure 1 . The problem is to define a natural
operation on the set of 3-switches which turns this set into a group.
1
1. Th e genera l notio n o f a group 0 3
*1 P l
h-v^—f 1^—^/'T H P 1T v x T l
v^xTT
2*
3J
F
L
2
3
i^Sc—P 2 i*>^—P 2
1—"s^ i 2 i j 5 v T 2
1—^—P
nx n 2 n 3 n 4 n 5 n 6
F i g u r e 1 . 3-switche s
Solution. A natural operatio n o n the set of switche s

is concatenation. T o concatenate tw o switches mean s
to connec t th e inputs o f one o f them t o the outputs of
another. Fo r example , if we concatenate th e switche s 11 2
and n 4 , inpu t numbe r 1 of II2 wil l go through th e inpu t
number 3 of II4 t o the outpu t numbe r 2 , so that i n the
result w e have tha t inpu t 1 is connected t o output 2 .
Similarly, inpu t 2 goes int o outpu t 1 and input 3 goes
into outpu t 3 . Thi s is the sam e patter n tha t w e have for
the switc h 1 1 6 . I n this sens e w e ca n writ e tha t 1 1 21 1 4 =
UQ. Not e tha t thi s operatio n i s not commutative; fo r
example, U4II2 = II5.
The complet e multiplicatio n (or , mor e exactly , con -
catenation) tabl e fo r the set of 3-switches look s a s fol-
lows:
ni n2 n3 n4 n5 n6
III ni n2 n3 n4 n5 n6
n2 n2 n3 ni n6 n4 n5
n 3 n3 ni n2 n5 n6 n4
n4 n 4 n5 n6 ni n2 n3
n5 n 5 n6 n4 n3 ni n2
n6 n 6 n4 n5 n2 n3 ni
It turn s ou t that th e set of switches wit h thi s mul -
tiplication tabl e form s a group. Bu t ho w ca n on e prov e
this? Usin g the direct procedure of verifying al l the grou p
axioms, just fo r axio m numbe r 2 (associativity ) on e ha s

to check 6 3 = 21 6 equalities 1^(11,11*) = (UiUj)U k. For -
tunately, ther e is a less tedious way to do all these checks.
Note tha t i f yo u replac e th e letter s III , II2 , II3 , II4, II5,
116 in th e multiplicatio n tabl e o f th e switche s b y id , R,
R2, 5 a , S c, St, respectively an d swa p th e tw o las t row s
and th e tw o last column s o f the tabl e obtained , the n yo u
will ge t th e multiplicatio n tabl e fo r th e grou p D 3 (se e
p. 92) . Thi s mean s tha t th e concatenatio n o f 3-switche s
and th e compositio n o f th e movement s i n th e grou p D 3
establish exactly the same relations between the element s
of th e respectiv e sets . Therefore , th e operatio n o f con -
catenation i n th e se t o f switche s ha s al l th e propertie s
that th e compositio n o f transformation s has : i t i s asso -
ciative, ther e i s a neutra l elemen t (th e switc h IIi) , an d
every switc h ha s a n inverse . Thi s implie s tha t th e se t o f
3-switches form s a grou p wit h respec t t o concatenation .
The mathematica l conten t o f the previous example consist s i n th e

description o f all possible one-to-on e mapping s o f the se t {1 , 2, 3} into
itself, or , i n anothe r terminology , al l transformations o f thi s set .
A transformatio n o f th e se t {1 , 2 , . . ., n} i s calle d a permutation
of n elements , or a permutation of degree n. A permutation tha t take s
1 int o ii , 2 into i^, . . . , n int o i n i s denote d a s
12 .. . n
hh .. . i n
There ar e n ! permutation s o f degre e n i n all , an d the y for m a group ,

denoted b y S n.
There ar e eve n tw o differen t natura l way s t o defin e th e grou p
structure i n the set of all permutations o f a given degree. Th e first on e
is t o trea t permutation s exactl y lik e transformation s an d defin e th e
product o~\&2 o f th e tw o permutation s o\ an d G2 a s th e compositio n
G\ o cj2 o f th e tw o mappings . W e recal l tha t th e compositio n G\ O G^
is obtaine d b y performin g firs t th e transformatio n 5 2 an d the n th e
1. T h e g e n1
era l notio n o f a grou p 0 5
transformation s i . Accordin g t o thi s definition , w e wil l hav e
/l2 3 \A 2 3 \ _ / l2 3 \
VI 3 2) ° \3 2 l) ~ \2 3 l) '
Another wa y t o defin e th e produc t o f th e tw o permutation s G\

and (7 2 is t o first perfor m o\ an d the n o<i — exactl y a s w e define d th e
concatenation o f switche s i n Proble m 34 .
There ar e tw o school s o f mathematicians : on e maintain s t h a t
t h e produc t o f permutation s shoul d b e define d a s G\ o cr 2, t h e other ,
t h a t i t shoul d b e define d a s G2 ° 0"i - Multiplicatio n table s fo r S n
adopted b y th e tw o school s diffe r b y a reflectio n i n th e mai n diagonal .
Actually, th e tw o viewpoint s ar e no t s o fa r apart : afte r studyin g th e
next section , yo u wil l b e abl e t o prov e t h a t th e tw o permutatio n
groups resultin g fro m th e tw o definition s ar e i n fac t isomorphic.
Here ar e som e mor e problem s wher e differen t group s appear , im -
plicitly o r explicitly .
E x e r c i s e 7 3 . O n a blackboard , severa l circles , square s an d triangle s
are drawn . Yo u ar e allowe d t o eras e an y tw o figures an d replac e
them b y a ne w figure followin g th e rule :
— tw o circle s mak e a circle ;
— tw o square s mak e a triangle ;
— tw o triangle s mak e a square ;
— a circl e an d a squar e mak e a square ;
— a circl e an d a triangl e mak e a triangle ;
— a squar e an d a triangl e mak e a circle ;
Prove tha t th e shap e o f th e las t figure tha t wil l remai n doe s
not depen d o n th e orde r i n whic h th e replacement s ar e made .
E x e r c i s e 7 4 . Conside r th e rationa l algebrai c expression s i n on e vari -

able, i.e. , quotient s o f tw o polynomial s i n x wit h rea l coefficients .
If A an d B ar e tw o suc h expressions , the n w e can for m th e super-
position A * B b y substitutin g B instea d o f x int o A. Prov e tha t
the se t < $ of al l expression s tha t ca n b e obtaine d b y substitution s
from A± ~ 1 — x an d A2 — 1/x form s a group , an d find it s order ,
its lis t o f element s an d it s multiplicatio n table .
E x e r c i s e 7 5 . Fin d a rational expression .£? , different fro m a constant ,

such tha t B * A\ = B * A2 = B, wher e A± = 1 — x, Ai — 1 /x.
OO P
2. Isomorphis m
When workin g o n Proble m 34 , w e deduce d al l th e propertie s o f th e
concatenation fro m th e similar propertie s o f the compositio n o f trans-
formations, usin g th e fac t tha t bot h operation s hav e th e sam e inne r
structure. Th e precis e notio n suite d t o characteriz e suc h situation s i s
isomorphism.
Definition 1 4 . Tw o group s G an d H ar e sai d t o b e isomorphic, i f

there i s a one-to-on e correspondence , denote d b y '<->' , betwee n th e
elements o f G an d th e element s o f H, suc h tha t g\ «- » h\ an d g 2 *- * h2
always imply gig 2 <- > h\h2. On e can also say that thi s correspondenc e
respects, or agrees with, th e grou p operation s i n bot h groups .
A more exact wa y to state the definition o f an isomorphism is : th e

groups G an d H ar e isomorphic , i f there exist s a one-to-on e mappin g
p : G —> H suc h tha t
(30) <p{gi92) = ^{9

1 )^(92)
for an y element s g\ an d g 2 o f G. On e ca n als o sa y tha t (p is a n

isomorphism o f th e grou p G ont o th e grou p H.
At firs t sigh t i t seem s tha t th e secon d versio n o f th e definitio n i s
different fro m th e first, becaus e the two groups do not ente r symmetri -
cally. I n reality, however , th e two groups have equal rights, because, if
p is an isomorphis m o f G onto H, the n p~ l wil l be a n isomorphis m o f
H ont o G. Indeed , denotin g hi = <p(gi), h 2 = ^(#2 ) an d applyin g p~ l
to bot h side s o f equatio n (30) , w e ge t p~ 1 {hi)p~1 (h2) =
p' 1 ^^).
2. I s o m o r p h i s m 107
The mos t direc t wa y t o establis h th e isomorphism , especiall y fo r

finite groups , i s t o explicitl y indicat e al l pair s o f correspondin g el -
ements an d the n chec k t h a t , whe n al l element s o f on e grou p i n it s
multiplication tabl e ar e replace d b y thei r counterpart s fro m th e othe r
group, w e obtai n th e multiplicatio n tabl e o f th e secon d group . (O f
course, i t migh t b e necessar y t o chang e th e orde r o f column s an d row s
in th e tabl e obtaine d t o mak e i t literall y coincid e wit h th e multiplica -
tion tabl e o f th e secon d grou p a s i t wa s given. ) Not e t h a t w e followe d
this procedur e i n Proble m 34 .
E x e r c i s e 76 . Chec k whethe r th e followin g correspondenc e i s a n iso -
morphism betwee n th e grou p o f 3-switche s an d th e symmetr y
group o f the equilatera l triangl e D3 : i d <- * 111, R2 <-> II2 , R «- > II3,
Sb <- > ru, s c «- > lis, s a <- > n 6 .
This exercis e lead s t o a n importan t observation : i f tw o groups. G
and H ar e isomorphic , th e isomorphis m H i s no t i n genera l
unique. I n particular , ther e migh t exis t isomorphism s o f a grou p ont o
itself, differen t fro m th e identit y transformation .
E x e r c i s e 7 7 . Fin d al l isomorphism s o f th e grou p D3 ont o itself .
The notio n o f a n isomorphis m illuminate s th e meanin g o f som e

analogies t h a t a n observan t reade r migh t hav e notice d i n th e mate -
rial o f th e previou s chapters . W e ca n no w stat e t h e m a s clear-cu t
problems.
E x e r c i s e 78 . Prov e tha t th e se t o f al l point s o f th e plan e wit h th e

operation o f additio n ove r a fixed pol e (se e p . 1 0 ) form s a grou p
isomorphic t o th e additiv e grou p o f plane vectors . Als o prove tha t
assigning t o a poin t (o r a vector ) th e pai r o f it s coordinate s i n a
certain basi s establishe s a n isomorphis m o f th e respectiv e grou p
and th e grou p o f pairs o f rea l number s wit h th e operatio n define d
by th e rul e
(ai,6i) + (02,62 ) = (a i +a 2 ,&i + 62) .

E x e r c i s e 7 9 . Prov e that th e se t o f vertices o f a regular hexago n wit h
the multiplicatio n describe d i n Proble m 7 (p . 26 ) form s a cycli c
group isomorphi c t o th e grou p C§ o f rotation s wit h a commo n
centre throug h angle s whic h ar e multiple s o f 60° .
E x e r c i s e 8 0 . Prov e tha t th e se t {circle , triangle , square } wit h th e

operation define d i n Exercis e 73 , i s a grou p isomorphi c t o th e
cyclic group C3. Ho w many different isomorphism s between these

groups ar e there?
Exercise 8 1 . Prov e that th e grou p o f rational algebrai c expression s
defined i n Exercise 7 4 is isomorphic to the dihedra l grou p D3.
The results of Exercises 79 and 8 0 are generalized b y the followin g
assertion: any two cyclic groups of the same order are isomorphic.
Indeed, le t g be the generator o f the first grou p G, and h the generato r
of th e secon d grou p H. Defin e th e mappin g</ ? : G — • H b y th e rul e
ip(gk) — hk. Th e la w o f multiplicatio n o f power s (29 ) hold s i n eithe r
group an d implie s tha t <p i s a n isomorphism :
<p(9k9l) = <fi(9 k+l) = h k+l = h khl = <p(g kMgl).
It i s likewis e clea r tha t a grou p isomorphi c t o a cycli c grou p i s itsel f
cyclic, becaus e th e imag e o f a generato r unde r a n isomorphis m wil l
also b e a generator .
If w e ar e intereste d i n th e inne r structur e o f a group , w e ca n
forget abou t th e natur e o f element s i t consist s of , keepin g trac k onl y
of the propertie s o f th e grou p operation .
Definition 1 5 . A n abstract group i s a clas s o f al l group s whic h ar e
isomorphic betwee n themselves .
For example , al l cycli c group s o f orde r n , suc h a s th e grou p o f

rotations C n o r th e grou p o f comple x n-t h root s o f unity , ar e repre -
sentatives or , i n Buddhis t terminology , incarnations, o f on e an d th e
same abstrac t cycli c grou p o f orde r n . I n th e sam e way , th e dihedra l
group D3 , the permutatio n grou p 53 , the grou p o f switches (Proble m
34) an d th e grou p o f rationa l expression s (Exercis e 74 ) ar e al l repre -
sentatives o f on e an d th e sam e abstrac t group . Late r (pag e 1 36 ) w e
will explai n ho w t o defin e a n abstrac t grou p wit h a give n structur e
(set o f relation s betwee n generators) .
We now pass t o th e followin g genera l problem : given two groups,
decide whether they are isomorphic or not. Usually , i t take s mor e
efforts t o establis h isomorphis m tha n t o establis h non-isomorphism ,
because, i n th e first case , on e normall y ha s t o construct th e isomor -
phism, whil e i n th e secon d case , i t i s often enoug h t o find som e prop -
erty whic h mus t b e preserve d b y a n isomorphism , bu t whic h distin -
guishes th e group s unde r study . Her e i s a shor t lis t o f som e simpl e
2. I s o m o r p h i s m 109
properties whos e coincidenc e fo r tw o group s i s a necessar y conditio n

for thei r isomorphism :
(1) The order of the group. Group s t h a t hav e a differen t numbe r

of element s canno t b e isomorphic .
E x e r c i s e 8 2 . I s the grou p o f al l integer s wit h additio n iso -
morphic to the group of all even numbers wit h addition ?
(2) Commutativity. A commutativ e grou p canno t b e isomorphi c
to a non-commutativ e group .
(3) Cyclicity. A cycli c grou p canno t b e isomorphi c t o a non -
cyclic group .
E x e r c i s e 8 3 . Ar e ther e an y pair s o f isomorphi c group s i n
the lis t Ci , £>i , C 2 , D 2, C 3 , A$ , . . . ?
(4) The orders of elements. T h e numbe r o f element s o f orde r
n i n on e grou p mus t b e equa l t o th e numbe r o f element s
of orde r n i n th e othe r group , becaus e th e order s o f corre -
sponding element s ar e th e same .
We wil l onl y prov e th e las t item , becaus e i t i s somewha t mor e

complicated t h a n t h e others .
To begi n with , not e t h a t unde r a n isomorphism , t h e uni t element s
of th e group s correspon d t o eac h other . I n fact , i f e i s th e uni t o f th e
group G an d </ ? : G — > H i s a n isomorphism , the n e e = e implie s
(f(e)(f(e) — </?(e) , which , upo n multiplicatio n b y th e elemen t o f H
inverse t o </?(e) , lead s t o th e equalit y ip(e) — e', wher e e' i s th e uni t o f
H. No w le t g b e a n elemen t o f G t h a t ha s orde r n i n G. B y definition ,
gn ~ e , whenc e (f(g) n = e' , i.e. , th e orde r o f h = <p(g) doe s no t excee d
n. A n invers e argumen t show s t h a t th e orde r o f g canno t excee d th e
order o f h. Therefore , th e tw o order s ar e equal .
For example , t o distinguis h betwee n th e group s CQ an d D3 , i t
is enoug h t o us e an y o f th e thre e criteri a 2-4 . Firs t o f all , CQ i s
commutative, bu t D 3 i s not . Also , CQ is cyclic, b u t D 3 i s not . Finally ,
CQ ha s on e elemen t o f orde r 1 , on e elemen t o f orde r 2 , tw o element s
of orde r 3 an d tw o element s o f orde r 6 , whil e D 3 ha s on e elemen t o f
order 1 , thre e element s o f orde r 2 an d tw o element s o f orde r 3 .
T h e lis t o f propertie s t h a t ar e necessar y fo r tw o group s t o b e
isomorphic i s virtuall y infinite , becaus e i t contain s an y propert y o f
the grou p whic h ca n b e formulate d i n term s o f th e grou p operatio n

without referrin g t o th e specifi c natur e o f th e element s o f th e group .
We will, however, pas s to the second half of the isomorphism problem .
Suppose tha t w e ar e give n tw o group s G an d H an d w e canno t fin d
any intrinsi c propert y tha t distinguishe s the m fro m eac h other. The n
the conjectur e arise s tha t th e group s ar e isomorphic . T o prov e this ,
one mus t construc t a n isomorphis m / : G—> • H betwee n G an d H.
How ca n thi s b e done ?
First w e recall that , i f e and e' ar e the uni t element s i n G an d H
respectively, the n w e have (p(e) = e''. Further, i f f(g) = h for a certai n
pair o f element s g E G an d h e H, the n b y repeate d applicatio n o f
(30) w e can deriv e tha t <p(g k) = h k fo r an y natura l k.
Exercise 84 . Prov e the equality p(g k) = h k fo r negativ e values of k.
We thus se e that, i f the mapping ip is defined o n a certain elemen t
g of G , the n i t i s als o uniquel y define d o n th e whol e subgrou p gener -
ated b y g. Quit e similarly, i f the images of several element s #i, . . . , g n
of the group G are known, then one can uniquely determin e th e imag e
of an y elemen t expressibl e i n term s o f g\ , . . . , g n. I f thes e element s
are generators o f G , the n th e value s <p(gi) = /ii , . . . , <p(gn) — h n
completely determin e th e mappin g (p. I n th e cas e o f tw o generator s
we ca n writ e th e correspondin g formul a a s follows :
k l k
(31) ?(&&... g -g 2-) = h >h!}...hk-h!i-.
Thus, i f th e grou p G i s generate d b y tw o element s gi an d g<2,

then, t o construc t a n isomorphis m ip : G — > H, w e mus t defin e it s
values <p(g±) = /&i , ^(^2) = ^ 2 an d the n exten d th e mappin g t o al l of
G accordin g t o relatio n (31 ) . Bu t ho w shoul d th e element s hi, hi b e
chosen? I t i s clea r tha t the y ough t t o b e a se t o f generator s o f th e
group iJ , the y mus t hav e the same respective order s as g\ an d #2 , and
they mus t satisf y al l the relation s that g\ an d g^ satisfy. Fo r example ,
if g\g\ — e, the n w e mus t hav e h\h\ = e'. Thes e observation s allo w
us t o gues s a 'candidate ' fo r th e isomorphis m ip. Afte r th e mappin g
is constructed , on e ha s t o verif y tha t i t i s really a n isomorphism .
1 \/ 3
Problem 35 . Let e be the complex number — - + — -i (note that
s3 = 1 ). Consider two functions of a complex variable F\(z) = ez,
2. Isomorphis m 111
F2(z) — ~z- Prove that the set of all functions that can be obtained
from F\ and F2 by superposition forms a group isomorphic to the
dihedral group D3 .
Solution. W e successivel y fin d tha t
F3(z) = F l(F2(z)) = F^z) = ez,

F4(z) = F 2(F2(z)) = F 2(z) = z,
F5(z) = F 1 (F1 (z)) = F 1 (ez)=e2z,
F6(z) = F 2(F
1 {z)) = F 2{ez)=ez = e 2z.
A straightforward chec k shows that furthe r applicatio n of

F\ an d F 2 t o thes e expression s doe s no t lea d t o an y ne w
functions. Thus , th e se t o f si x function s Fi , . . . , FQ is
closed under superposition. Th e inverse of every functio n
belonging to this list als o belongs to this list. Thi s prove s
that wha t w e have is a group. Th e neutral elemen t i s th e
identity functio n F4 , the function s F 2l F3 , FQ have orde r
2, an d th e function s Fi , F 5 hav e orde r 3 . Th e grou p i s
not commutativ e since , fo r example , F\{F 2{z)) — Fs(z),
while F 2(Fi(z)) = FQ(Z). Thes e observation s sugges t
that ou r grou p G i s likely t o b e isomorphi c t o D%.
To construct a n isomorphis m ip : G —> D3 , note tha t
G b y definitio n ha s tw o generator s F\ an d F 2, whos e
orders ar e 3 an d 2 . I n D 3 w e ca n als o fin d a syste m o f
two generator s wit h order s 3 an d 2 : a rotatio n an d a
reflection. Fo r example , se t
Fi <- • R, F 2 <- • Sa
(in th e notatio n o f p . 92) . The n
F 3 <- > Sc , F 4 ^ id , F 5 ^ R 2, F 6 ^ S h.
Replacing ever y elemen t o f D3 b y th e correspondin g el -

ement o f G i n th e multiplicatio n tabl e o f D 3 (p . 92) , w e
will obtai n th e followin g table :
112 4. A r b i t r a r y G r o u p s
F4 Fi F5 F2 F6 F3
F4 Fi Fx F5 F2 F6 Fs
Fi Fx F5 F4 F6 F3 F2
F5 F5 F4 Fx F3 F2 F6
F2 F2 F3 F6 F4 F5 Fx
F6 F6 F2 F3 Fx F4 F5
F3 F3 F6 F2 F5 Fx F4
which, a s on e ca n easil y check , i s th e correc t multiplica -
tion tabl e fo r G.
T h e isomorphis m i s thu s established . Ther e is , how -
ever, a mor e natura l wa y t o find a n isomorphis m be -
tween th e tw o group s i n question . Indeed , le t u s re -
call t h a t a functio n o f a comple x variabl e ca n b e viewe d
as a n analytica l representatio n o f plan e transformations .
In particular , th e functio n F\(z) = ez correspond s t o
the rotation s aroun d 0 throug h 1 20° , whil e t h e functio n
F2(z) = z , t o a reflectio n i n th e rea l axi s (axi s a i n Figur e
2).
If w e assig n t o ever y functio n obtaine d fro m F\ an d
F2 b y superposition s th e correspondin g plan e transfor -
mation, w e wil l obtai n a grou p isomorphism. 1
T h e isomorphis m constructe d b y th e secon d metho d i s called nat-

ural. A natura l isomorphis m reveal s th e reaso n wh y t h e tw o group s
are isomorphic .
E x e r c i s e 8 5 . Indicat e a natura l isomorphis m
(1) betwee n th e grou p o f 3-switches (se e Problem 34 ) an d th e
group D 3 ;
(2) betwee n th e grou p o f rational algebrai c expression s o f Ex -
ercise 7 6 an d th e grou p D3.
Note tha t th e term s 'superposition ' an d 'composition ' actuall y hav e th e sam e
meaning, onl y th e firs t on e i s use d i n analysi s fo r functions , whil e th e secon d on e i s
used i n geometr y fo r transformations .
2. Isomorphis m 113
Figure 2 . Equilatera l triangl e i n th e comple x plan e
We hav e t o admi t tha t th e naturalit y w e ar e talkin g abou t i s b y

no mean s a stric t mathematica l notion , i t rathe r bear s a heuristi c
character. I n a certai n sense , an y isomorphis m i s natural , but , i n
order t o understand why is it natural , peopl e ofte n hav e t o develo p a
special mathematica l theory . A ques t fo r natura l isomorphism s tha t
explain th e reaso n wh y simila r object s appea r i n differen t area s o f
mathematics i s a powerfu l impetu s tha t foster s th e developmen t o f
knowledge.
Here, w e hav e use d th e genera l wor d 'knowledge ' o n purpose ,
to emphasiz e th e fac t tha t th e notio n o f isomorphis m i s importan t
not onl y i n mathematics , bu t i n an y are a o f thinking . T o ge t a n
idea o f this genera l meanin g o f 'isomorphism' , w e invite the reade r t o
contemplate ove r th e followin g historica l example .
In 1 970 , on e o f th e problem s o f th e entranc e examinatio n se t
for th e Gelfan d Correspondenc e Mathematica l Schoo l i n Mosco w wa s
published b y tw o majo r journal s i n differen t formulations .
One o f th e journals state d th e proble m lik e this :
One o f thre e gangsters , know n i n cit y M unde r th e

names o f Archie , Bos s an d Wesley , ha s stole n a ba g
of money . Eac h o f the m mad e thre e declarations :
• Archie :
— I di d no t stea l th e bag .
— O n th e da y o f th e thef t I wa s no t i n th e
city.
— Wesle y stol e th e bag .
• Boss :
— Wesle y stol e th e bag .
— Eve n i f I stol e it , I woul d no t confess .
— I hav e lot s o f money .
• Wesley :
— I di d no t stea l th e bag .
— I'v e bee n lon g lookin g fo r a goo d bag .
— Archi e tol d th e trut h tha t h e wa s no t i n
the city .
During th e investigatio n i t wa s foun d tha t tw o dec -
larations o f eac h gangste r wer e tru e an d on e false .
Who stol e th e bag ?
Another journa l propose d th e followin g proble m (name s men -

tioned belon g t o protagonist s o f Russia n fol k tales) :
The Kin g learned tha t somebod y ha s killed the ruth -

less Dragon . H e kne w tha t thi s coul d onl y b e don e
by one o f the thre e famou s warriors : Ily a Muromets ,
Dobrynia Nikitic h o r Alyosh a Popovich . The y wer e
summoned t o the Kin g an d eac h o f them spok e thre e
times. Her e i s what the y said .
• I . M. :
— I di d no t kil l th e dragon .
— O n tha t da y I wa s travellin g abroad .
— A . P . di d that .
• D . N. :
— A . P . di d that .
— Eve n i f I kille d him , I woul d no t confess .
— Ther e ar e man y evi l spirit s stil l alive .
• A . P. :
Licensed to Penn St Univ, University Park.— I di
Prepared d no
on Tue t 18:16:11
Mar 12 kil l EDT
th e2019for
dragon . IP 132.174.254.159.
download from
2. Isomorphis m 115
— I'v e bee n lon g lookin g fo r a nic e fea t t o

do.
— I t i s true tha t I . M . wa s abroad .
The Kin g foun d ou t tha t eac h o f th e thre e warrior s
twice tol d th e trut h an d onc e lied . Wh o kille d th e
dragon?
It i s easy t o se e that , althoug h th e tw o problem s ar e abou t quit e
different things , thei r logica l structure i s the same . Her e is a glossar y
of names, thing s an d action s that correspon d t o eac h other i n the tw o
problems:
Archie Ily a Muromet s
Boss Dobryni a Nikitic h
Wesley Alyosh a Popovic h
bag drago n
to stea l t o kil l
to leav e th e cit y t o g o abroa d
If, i n th e statemen t o f the firs t problem , al l th e significan t word s
are replace d b y thei r counterpart s fro m th e right-han d column , th e
result almos t coincide s wit h th e statemen t o f th e secon d proble m
— wit h th e exceptio n o f Boss' s thir d statement , whic h i s actuall y
irrelevant fo r th e solutio n o f th e problem . I n thi s sense , th e tw o
problems ar e isomorphic .
This isomorphis m ca n b e use d a s follows . I f yo u solv e th e firs t
problem an d find tha t th e answe r i s 'Boss' , the n yo u d o no t hav e t o
solve th e secon d problem : th e correc t answe r i s give n b y th e wor d
that correspond s t o Bos s i n our glossary , namel y 'Dobryni a Nikitich' .
F i g u r e 3 . Isomorphis m
In th e sam e wa y on e ca n us e th e isomorphis m o f groups : i f G i s

isomorphic t o H, the n ever y assertio n abou t G that ca n b e state d i n
terms o f th e grou p operatio n wil l als o hol d fo r th e grou p H, afte r a n
appropriate translation .
Another, mor e simple and direct , applicatio n o f the group isomor -
phism i s th e computatio n o f th e produc t o f element s i n on e group ,
using th e produc t i n another , provide d tha t th e secon d multiplica -
tion i s less difficul t an d time-consuming . Mor e exactly , i f H
is a n isomorphis m betwee n th e group s (G , *) an d (ii , o), the n th e
*-product ca n b e compute d b y th e formul a
(32) gi*g2 = (p~ l(<p(gi) o tp(g 2))-
A classica l exampl e o f thi s kin d o f computation s i s provide d b y

logarithms, invente d b y J . Napier 2 (earl y seventeent h century) , wh o
sought to replace the multiplication o f numbers by a simpler operatio n
— addition . Denotin g th e decima l logarith m o f x b y lgx , w e hav e
the identit y
lgXl+l x
(33) x 1 x2 =
1 0 z\
which i s a particula r cas e o f th e genera l relatio n (32) . Th e isomor -
phism tha t make s i t possibl e t o comput e b y usin g logarithm s i s th e
isomorphism l g : (R + , • )— > (R , +) o f th e grou p o f positiv e rea l num -
bers wit h multiplicatio n ont o th e grou p o f al l rea l number s wit h ad -
dition. Th e tw o basi c propertie s o f the logarithmi c function , namel y
(1) i t i s on-to-one o n th e set s specified , an d
(2) i t satisfie s th e identit y
lg(xix 2 ) = lgx i + l g £ 2 ,
mean precisel y tha t i t establishe s a n isomorphis m betwee n th e tw o
groups.
Exercise 86 .
(1) Fin d th e analo g o f (33) , i f th e decima l logarith m y = lg x i s
replaced b y the Napie r functio n y = A\gx + B.
Actually, logarithm s i n the contemporar y sens e o f the wor d wer e introduce d an d

tabulated b y hi s disciple G . Briggs ; Napie r himsel f use d a function y = Algx-\-B wit h
some constant s A an d B.
2. I s o m o r p h i s m 117
(2) Fin d a group operatio n * on th e se t o f al l real number s suc h tha t

the Napie r functio n give s a n isomorphis m o f (K+ , •) onto (R , *).
T h e secon d hal f o f th e las t exercis e i s a particula r cas e o f th e

so-called transition of structure. Her e i s wha t w e mea n b y t h a t .
Let G b e a group wit h operatio n A an d H a se t wit h n o operation .
Suppose t h a t a one-to-on e mappin g ip : H — » G i s given . The n i t i s
possible t o carry the group operation from G to H along <p b y th e
formula
-1
hi V h
2= ^ ( ^ ( ^ i ) A ^(^2)) .
We hav e actuall y use d thi s method :
• t o deriv e th e additio n o f point s fro m th e additio n o f vector s

(Chapter 1 ) ;
• t o defin e th e unusua l grou p operatio n ove r th e rea l number s
x^y — x -V y — 1 (Proble m 33) . Thi s operatio n i s obtaine d
from th e ordinar y addition , whic h i s carrie d ove r fro m on e
copy o f R int o anothe r alon g th e mappin g ip(x) = x — 1 .
Indeed, x*y = ip~ 1 (ip(x) + <£>(y) ) = ((x- l) + (y— 1 ) ) + 1 =
x-fy — 1 .
x ~\- v
T h e operatio n x * y = t h a t appeare d i n Exercis e 7 1 ha s
l-xy
t h e sam e origin . W e hav e trie d t o perfor m th e transitio n o f th e grou p
structure (R , + ) alon g th e mappin g <p(x) = t a n x , bu t w e wer e no t
quite successful , becaus e thi s mappin g i s no t one-to-one . However ,
for an y se t M C R whic h i s a n additiv e grou p an d ha s th e propert y
t h a t th e value s o f th e tangen t i n th e point s o f M ar e al l different , th e
set o f thes e value s form s a grou p wit h respec t t o th e operatio n * .
E x e r c i s e 8 7 . Prov e that :
(1) A s suc h a se t M on e ca n tak e th e se t o f al l multiple s o f a rea l
number a tha t i s incommensurabl e wit h TT.
(2) A se t M wit h th e require d propertie s canno t contai n an y ope n
interval o f th e rea l axis .
Exercise 88 .
(1) Wha t operatio n o n rea l number s i s th e resul t o f transitio n o f
addition alon g th e mappin g x i- » x 3 ?
(2) Ho w wa s th e operatio n x * y = xy — x — y + 2 (Exercis e 71 )

obtained?
In terms o f the transitio n o f structure, th e notio n o f isomorphis m
can b e formulate d a s follows : th e mappin g cp : G— > H i s a n isomor -
phism o f groups , i f the grou p operatio n o f H carrie d ove r t o G alon g
ip coincides wit h th e grou p operatio n o f G.
3. Th e Lagrang e theore m
In thi s sectio n w e will state an d prov e th e ver y first theore m o f grou p
theory, which was found b y the French mathematician Lagrang e in the
late eighteenth century , eve n before th e notio n o f group was explicitl y
introduced i n mathematic s i n th e nineteent h centur y b y E . Galois .
Theorem 7 (Lagrange) . The order of a subgroup of a finite group
is always a divisor of the order of the whole group.
Since every element o f a group generate s a cyclic subgroup whos e

order i s equa l t o th e orde r o f thi s element , w e obtain , i n particular ,
that the order of a finite group is divisible by the order of each element.
The reade r migh t hav e notice d thi s la w i n th e example s considere d
above (group s C1 2 , £> 3, etc.).
To prove the theore m o f Lagrange i n the general setting , w e shall
use th e importan t constructio n o f the coset decomposition o f a grou p
over a subgroup .
Let G b e a grou p o f orde r n , an d H a subgrou p o f G o f orde r
m: H = {/ii , /1 2 , • • •, hm}. Sinc e ever y subgrou p contain s th e uni t
element o f the group, we can assume that hi = e . Choos e an arbitrar y
element g of G tha t doe s no t belon g t o H, an d conside r th e se t
gH = {gh 1 ,gh2,...,ghTn}
obtained b y multiplyin g al l th e element s o f H b y on e an d th e sam e

element g on th e left . Th e se t gH i s calle d a left coset of G over H.
It ha s tw o importan t properties :
(1) \gH\ — \H\, i.e. , gH ha s the sam e numbe r o f elements a s H\
(2) gH n H — 0, i.e. , th e set s gH an d H d o no t hav e commo n
elements.
3. Th e Lagrang e theore m 119
To prov e propert y (1 ) , w e hav e t o sho w tha t al l th e element s o f

the lis t ghi,gfi2, ... , #/im ar e distinct . Indeed , i f w e ha d ghi = ghk,
then, afte r multiplyin g thi s equatio n b y g~ x o n th e left , w e woul d
obtain hi — h^.
To prov e (2) , suppos e tha t hi = gh^. Thi s implie s tha t g ca n b e
expressed a s hih^ 1 an d must , therefore , belon g t o th e subgrou p H,
contrary t o th e supposition .
The secon d propert y ha s the followin g generalization : ifg\H is a
coset and g2 £ G an element of the group that does not belong to g\H,
then the two cosets g\H and g^B. do not have common elements, or ,
in othe r words , two cosets either coincide or are disjoint. I n fact , i f
there wer e a commo n element , w e woul d hav e g\hi = g^h^ henc e
#2 = gihihfr 1 and , sinc e hih^ 1 E H, thi s woul d impl y tha t # 2 £ 9\H
and therefor e g2H ' = g\H.
Now, th e proces s o f decomposin g th e grou p G int o lef t coset s
over th e subgrou p H ca n b e describe d a s follows . I f th e subgrou p
H coincide s wit h th e entir e grou p G , the n th e cose t decompositio n
consists o f only one set, H. Otherwise , choos e a n element g\ £ H an d
consider th e cose t g\H. I f H U g\H — G, th e proces s terminates . I f
not, w e choose a g<i £ G whic h belong s neithe r t o H no r t o g\H an d
thus obtai n thre e pairwis e disjoin t coset s H, g\H an d #2# -
Since th e grou p i s finite, thi s proces s eventuall y terminates , an d
we obtain th e require d decompositio n
G = HUgÛ-'-UgkH,
where eac h o f th e liste d subset s ha s m element s an d the y ar e al l
pairwise disjoint .
Therefore, th e numbe r n o f element s i n th e grou p i s divisibl e b y
the numbe r m o f element s i n th e subgroup . Th e theore m i s proved .
Let u s remark that , similarl y t o the lef t cose t decomposition , on e
can als o consider th e righ t cose t decomposition . I n general , thes e tw o
decompositions d o no t coincide , an d w e wil l discus s thi s questio n i n
the nex t chapter .
Figures 4 an d 5 show th e lef t cose t decomposition s o f th e grou p
Ds ove r a subgrou p o f orde r 3 and a subgrou p o f orde r 2 .
>sa >sb
>R
Figure 4 . Firs t cose t decompositio n o f th e grou p D%
Figure 5 . Secon d cose t decompositio n o f th e grou p D 3
Exercise 89 . Fin d al l subgroups o f D 3.

The Lagrang e theore m implie s th e followin g importan t fact .
Problem 36 . Prove that every finite group whose order is a prime
number is cyclic.
Solution. Le t G b e a grou p o f prim e orde r p an d g a n
arbitrary elemen t o f G differen t fro m th e uni t elemen t
e. Denot e b y H th e subgrou p o f G generate d b y g. Th e
order o f H i s a t leas t 2 , sinc e i t contain s e an d g. Th e
only diviso r o f the prim e numbe r p which i s greater tha n
1 i s p itself . Therefore , th e orde r o f H i s p an d H = G.
The grou p G i s thus generate d b y on e elemen t g.
As a direc t consequenc e o f th e assertio n jus t prove d w e obtai n

the followin g fact : any group of prime order is commutative.
We shal l no w discus s som e application s o f grou p theory , i n par -

ticular, Lagrange' s theorem , t o arithmetic .
The simplest grou p that w e come across in arithmetic i s the grou p
Z o f al l integer s unde r addition . Sinc e th e grou p operatio n i s addi -
tion, instea d o f power s o f a certai n elemen t w e wil l spea k abou t it s
multiples, i.e. , element s tha t ar e obtaine d b y successivel y addin g th e
given elemen t t o itself . Th e grou p Z i s cycli c wit h generato r 1 , sinc e
every intege r i s a multipl e o f 1 : n = n • 1 .
Exercise 90 . I s there another generato r i n the grou p Z?
As in any group, every element n of Z generates a subgroup. Thi s
subgroup consist s o f al l multiple s o f n , an d w e denote i t b y nZ .
Exercise 91 . Prov e that every subgroup o f Z ha s the for m nL fo r a
suitable n.
This simpl e resul t alread y provide s a basis fo r usefu l application s
in numbe r theory . A s a n example , w e wil l giv e a shor t proo f o f th e
following well-know n fact : if a and b are mutually prime numbers,
then there exist two integers x and y such that ax + by = 1 .
Indeed, le t H b e th e subgrou p i n Z generate d b y th e tw o give n
numbers a and b. B y definition , H — {ax + by\x,y G Z} (le t u s not e
that i n multiplicativ e notatio n thi s expressio n woul d b e writte n a s
axby). Accordin g t o Exercis e 91 , we ca n find a natura l n suc h tha t
H = nL. Sinc e th e subgrou p H contain s th e element s a an d 6 , bot h
of the m ar e divisibl e b y n . But , sinc e the y ar e mutuall y prime , n
must b e equal t o 1 . Therefore , th e numbe r 1 belongs t o H an d henc e
can b e writte n a s ax + by.
Lagrange's theore m doe s not directl y appl y t o the pai r consistin g
of the group Z and the subgroup nZ, because these groups are infinite .
However, th e constructio n o f th e cose t decompositio n o f th e grou p
over the subgroup doe s make sense and lead s to the important notion s
of residu e classe s an d modula r arithmetic .
Take, fo r example , n — 3. I f w e ad d 1 to al l th e element s o f th e
subgroup 3 Z (i.e. , multiples o f 3), we get th e se t o f all whole number s
that hav e remainde r 1 in divisio n b y 3 . Likewise , addin g 2 to al l th e
elements o f th e subgroup , w e ge t th e se t o f al l number s tha t hav e
remainder 2 i n divisio n b y 3 . Sinc e ther e ar e n o othe r remainder s
in divisio n b y 3 , w e se e tha t th e se t o f al l integer s split s int o th e 3

classes tha t w e obtained . Thi s i s th e coset decomposition of Z over
3Z. A visua l representatio n o f this decompositio n i s shown i n Figur e
6. Sinc e the sets 3Z, 3Z+1 an d 3Z+ 2 ar e infinite, onl y a few element s
of eac h cose t ar e listed .
. , - 6 , - 3 , 0, 3, 6, . . . 3Z
-5, - 2 , 1 , 4 , 7 , . . . 3 Z + 1
- 4 , - 1 , 2 , 5 , 8 , .. . 3 Z+2
Figure 6 . Cose t decompositio n o f th e grou p Z ove r 3 Z
The coset s o f Z ove r raZ ar e calle d residue classes over m. Th e

class o f al l integer s wit h remainde r k i n divisio n b y m , i s convention -
ally denote d b y k. Ther e ar e m residu e classe s i n all ; 0 , 1 ,... , m — 1 .
For example , ther e ar e 3 classes modul o 3 : 0 , 1 , 2 (se e Figur e 6) .
Looking a t th e figure, on e ca n se e tha t th e su m o f tw o num -
bers alway s belong s t o on e an d th e sam e class , whic h i s determine d
only b y th e classe s o f th e tw o number s unde r stud y an d i s indepen -
dent o f the particula r choic e o f the representative s withi n th e classes .
For example , takin g th e representative s 1 , —2 , 7 o f th e clas s 1 an d
representatives —4 , 5 , 8 o f th e clas s 2 , w e se e tha t th e thre e sum s
1 + (-4 ) = - 3 , (-2 ) + 5 = 3 , 7 + 8 = 1 5 belong t o on e an d th e sam e
class 0 .
In general , th e identit y
(mx + k) + (my + I) = m(x + y) + (k + I)
allows us to defin e th e operatio n o f addition o f residue classe s over m:
the su m o f th e classe s k an d I is th e clas s tha t contain s al l th e sum s
fc+Z, wher e k is a representative o f the clas s k an d I is a representativ e
of the clas s /. Fo r example, i f m = 3 , we obtain th e followin g additio n

table i n th e se t o f residue s modul o 3 :
+ 0 T 2
0 0 T 2
T T 2 0
2 2 0 T
One ca n se e fro m thi s tabl e tha t th e residu e classe s modul o 3
form a cycli c grou p o f orde r 3 .
Exercise 92 . Prov e that the residue classes over an arbitrary numbe r
m for m a cyclic group of order m.
Can we also define the multiplication o f residue classes in a similar
way? Yes . Indeed , conside r tw o classe s /c , I. Th e produc t o f tw o
arbitrary representative s mx + k an d my + I o f thes e tw o classes ,
equal t o m{mxy + xl - f ky) + kl, ha s th e sam e remainde r i n divisio n
by m a s the number kl. Thi s remainder doe s not depen d o n the choic e
of th e representatives . Therefore , th e operatio n i s correctl y define d
in th e se t o f residu e classes .
Here i s the multiplicatio n tabl e o f residue classe s modul o 3 :
X 0 T 2
0 0 0 0
T 0 T 2
2 0 2 T
It i s evident tha t thi s multiplication doe s not satisf y al l the grou p
axioms, because it has one row and one column that entirel y consist of
zeroes, whereas in the multiplication tabl e o f a group no two element s
may coincide . However , th e smalle r tabl e tha t remain s afte r th e zer o
row an d colum n ar e deleted ,
X T 2
T T 2
2 2 T
does obe y al l th e grou p law s — i t represent s a cycli c grou p o f orde r

2.
Exercise 93 . D o al l th e nonzer o residue s modul o 6 for m a multi -
plicative group?
This exercis e suggest s th e followin g conclusion : to form a multi-

plicative group out of residues over ra it is reasonable to choose only
numbers mutually prime with ra. Fo r example , i f m = 6 , then th e ele-
ment 4 , which ha s a commo n multipl e 2 with th e numbe r 6 , become s
0 afte r multiplicatio n b y 3 . Bu t 0 cannot belon g t o a group , becaus e
it make s a whol e ro w o f zeroe s i n th e multiplicatio n table !
We shal l no w prov e th e followin g importan t fact : the set of all
residue classes k modulo m, such that the number k is mutually prime
with m, forms a multiplicative group.
Indeed, i f tw o number s ar e mutuall y prim e wit h ra, the n thei r
product i s mutually prime, too; this means that th e operation is closed
on th e give n set . Th e associativit y follow s fro m th e associativit y o f
ordinary multiplicatio n o f numbers . Th e clas s 1 i s mutuall y prim e
with ra an d plays the role of the unity. Th e only non-evident propert y
that w e must chec k i s tha t ever y residu e clas s i n th e se t unde r stud y
has a n inverse . I n othe r words , fo r ever y a , mutuall y prim e wit h ra,
there mus t exis t a numbe r x , mutuall y prim e wit h m an d suc h tha t
ax = 1 (modra). Th e las t formul a i s rea d alou d a s Hhe numbers ax
and 1 are congruent modulo m\ whic h means , b y definition , tha t ax
has remainde r 1 in divisio n b y ra. Thi s ca n b e rephrase d a s follows :
there exists an integer y such that ax + my = 1 . No w recal l tha t w e
have alread y prove d thi s fact , state d i n this for m (a s a corollar y fro m
Exercise 91 ) . W e did no t mentio n tha t x wil l be mutuall y prim e wit h
ra, bu t thi s i s evident .
Exercise 94 . Conside r th e se t o f tw o residu e classe s {2,4 } modul o
6. Doe s it mak e a group with respec t t o multiplication ?
For an y give n ra, w e shall denot e b y ZJ ^ the multiplicativ e grou p

constituted b y all residue classes modulo ra whic h ar e mutually prim e
with ra. Th e orde r o f thi s group , i.e . th e numbe r o f al l suc h residu e
classes, i s equa l t o </?(ra) , the Eule r functio n o f ra. W e hav e encoun -
tered thi s functio n whe n w e discusse d th e numbe r o f generator s o f a
cyclic grou p (se e pag e 89) . W e ar e no w read y t o appl y th e Lagrang e

theorem.
Note tha t i n an y grou p th e followin g identit y holds : g m ~ e ,
where m i s th e orde r o f th e group , g a n arbitrar y elemen t an d e
the unity . Indeed , i f k i s th e orde r o f # , the n b y Lagrange' s theore m
m — kl fo r a n appropriate intege r / , and w e have g 1 71 = (g k)1 = e l — e.
In th e case o f the grou p Z ^ thi s implie s th e followin g theorem .
Theorem 8 (Euler) . If a is a number mutually prime with m, then
a^>{rn) = 1 ( m o d m ) j
where (p(m) is the Euler function of m, i.e. the number of integers

between 1 and m mutually prime with m.
In the cas e when m = p is a prime number , w e have ip(p) = p — 1 ,

and th e Eule r theore m take s th e shap e o f th e followin g fact , know n
as Fermat y s little theorem.
Theorem 9 . If p is a prime number, then
v l
(34) a ~ = 1 (modp )
for any integer a not divisible by p.
Historical remark. Neithe r Ferma t no r Euler use d group theoreti c

considerations explicitl y t o prov e thei r theorems . I t wa s onl y i n th e
early nineteent h centur y tha t grou p theor y cam e int o bein g i n th e
work of E. Galois . However , bot h Ferma t an d Eule r implicitly di d us e
such notion s as , fo r example , cose t decompositio n o f residu e classes .
These investigation s becam e on e o f th e source s fro m whic h grou p
theory wa s born . Th e explici t applicatio n o f grou p theor y notion s
and theorem s make s th e arithmetica l fact s cleare r an d le t u s devis e
far-reaching generalizations .
To conclude thi s chapter , w e propose tw o problems i n elementar y
number theor y tha t ca n be solved usin g residues an d Euler' s theorem .
Exercise 95 , Prov e tha t th e equatio n x 2 = 3y 2 + 8 has n o intege r
solutions.
Exercise 96 . Wha t ar e the tw o last digit s of the number 2003 2004?
Chapter 5
Orbits an d Ornament s
Topics that wil l be touched upo n in this chapter include group actions,
orbits, invariants , an d ornaments .
Transformation groups , b y thei r definition , act o n certai n sets .
Thus, the group of movements of the plane acts on (th e set of all points
of) th e plane. Th e permutation grou p 63 acts on the set {1 ,2,3} . Th e
ability t o transfor m th e set s i s s o inheren t i n th e notio n o f a grou p
that i t i s als o preserve d fo r arbitrar y abstrac t groups . T o giv e th e
exact definitio n o f th e grou p action , w e shal l nee d th e notio n o f a
homomorphism.
1. Homomorphis m
The notion of a homomorphism generalize s that o f an isomorphism. A
homomorphism i s defined b y the sam e operation-preserving property ,
only without th e requiremen t tha t i t mus t b e a one-to-one correspon -
dence.
Definition 1 6 . A homomorphism fro m a grou p G int o a grou p H i s

a mappin g ip : G— > H suc h tha t
(35) <p(ab) = ip{a)<p(b)
for al l a,b £ G (th e grou p operatio n i s here written a s multiplication ,

but o f course , i t ma y hav e a n arbitrar y nature) .
128 5. Orbit s an d Ornament s
An isomorphism i s thus a one-to-one homomorphism. Som e prop-

erties of isomorphisms generaliz e to arbitrary homomorphisms . Thus ,
the imag e of the unit e G G under a homomorphism i s always the uni t
e! G H. Also , fo r an y elemen t g G G w e alway s hav e
(36) v^ 1
)= <p(fl) _1 -
Both o f thes e equalitie s easil y follo w fro m (35) .
Note tha t fo r an y tw o group s G an d H th e mappin g tha t take s
all th e element s o f G int o th e uni t o f H i s a homomorphism , calle d
the trivial homomorphism.
If (f : G— » G' i s an isomorphis m an d G' i s a subgroup o f a bigge r
group H , the n w e can vie w t p a s a mappin g fro m G int o H whic h is ,
of course, a homomorphism . Suc h homomorphism s ar e referre d t o a s
injective, o r monomorphic homomorphisms. The y ar e characterize d
by th e propert y tha t n o two differen t element s o f G g o into th e sam e
element o f H unde r <p.
The imag e o f G b y th e trivia l homomorphis m i s th e trivia l sub -
group (consistin g o f onl y on e element) . Not e tha t th e imag e o f G
by an y homomorphis m tp : G — » H i s a subgrou p o f H. Th e mos t
interesting clas s o f homomorphisms , i n a certai n sens e dua l t o th e
class o f monomorphisms , consist s o f epimorphic, o r surjective homo-
morphisms. A homomorphis m if : G— » H i s surjective, i f it s imag e
coincides wit h th e whol e of H, or , whic h i s the same , i f every elemen t
of H i s the imag e o f som e elemen t o f G.
We wil l conside r severa l example s o f surjectiv e homomorphisms .
Problem 37 . Construct a surjective homomorphism of the group of
integers Z with the operation of addition onto the group of residue
classes Z m .
Solution. Th e solutio n i s ver y simple : th e mappin g p
that take s ever y numbe r a int o a , th e clas s o f a modul o
m, i s th e require d epimorphism . I t preserve s th e addi -
tion, becaus e b y definitio n o f th e grou p Z m (se e p . 1 22 )
we hav e
p(a + 6 ) = a + b = a + b = p(a) - f p(b).
Also, th e mappin g p i s obviously surjective .
1. H o m o m o r p h i s m 129
Note i n passin g t h a t p als o preserve s th e product :

p(ab) — p(a)p(b), s o i t i s a homomorphis m wit h respec t
to multiplicatio n — an d yo u probabl y use d b o t h o f thes e
properties whil e solvin g Exercis e 95 .
E x e r c i s e 97 . Fo r wha t value s o f m an d n doe s ther e exis t a homo -

morphism o f Z m ont o Z n ?
P r o b l e m 38 . Find a homomorphism of S^ onto S3 .
S o l u t i o n . Recal l (1 ) t h a t S n denote s th e permutatio n

group o f n elements . Thus , S4 consist s o f 2 4 permuta -
tions o f th e se t o f cardinalit y 4 , whil e 5 3 i s mad e u p o f
6 permutation s o f th e se t o f cardinalit y 3 . W e hav e see n
earlier (Exercis e 74 ) t h a t £ 3 i s isomorphi c t o t h e grou p
of function s generated b y l/x an d 1 — x. Therefore ,
it i s enoug h t o construc t a homomorphis m o f 6 4 ont o .
This ca n b e don e a s follows .
Consider th e expressio n i n fou r variable s
a— c a — d
b — c ' b — d'
which i s called th e double ratio o f a, 6 , c and d. I f th e fou r
letters a , 6 , c, d i n thi s expressio n ar e permuted , th e valu e
of th e doubl e rati o i s changed . I t i s remarkabl e t h a t th e
new valu e ca n alway s b e expresse d i n term s o f x alone .
For example , afte r th e permutatio n I I (i.e. ,

\b c d aj
when a \—> 6 , b 1—> c , c 1—> d , d 1 — > a) w e obtai n
E x e r c i s e 9 8 . Fo r eac h o f th e 2 4 permutation s o f th e
letters a , fr,c, d, find th e expressio n o f th e doubl e ra -
tio afte r th e permutatio n i n term s o f its initia l valu e
x.
A diligen t reade r wh o ha s solve d thi s exercis e wil l
recall t h a t th e se t o f 6 function s obtaine d i s exactl y th e
group o f function s t h a t w e kno w a s $ . Denot e th e ra -
tional functio n t h a t correspond s t o th e p e r m u t a t i o n a
by f a(x). T o sho w tha t th e mappin g f : a \-> fa(x) i s

a homomorphism , w e hav e t o chec k tha t f TG = f T o f a.
Indeed, i f
a (a) - or(c) a{a) - a(d)
Ja[ ]
cr{b)-a(c) ' a(b)-a(d) V '
then evidentl y
_ rcr(a ) - ra(c ) ra(a ) - rcr(d ) _
/r* W - r c j ( 6 ) _ r ( j ( c ) • T(j{b) _ ra{d) ~ JAV),
and thu s f Ta{x) = f r(fG(x)).
Now we would lik e to dra w you r attentio n t o th e followin g inter -

esting observation . Th e fac t tha t / i s a homomorphism , i.e . satisfie s
the relatio n f Ta — fr o fa, greatl y simplifie s th e solutio n o f Proble m
38. I n fact , t o prov e tha t th e imag e o f / coincide s wit h th e se t ,
we do not hav e t o chec k al l th e 2 4 permutations o f the fou r letter s a s
suggested i n th e exercis e above . I t i s sufficien t t o chec k 3 transposi -
tions, a <- > 6, b <-> c an d c «-• d, whic h generat e th e grou p 54 . Thes e
transpositions correspon d t o the function s 1 /x, 1 — x an d 1 /x, respec -
tively, an d w e kno w tha t th e tw o function s 1 /x an d 1 — x generat e
the grou p 3>.
Exercise 99 . Assig n + 1 t o ever y movemen t tha t preserve s orienta -
tion an d — 1 to every movemen t tha t change s orientation. Chec k
that thi s assignment i s a homomorphism o f the group of all plane
movements onto the group of 2 elements {+1 ,-1 } wit h the oper -
ation o f multiplication .
Problem 39 . Construct a homomorphism of the group G of all

proper plane movements onto the group T of complex numbers whose
modulus is 1 .
Solution. Recal l (8 ) tha t an y prope r plan e movemen t
can be written analyticall y a s a complex functio n f(z) —
pz + a , wher e \p\ = 1 . W e defin e th e mappin g (p : G—>
T b y <p(f) = p. Le t u s chec k tha t thi s mappin g i s a
group homomorphism . Indeed , th e compositio n o f tw o
movements, / an d g define d b y g(z) — qz + 6 , is
a
9{f{z)) = (Pz + a) + b = qpz + (aq + 6) ,
2. Q u o t i e n t g r o u p 131
where th e coefficien t o f z i s qp. Therefore , <p(gf) =

<p(g)<p(f)-
Geometrically, th e assertio n jus t prove d implie s t h a t whe n tw o

rotations, eve n wit h differen t centres , ar e multiplied , th e angle s o f
rotation ar e adde d together . I n particular , th e movemen t R^ o R^
is alway s a paralle l translation .
To solv e th e nex t exercise , yo u wil l fin d th e followin g fac t useful :
the angle between the lines I and m is equal to the angle between R^(l)
and R^(m). Indeed , rotation s preserv e th e angles ; henc e th e angl e
between I an d m equal s th e angl e betwee n R^(l) an d R^(m) — an d
the tw o line s R^(m) an d i t ^ ( r a ) ar e parallel .
E x e r c i s e 1 0 0 . Le t BE an d CF b e the altitude s o f the triangl e ABC ,
and O th e centr e o f th e circumscribe d circle . Prov e tha t AO _ L
EF.
E x e r c i s e 1 0 1 . Th e determinant o f a matri x I 7 I i s th e numbe r

\c d)
ad — be. Th e product o f two matrice s ( , j an d ( ) is th e
\c d) \p q)
matrix ( , . , I . Prov e tha t th e se t o f al l matrice s
\cm + dp cn + dqj
with non-zero determinant form s a group, and th e determinant i s a
homomorphism o f this grou p ont o th e grou p o f non-zer o number s
with multiplication .
2. Quotien t grou p
Let u s loo k a t th e grou p o f residu e classe s Z m an d th e homomorphis m
Z— > Zm fro m a mor e genera l viewpoint . T h e grou p Z m consist s o f
the coset s o f t h e grou p Z modul o th e subgrou p m Z . Thi s fac t ca n
also b e state d a s iLirn IS
the quotient group of Z modulo raZ, whic h i s
written a s Z m = Z / r a Z .
Now le t G b e a n arbitrar y grou p an d H a subgroup o f G. Conside r
the se t o f coset s i n G modul o H an d t r y t o mak e i t int o a group , usin g
the constructio n o f Z m a s a n example . T h e firs t questio n t h a t arise s
is: wha t coset s shoul d w e use , lef t coset s eH, g\H, g2H, . . .o r righ t
cosets He, Hgi, Hg2, . . . ? Thi s questio n doe s no t aris e fo r th e grou p
Z,tosinc
Licensed Penn St e i tUniversity
Univ, i s commutative . 12 18:16:11 EDT 2019for download from IP 132.174.254.159.
Park. Prepared on Tue Mar
Recall tha t th e lef t cose t gH i s made u p o f all products gh wher e

g G G i s fixed an d h range s ove r H, wherea s th e righ t cose t Hg
consists o f al l product s hg. I f th e grou p i s no t commutative , the n
in genera l gH ^ Hg. Fo r example , tak e th e grou p o f plan e move -
ments Ds a s th e whol e grou p G , an d th e subse t H — {id , Sa} a s th e
subgroup. Then , usin g th e multiplicatio n tabl e (p . 92) , we find tha t
RH = {Roid,RoS a} = {R,S b},
HR = {idoR,S aoR} = {R,S c}.
However, for the subgroup G 3 = {id , R, R 2} bot h the left an d the right
coset decomposition s consis t o f th e sam e tw o classes : th e subgrou p
G3 itself an d it s complemen t D% \ G3 .
Definition 1 7 . A subgrou p H o f a grou p G i s calle d normal, i f fo r
any elemen t g G G th e tw o coset s gH an d Hg coincide .
If H C G i s normal, the n th e followin g trai n o f equalitie s i s true:

(giH)(g2H) = gi(Hg 2)H = 9l (g2H)H = g ld2H,
meaning tha t i f yo u choos e an y elemen t o f th e cose t g\H an d an y

element o f th e cose t g 2H, the n thei r produc t wil l belon g t o on e an d
the sam e cose t (gig 2)H. Therefore , i n th e se t o f cosets ove r a norma l
subgroup ther e i s a well-define d multiplication : cfig 2 = gig 2, wher e
the ba r ove r a lette r denote s th e cose t o f a n elemen t ove r th e give n
normal subgroup : g = gH = Hg. I t i s eviden t tha t thi s operatio n i s
associative; th e rol e o f uni t elemen t i s playe d b y th e cose t H = eiJ ,
and th e elemen t invers e t o gH i s g~ lH. Th e se t o f coset s H, g\H,
g2H, . . . thus acquire s th e structur e o f a group .
Definition 1 8 . Le t G b e a grou p wit h a norma l subgrou p H. Th e

quotient group of G over H, denote d b y G/H, i s the se t o f al l coset s
over H wit h th e produc t o f tw o coset s give n b y th e rul e gig 2 = g\g 2.
The structur e o f th e quotien t grou p G/H ca n b e rea d of f th e

multiplication tabl e o f G , i f the element s i n th e first ro w an d an d th e
first colum n o f th e tabl e ar e arrange d b y cosets . Look , fo r example ,
at th e multiplicatio n tabl e o f th e grou p D% (p. 92) . Yo u ca n clearl y
see tha t th e tabl e i s split int o 4 blocks o f th e siz e 3 x 3 each . Denot -
ing th e bloc k tha t correspond s t o rotation s b y R an d th e bloc k tha t
2. Quotien t grou p 133
corresponds t o reflection s b y 5 , w e can represen t th e bloc k structur e

of th e whol e tabl e a s
R S
R R R
S S R
This tabl e define s a cycli c grou p o f orde r 2 isomorphi c t o Z2 .
Thus, w e can writ e D3/C 3 = Z2 , where th e sig n = stand s fo r isomor -
phism.
Problem 40 . Let G be the group of proper movements of the plane,
consisting of all rotations and all translations. Denote by N the sub-
group of rotations around a fixed point A and by K the subgroup of
all translations. Determine whether these subgroups are normal and,
if appropriate, find the structure of the quotient groups.
Solution. Th e conditio n o f normalit y gH = Hg ca n b e
rewritten a s gHg~ l = H, whic h mean s tha t a subgrou p
H i s norma l i f an d onl y if , togethe r wit h ever y elemen t
of it , i t contain s al l th e conjugat e elements . Sinc e con -
jugation i s lookin g a t thing s fro m a differen t viewpoin t
(see p . 82) , a subgrou p i s norma l wheneve r i t look s th e
same fro m an y standpoint . I t i s clea r that , fo r a perso n
living i n th e plane , th e se t o f al l translation s look s th e
same, whateve r hi s positio n ma y be . However , th e se t
of rotation s wit h a fixe d centr e A look s differen t fo r a
person place d a t A an d fo r a perso n place d somewher e
else. Therefore , th e subgrou p K i s normal, bu t N i s not .
A mor e rigorou s proo f o f thi s fac t ca n b e obtaine d
using th e resul t o f Exercis e 5 8 (p . 234) . Th e movemen t
conjugate t o a translatio n i s always a translation ; there -
fore K i s normal. Th e movemen t conjugat e t o a rotatio n
around A b y a movemen t / i s a rotatio n aroun d f(A);
therefore N i s no t norma l
Since K C G is normal , th e quotien t grou p G/K i s
defined. T o understan d it s structure , conside r th e ho -
momorphism ip : G— > T studie d abov e i n Proble m 39 .
We clai m tha t tw o element s o f G hav e th e sam e imag e
under cp i f an d onl y i f the y belon g t o th e sam e cose t o f

G wit h respec t t o K. Indeed , le t <p(f) = ip(g) = p G T.
If p = 1 , the n bot h / an d g ar e translation s an d belon g
to K. I f p = co s a + 2 sin a ^ 1 , the n bot h / an d g ar e
rotations throug h th e angl e a , say , / = R% and g = R^.
In thi s cas e / = R<% = R% o (i?~a o ij«) G # #, becaus e
jRâ o i?^ i s a translation . Conversely , i f / an d g belon g
to on e cose t ove r K, the n / = g o h an d </?(/ ) = (p(g).
The propert y tha t w e hav e just prove d implie s tha t
(p establishes a one-to-on e correspondenc e betwee n th e
sets G/K an d T , whic h w e ca n denot e b y (p. Th e map -
ping ip obviously agree s wit h th e grou p operation s an d
thus yield s a n isomorphis m ip : G/K — > T. Th e quo -
tient grou p G/K i s thus th e sam e thin g a s th e grou p o f
complex number s wit h uni t modulus , o r th e grou p o f al l
rotations wit h a commo n centre .
Note tha t th e cose t decompositio n o f G ove r K ha s
a simple geometric meaning: ever y coset consist s of rota -
tions through th e same angle around a n arbitrar y centre .
The subgrou p K itsel f (th e se t o f al l translations ) i s th e
unit cose t — i t correspond s t o th e uni t o f th e grou p T
under (p.
Generalizing thi s argument , w e arriv e a t th e followin g assertion ,

called th e first homomorphism theorem.
Theorem 1 0 . If (p is a homomorphism of a group G onto a group

H and K C G is its kernel, i.e., the set of all elements of G that go
into the unit of H under cp, then G/K = H.
Note tha t th e kerne l K o f any homomorphis m (p i s alway s a nor -

mal subgroup i n G, s o that th e quotient G/K i s correctly defined . In -
deed, i fA ; € If , the n <p(k) = e; hence (p(gkg~ 1 ) = (p(g)ip(k)(p(g~ 1 ) =
(p(g)e(p(g~1) = e , whic h mean s tha t gkg~ 1 G K.
The firs t homomorphis m theore m implie s tha t th e order s o f th e
three group s G , H an d K ar e relate d b y th e simpl e equalit y \G\ =
\H\ • \K\. I n particular, w e have the followin g corollary : if there exists
2. Q u o t i e n t g r o u p 135
a homomorphism of finite groups G —> H, then the order of H is a

divisor of the order of G.
E x e r c i s e 1 0 2 . I s ther e a homomorphis m (a ) o f D 3 ont o Z 2 ? (b ) o f

£>3 onto Z3 ?
P r o b l e m 4 1 . Find the kernel of the homomorphism 3>

discussed in Problem 38.
S o l u t i o n . Le t K C 5 4 b e th e kerne l o f ip. Sinc e p

is a surjectiv e homomorphis m (i.e. , it s imag e coincide s
with th e entir e grou p ) , th e group s S±/K an d $ ar e
isomorphic an d henc e \K\ — 2 4 : 6 = 4 . I t i s eas y t o
check t h a t th e fou r permutation s
ab c d\ ( a b e d
ab c d) \b a d c
ab c d\ ( a b e d
cd a b) ' \d c b a
leave invarian t th e expressio n
a— c a —d
b—c b —d
The multiplicatio n tabl e fo r thes e fou r element s co -
incides, u p t o th e choic e o f notation , wit h th e multipli -
cation tabl e o f th e grou p D2. T h u s K = D2 , an d th e
homomorphism theore m i n ou r exampl e ca n b e writte n
as S A/D2 ^ D 3.
E x e r c i s e 1 0 3 . Fin d th e imag e an d th e kerne l o f th e homomorphis m

of th e additiv e grou p o f function s / : R— > R int o itsel f give n b y
the formul a f(x) H ^ f(x) - f f(—x).
E x e r c i s e 1 0 4 . Le t S b e th e grou p o f rotation s o f th e plan e wit h a

fixed centre , an d C n it s cycli c subgrou p o f orde r n. Prov e tha t
S/Cn £* S.
E x e r c i s e 1 0 5 . Usin g the first homomorphis m theorem , represen t th e
group S o f th e previou s exercis e a s a quotien t grou p o f (E , + ).
3. Group s presente d b y generator s an d relation s

We ar e no w i n a positio n t o explai n th e constructio n o f a n abstrac t
group wit h a give n se t o f generator s an d definin g relations . Befor e
(page 9 3 in Chapte r 3 ) w e have alread y talke d abou t generator s an d
relations i n a given , alread y define d group . Th e curren t proble m i s
the inverse: no w we want t o define a group, starting from a n arbitrar y
set o f generator s an d relation s betwee n them .
This definitio n relie s o n th e notio n o f the quotien t grou p tha t w e
have jus t studie d an d th e notio n o f a fre e grou p tha t w e wil l no w
define.
Definition 1 9 . Le t S b e a n arbitrar y set , consistin g o f letter s i n a

certain alphabet . Th e free group over 5 , denote d b y F(S), consist s of
all monomial s ove r S (se e pag e 93) , th e grou p operatio n bein g give n
by writin g tw o monomial s sid e b y sid e an d usin g th e simplificatio n
rules s ksl — s k+l an d s° = 1 .
For example, i f S consist s of one element, the n F(S) i s the infinit e

cyclic group .
Definition 20 . Le t R b e a se t o f monomial s ove r S (eac h mono -

mial r i s understoo d a s a relatio n r = 1 between th e element s o f S).
The group with generators and defining relations R i s defined a s th e
quotient grou p F(S)/H(R), wher e F(S) i s the fre e grou p ove r S an d
H(R) i s th e minima l norma l subgrou p o f F(S) containin g al l rela -
tions belongin g t o R (i n othe r words , H(R) i s th e subgrou p o f F(S)
generated b y al l element s conjugat e t o th e element s o f R).
An abstract grou p with generators «si,... , sn an d relations r i , . . . ,

r m i s denote d b y
(si,...,sn|ri,...,rm)
(The relation s o n th e righ t ar e sometime s writte n simpl y a s monomi -
als ove r S, eac h monomia l r meanin g th e equalit y r = 1 ) .
For example , i t i s easy t o se e tha t
(a|an-l)
is the cycli c grou p o f orde r n.
4. Grou p action s an d orbit s 137
Problem 42 . Prove that the group
(a,b\ab=1 )
is the infinite cyclic group, isomorphic to the additive group of all

integers Z.
Solution. Th e fre e grou p F(a,b) consist s o f al l word s

aklbl1 ...a knbln o f arbitrar y lengt h an d wit h arbitrar y
integer exponents . T o obtai n th e quotien t group , suc h
words shoul d b e considere d modul o th e element s o f th e
subgroup H(R): i f x = yh o r x = hy, wher e h G H(R),
then x an d y belon g t o on e an d th e sam e coset . Now ,
a = (ab)b~ 1 ', therefore, a is equivalent t o 6 _ 1 , an d henc e
every wor d i n a an d b is equivalen t t o som e wor d i n b
only. W e se e tha t ever y elemen t o f th e quotien t grou p
F(S)/H(R) i s a powe r o f 6 , so that thi s grou p i s cyclic .
Note tha t i n ever y elemen t o f th e subgrou p H(R)
the su m o f al l exponent s o f a i s equa l t o th e su m o f al l
exponents o f b; therefore b n wit h w / 0 canno t belon g
to H(R), whic h mean s tha t al l power s o f b are different .
The grou p F(S)/H(R) i s thus infinite .
Remark. I n th e abov e argument , th e generator s a an d b pla y

symmetrical roles , becaus e th e produc t ba i s conjugat e t o ab: ba —
a~1(ab)a an d therefor e belong s t o th e subgrou p H(R) — thi s i s on e
of th e reason s wh y i n Definitio n 2 0 th e subgrou p H(R) i s se t t o b e
the normal subgrou p generate d b y R.
Exercise 1 06 . Wha t i s the group presented b y
(a,6|a 2 = 1 , b n = 1 , aba^b" 1
)?
Exercise 1 07 . Prov e tha t
(a,b | aba = bob) = (x,y | x —y ) .
4. Grou p action s an d orbit s

Group action s ca n b e define d i n term s o f homomorphism s a s follows .
Definition 21 . Le t G be a grou p an d X a set . A n action o f G on X

is a homomorphis m o f G int o th e grou p o f transformations o f X (se e
p. 76) :
T:G->Tr(X).
The imag e o f a poin t x e X unde r th e actio n o f th e transforma -
tion T g i s T g(x), whic h i s ofte n denote d b y gx fo r short .
We shoul d lik e t o stres s fro m th e beginnin g tha t a give n grou p
G ca n ac t o n a give n se t X i n a variet y o f differen t ways , becaus e i n
general ther e ma y exis t man y differen t homomorphism s G — > Tr(X) .
For example , th e symmetr y grou p o f a n equilatera l triangl e D% acts
on the plan e by definition. However , th e actio n depend s o n the choic e
of th e equilatera l triangl e i n question , o r mor e exactly , o n th e choic e
of it s centr e an d thre e symmetr y axes .
Figure 1 shows a n equilatera l triangl e wit h centr e O an d symme -
try axe s a, 6 , c.
Figure 1 . Actio n o f th e grou p D% on th e plan e
Under th e actio n o f an y elemen t o f D3 , the origi n (poin t O) goe s

into itself . A poin t lyin g o n on e o f the line s a, 6 , c and differen t fro m
O give s thre e differen t point s (includin g itself) . An y othe r poin t o f
the plan e give s si x differen t point s unde r th e actio n o f th e group .
Definition 22 . Th e se t o f al l point s obtaine d fro m a give n poin t x
under th e actio n o f a grou p i s calle d th e orbit o f thi s point :
0(x) = {T
Licensed to Penn St Univ, University Park. Prepared on Tue Mar 12 18:16:11 EDT g
(x)\geG}.
2019for download from IP 132.174.254.159.
4. Grou p action s an d orbit s 139
The cardinalit y o f th e se t 0{x) i s called th e length o f th e orbit .
In th e previou s exampl e w e had orbit s o f three differen t kinds : o f

lengths 1 , 3 and 6 .
It i s clear tha t an y poin t alway s belongs to it s own orbit, becaus e
it goe s into itself unde r th e actio n of the neutra l elemen t o f the group .
A poin t whos e orbi t consist s o f onl y on e poin t i s calle d a fixed point
of th e action . I n ou r exampl e th e onl y fixe d poin t wa s th e poin t O.
Look a t Figur e 1 . Th e orbi t o f the poin t A — A\ consist s of thre e
points Ai, Ai, As. I f yo u tak e A<i o r As a s th e initia l point , yo u wil l
get th e sam e orbit . Thi s i s a manifestatio n o f th e genera l property :
the orbit of any point belonging to the orbit of a point x coincides with
O(x). Indeed , i f y G 0(x), the n y — hx fo r a certai n grou p elemen t
h G G. The n 0(y) = {gy\g G G} = {ghx\g G G}, bu t th e axiom s
of th e grou p impl y that , fo r an y fixe d elemen t h G G, th e se t o f al l
products gh, wher e g range s throug h G , coincide s wit h th e se t G .
This observatio n implie s th e followin g importan t fact : any two
orbits either coincide or do not have common elements. I n fact , i f
two orbit s 0(x) an d 0(y) hav e a commo n elemen t z, the n 0(x) —
0(z) = 0(y).
The meanin g o f th e wor d orbit an d th e splittin g o f th e se t int o
orbits ca n b e ver y clearl y see n fo r th e actio n o f th e circl e 5 , under -
stood a s th e multiplicativ e grou p o f comple x number s wit h modulu s
1, on th e comple x plan e b y mean s o f multiplicatio n (Figur e 2) . Thi s
action ha s on e fixed poin t (th e numbe r 0) , while th e res t o f the plan e
splits int o concentri c circle s whic h ar e th e orbits .
Exercise 1 08 . Le t Dz b e the symmetr y grou p o f an equilatera l tri -
angle in the complex plane, whose centre i s at th e origin and on e
of whos e symmetr y axe s i s th e x-axis . Conside r th e actio n o f
this grou p o n th e finit e se t {0,1 , - 1, 2 , - 2, iy/3, -iy/3,4, - 4 , 2 +
2iy/3, 2 - 2iy/3, -2 + 2iV3, -2 - 2iy/3}. Fin d th e orbit s o f thi s
action.
Problem 43 . Define a natural action of the group of rational expres-

sions < £ (Exercise 74) on a suitable set.
Solution. A rationa l expressio n i n on e variabl e ca n b e

considered a s a function, i.e . a s a mapping of the real line
140 5. O r b i t s a n d O r n a m e n t s
Figure 2 . Orbit s o f the grou p S
into itself. Unfortunately , thes e mapping s ar e no t define d

at al l point s o f th e rea l lin e R , becaus e th e denominator s
sometimes becom e zero . Ther e ar e tw o natura l way s t o
mend thi s situation . On e i s t o exclud e th e tw o point s
0 an d 1 fro m R (not e t h a t thes e ar e th e onl y value s o f
x wher e th e denominator s ma y vanish) . Th e othe r i s t o
add on e extr a poin t o o t o R an d exten d th e function s t o
this poin t accordin g t o th e rul e
Fi F2 ^3 F4 F, F6
0 oo 1 0 1 OO 0
1 1 0 1 oo 0 oo
oo 0 oo oo 0 1 1
This tabl e define s a genuin e actio n o f th e grou p $
on th e se t R U oo. I t shows , i n particular , t h a t t h e thre e
points 0 , 1 an d o o for m on e orbit , an d t h a t th e actio n
of $ o n thi s orbi t define s th e isomorphis m o f wit h th e
permutation grou p o n thre e symbol s S3 .
E x e r c i s e 1 0 9 . Fin d on e mor e 3-elemen t orbi t o f thi s action , an d

prove tha t al l th e remainin g orbit s consis t o f 6 elements .
E x e r c i s e 1 1 0 . Function s Fi G $ ca n als o b e considere d a s function s
of a complex variable . Therefore , th e actio n o f the grou p 3 > ca n b e
5. E n u m e r a t i o n o f o r b i t s 141
prolonged t o th e se t CUoo . Fin d a 2-elemen t orbi t o f thi s actio n

and prov e tha t al l th e othe r orbits , excep t thi s on e an d th e tw o
3-element orbit s indicate d above , consis t o f 6 elements .
5. Enumeratio n o f orbit s
Consider on e mor e exampl e o f a grou p action .
Problem 44 . Let Q be a cube in space and G the group of all space

rotations that take Q into itself (i.e., the proper symmetry group of
Q). Make up the list of all elements of Q and describe the action of
this group on the set of faces of the cube.
S o l u t i o n . Apar t fro m th e identity , th e grou p G con -

tains:
• 6 rotation s throug h 1 80 ° aroun d th e line s t h a t g o
through th e midpoint s o f tw o paralle l edge s (lik e
A A' i n Figur e 3) ,
• 3 rotation s throug h 1 80 ° aroun d th e line s connect -
ing th e midpoint s o f tw o opposit e face s (lik e BB'),
• 6 rotation s throug h 90 ° aroun d th e sam e lines , an d
• 8 rotation s throug h 1 20 ° aroun d th e line s t h a t con -
tain a pai r o f opposit e vertice s o f th e cub e (lik e
CC').
The grou p G thu s consist s o f 2 4 elements .
The natura l actio n o f th e grou p G o n t h e se t F o f
faces o f th e cub e i s transitive, i.e. , an y fac e ca n b e take n
into an y othe r b y a suitabl e elemen t o f th e group . I n
other words , th e se t F , whic h consist s o f 6 points , make s
one orbi t o f th e grou p action . Fo r an y fac e / E F ther e
are exactl y 4 rotation s t h a t preserv e it : th e identit y an d
the thre e rotation s aroun d th e lin e passin g t h r o u g h th e
midpoint o f thi s face . Not e t h a t 6 • 4 give s 24 , t h e orde r
of th e group .
E x e r c i s e 1 1 1 . Describ e th e actio n o f G o n th e se t E o f edge s an d

on th e se t V o f vertices .
Figure 3 . Rotation s o f th e cub e
So far , w e hav e foun d transitiv e action s o f th e grou p G o n th e

sets o f orde r 6 , 8 and 1 2 . Not e tha t al l thes e number s ar e divisor s of
24, th e orde r o f th e group . Th e numbe r 2 4 ha s som e mor e divisors ,
and i t turn s ou t tha t fo r ever y diviso r on e ca n construc t a se t o f
corresponding cardinalit y tha t consist s o f certai n geometri c element s
of the cub e an d o n whic h ou r grou p act s transitively . Figur e 4 shows
the set s o f 4 , 3 an d 2 element s endowe d wit h a natura l transitiv e
action o f ou r group : thes e ar e th e se t D o f bi g diagonals , th e se t M
of middl e line s an d th e se t T o f regula r tetrahedr a inscribe d i n th e
cube.
Definition 23 . A se t wit h a transitiv e actio n o f a grou p i s calle d a

homogeneous spac e o f this group .
In eac h o f thes e case s w e hav e a homomorphis m o f th e grou p G

into th e grou p o f transformation s o f th e correspondin g set .
Exercise 1 1 2 . Fo r whic h o f th e set s F , E, V, D, M , T i s this ho-
momorphism (a ) a n epimorphism? (b ) a n isomorphism ?
We ca n als o conside r th e actio n o f th e grou p G o n mor e compli -

cated object s tha t consis t o f severa l element s o f th e abov e sets .
5. Enumeratio n o f orbit s 143
Figure 4 . Homogeneou s space s o f th e symmetr y grou p o f th e cub e
Problem 45 . Describe the orbits of the action of our group G on the

set of all pairs (/ , e), where f G F is an arbitrary face and e G E is
an arbitrary edge of the cube Q.
Solution. Th e se t i n questio n ca n b e denote d a s F x E
(the Cartesia n produc t o f th e set s F an d E) . I t consist s
in all of 6-12 = 7 2 elements which come in three differen t
categories: th e edg e e can eithe r belon g t o th e fac e / (a s
in Figur e 5a) , hav e on e commo n verte x wit h / (a s i n
Figure 5b ) or , finally, hav e n o commo n point s wit h /
(Figure 5c) .
Figure 5 . Differen t face-edg e pair s
It i s clea r tha t a n edge-fac e pai r belongin g t o on e

of thes e type s canno t g o int o a pai r o f anothe r typ e un -
der an y movement . Le t u s prov e tha t an y tw o pair s o f
the sam e typ e ca n b e transforme d on e int o anothe r b y a
suitable movement . Suppos e tha t w e are give n two pair s
( / l j ^ i ) , (72,62 ) o f th e firs t typ e (edg e lie s i n th e face) .

First w e fin d a rotatio n whic h take s 7 i m t o 72 - Then ,
using th e 4 rotations t h a t preserv e thi s face , w e ca n mov e
e\ t o th e positio n o f e2 , an d t h a t ' s all . Th e tw o othe r
cases ar e treate d similarly .
We hav e thu s prove d t h a t th e se t F x E wit h th e
action o f th e grou p G consist s o f thre e orbits , typica l
representatives o f whic h ar e show n i n Figur e 5 .
E x e r c i s e 1 1 3 . Fin d th e numbe r o f orbit s an d indicat e thei r repre -

sentatives fo r th e actio n o f G o n th e followin g sets : (a ) V x F
(vertex-face pairs) , (b ) D x F (diagonal-fac e pairs) , (c ) E x E
(ordered pair s o f edges) .
Let u s no w generaliz e th e observation s t h a t w e hav e mad e i n th e

previous discussion . T o d o so , we will nee d th e notio n o f the stabilizer,
or stable subgroup, o f a point .
D e f i n i t i o n 2 4 . Give n a n actio n o f a grou p G o n th e se t X, th e

stabilizer o f a poin t x G l i s th e se t o f al l element s o f G t h a t preserv e
the poin t x:
St{x) = {geG\T g{x) = x}.
T h e stabilize r i s a subgrou p i n G. Indeed , i f b o t h g an d h ar e i n

St(x), the n w e hav e T gh(x) = T g(Th(x)) = T g(x) = x an d T g-i(x) =
T~1(x) = x.
T h e stabilize r o f a fixe d poin t coincide s wit h th e whol e grou p
G. Fo r example , loo k a t Figur e 1 . T h e stabilize r o f O i s th e whol e
group, th e stabilize r o f th e poin t A consist s o f tw o element s (identit y
and reflection) , whil e th e stabilize r o f th e poin t B i s trivia l (contain s
only th e identity) .
You ca n her e notic e th e sam e rul e t h a t w e sa w i n Proble m 44 :
the orde r o f th e stabl e subgrou p multiplie d b y th e lengt h o f th e cor -
responding orbi t alway s give s th e orde r o f th e whol e group :
(37) |0(s)|.|StOr)| = |G| .

To prove thi s fact , conside r th e lef t cose t decompositio n o f the grou p

G wit h respec t t o th e subgrou p H = St(x) :
G= gi HUg2HU.-.UgkH.
All th e element s o f th e sam e cose t ac t o n x i n th e sam e way : i f
h G H, the n T gh(x) — T g(Th(x)) = T g(x), whic h doe s no t depen d o n
the specifi c choic e of h. Conversely , i f two elements g,k G G move th e
point x t o th e sam e position , the n the y belon g t o on e an d th e sam e
coset. Indeed , ther e i s a n elemen t h G G suc h tha t k — gh. The n
Th{x) = T g-ik(x) = T^nix) = x,
which mean s tha t h G St(x) . Therefore , th e numbe r o f differen t
points i n th e orbi t o f x i s th e sam e a s th e numbe r o f coset s i n th e
decomposition G/St(x) , an d th e assertio n follows .
Formula (37 ) shows , i n particular , tha t th e stabilizer s o f al l th e
points tha t belon g t o th e sam e orbi t hav e th e sam e numbe r o f ele -
ments. I n fact, thes e subgroup s ar e alway s conjugate t o eac h othe r i n
the grou p G , an d henc e isomorphic .
To prove thi s fact , tak e tw o arbitrar y point s x an d y tha t belon g
to on e an d th e sam e orbit . The n ther e i s a grou p elemen t g suc h
that y = T g(x). W e clai m tha t th e subgroup s St(x ) an d St(y ) ar e
conjugate b y th e elemen t g. Indeed , i f h G St(rc), the n
Tghg-i(y) = T g(Th(T-\y))) = T g(Th(x)) = T g(x) = y,
which means that ghg~ l G St(y); and, since h is arbitrary, thi s implie s
that ^ S t ( x ) ^ - 1 C St(y) . Interchangin g th e role s o f x an d y i n th e
previous argumen t prove s th e invers e inclusion . Thu s ^ S t ( x ) ^ - 1 —
St(y), an d th e tw o subgroup s ar e indee d conjugate .
Problem 46 . How many different ways are there to paint the disk
divided into p equal parts using n colours ? The number p is assumed
to be prime. Two colourings are considered to be the same, if one of
them goes into another by a rotation of the disk.
Solution. W e dea l wit h th e actio n o f th e cycli c grou p

Cp o n th e se t o f al l possibl e n p colouring s o f th e disk ,
and w e ar e aske d t o find th e numbe r o f orbit s o f thi s
action. Th e lengt h o f the orbit , whic h i s always a diviso r
of th e orde r o f th e group , i n thi s exampl e ca n tak e onl y

two values : 1 an d p , becaus e p i s prime . A n orbi t o f
length 1 correspond s t o a colourin g whic h i s invarian t
under al l rotations, i.e . a colouring wher e the whol e dis k
is coloure d wit h on e colour . Th e tota l numbe r o f suc h
colourings i s n. Th e remainin g colourings , whos e tota l
number i s n p — n , spli t int o orbit s o f cardinalit y p , an d
the numbe r o f the orbits is (n p — n)/p. Th e total numbe r
of orbits, i.e. , th e numbe r o f different way s t o colou r th e
disk, i s thus (n p — n)/p + n.
Note that w e have proved, a s a byproduct, tha t th e number n p—n

is alway s divisibl e b y p , i f p i s prime . Thi s i s ye t anothe r proof o f
Fermat's littl e theore m (34) .
Exercise 1 1 4 . Tr y to solv e the sam e problem i n the cas e when p is
not necessaril y prime .
It i s rathe r difficul t t o solv e thi s exercis e b y a direc t argument ,
like th e on e w e use d i n th e previou s problem . However , ther e exist s
a genera l formul a tha t compute s th e numbe r o f orbit s fo r an y grou p
action — th e so-calle d Burnside formula, whic h w e shal l no w stat e
and prove .
If g G G i s a n elemen t o f th e grou p actin g o n a se t X , the n
we denot e b y N(g) th e numbe r o f fixe d point s o f th e correspondin g
transformation T g, i.e . th e number o f points x G M suc h that T g(x) =
x.
Theorem 1 1 (Burnside' s formula) . The number of orbits r is "the
arithmetic mean of the number of fixed points for all elements of the
group":
1 v ^
77*£^)>
\G\
gee
or, in other words, the sum of numbers N(g) for all the elements of
the group is \G\ times the number of orbits.
Proof. T o prove the formula, le t us think for a while about th e follow -

ing question: how many times does a given point x G M participate in
the total sum J2 geG
Licensed to Penn St Univ, University Park.
N(g) ?- Evidently, i t come s u p ever y tim e whe n
Prepared on Tue Mar 12 18:16:11 EDT 2019for download from IP 132.174.254.159.
an elemen t g preserves x. Th e answe r t o th e questio n i s thus | St(x)|.

Other point s belongin g t o th e orbi t 0(x) appea r i n the tota l su m th e
same numbe r o f times , becaus e al l th e stabilizer s hav e th e sam e car -
dinality. Therefore , th e contributio n o f thi s orbi t i s | St(x)| • |C?(x)|,
which, a s we know, is equal to |G| , the order o f the group. Sinc e every
orbit give s the sam e contributio n |G| , th e whol e su m i s \G\ times th e
number o f the orbits, an d this is exactly what w e wanted to prove. •
Let u s us e Burnside' s formul a t o solv e Proble m 4 6 onc e again .

Here the identity transformatio n ha s n p fixed points , while every non -
identity transformatio n ha s n fixed points . Therefore , th e numbe r o f
orbits i s r = (n p + (p — l)n)/p. I t i s easy t o se e that thi s resul t doe s
not diffe r fro m th e on e obtaine d before .
Problem 47 . Find the number of different necklaces made of 7 white

and 3 black beads.
Solution. I t i s natural t o treat tw o necklaces as equal, if

they diffe r onl y b y a rotation o r a reflection o f the circle .
Therefore, w e have to conside r th e se t X whos e element s
are al l possible disposition s o f 7 white an d 3 black bead s
in th e vertice s o f a fixed regula r 1 0-go n an d th e actio n
of th e dihedra l grou p Di o o n X. Th e proble m i s to find
the numbe r o f orbits o f this action .
Having Burnside' s formul a i n mind , le t u s calculat e
the numbe r o f fixed points i n X fo r ever y elemen t o f th e
group D\Q. Fo r th e identit y transformation , al l (3° ) =
120 points o f th e se t X ar e fixed.
A nontrivia l rotatio n canno t leav e an y necklac e in -
variant. Th e sam e i s tru e fo r th e reflection s o f typ e (a )
(see Figur e 6) , becaus e th e numbe r o f bead s o f eac h
colour i s odd . Bu t th e reflection s o f typ e (b ) d o hav e
invariant necklaces . Fo r ever y suc h reflectio n ther e ar e
2 - 4 = 8 suc h necklaces : first, on e o f th e bead s o n th e
symmetry axe s mus t b e whit e an d anothe r on e black ,
second, on e o f th e fou r symmetrica l pair s o f bead s mus t
be black . Th e numbe r o f reflections o f type (b ) i s 5 , an d
by Burnside' s formul a w e ge t r = (1 2 0 + 5 • 8)/2 0 = 8 .
Figure 6 . A necklace of type 3- 7
E x e r c i s e 1 1 5 . Unde r th e centra l symmetr y (rotatio n throug h 1 80° )

digits 0 , 1 and 8 are preserved , digit s 6 and 9 change places , whil e
all th e remainin g digit s los e thei r meaning . Ho w man y 5-digi t
numbers ar e centrall y symmetric ?
E x e r c i s e 1 1 6 . Solv e th e las t proble m i n th e cas e o f 6 whit e an d 4
black beads .
E x e r c i s e 1 1 7 . A dice i s a cub e marke d o n eac h sid e wit h number s
1 throug h 6 . Ho w man y differen t dic e ar e there ? (Tw o dic e ar e
regarded a s th e same , i f the y ca n b e s o rotate d i n spac e tha t th e
numbers o n correspondin g side s becom e equal. )
E x e r c i s e 1 1 8 . Ho w man y differen t way s ar e ther e t o pain t (a ) th e
vertices, (b ) th e edge s of a cube b y two colours (i.e . usin g n o mor e
than tw o give n colours) ? (Here , a s i n th e previou s exercise , onl y
proper rotation s shoul d b e considered. )
E x e r c i s e 1 1 9 . Ho w man y differen t hexagon s ca n b e inscribe d int o a
regular 1 5-gon ? (Tw o figure s ar e equal , i f the y coincid e afte r a
plane movement , no t necessaril y proper. )
6. Invariant s
T h e proble m abou t necklace s ca n b e solve d b y a simpl e direc t argu -
ment, withou t referrin g t o grou p theor y an d Burnside' s formula . T h e
most natura l solutio n ca n b e state d a s follows .
6. Invariant s 149
The thre e blac k bead s brea k th e circl e int o thre e parts , whic h
contain ra, n an d k white beads. Th e integers ra, n an d k ar e betwee n
0 an d 7 , inclusive , an d satisf y ra - f n + k = 7 . Not e tha t th e orde r i n
which thes e thre e number s appea r i s i n ou r cas e irrelevant , becaus e
rotations an d reflection s produc e an y o f the 6 possible permutations .
(For th e cas e o f 4 blac k bead s thi s observatio n woul d n o longe r b e
true!) Therefore , w e ca n assum e tha t ra < n < £; , an d th e proble m
is reduced t o th e enumeratio n o f al l triple s o f integers tha t satisf y al l
oo
the state d restrictions . Al l suc h triple s ca n b e foun d directly . Her e
they are , i n lexicographi c order : (0,0,7) , (0,1 ,6) , (0,2,5) , (0,3,4) ,
(1,1,5), (1 ,2,4) , (1 ,3,3) , (2,2,3) .
F i g u r e 7 . Invarian t o f a necklac e
Why i s i t tha t th e tripl e (ra , n, k) ca n serv e fo r th e enumeratio n

of orbits ? Becaus e i t satisfie s th e followin g tw o properties :
• first , i f tw o necklace s ar e th e sam e (belon g t o th e sam e or -
bit), the n th e correspondin g triple s ar e equal ,
• second , i f th e triple s o f tw o necklace s ar e equal , the n th e
necklaces themselve s ar e equivalent .
The firs t propert y ca n als o b e expresse d i n th e followin g way : th e
values o f m , n an d k ar e constan t o n th e orbit s o f th e action .
Consider th e genera l situation . Suppos e tha t th e grou p G act s
on th e se t X.
Definition 25 . A mappin g ip from X int o a certai n se t N i s calle d
an invariant of the action, i f th e value s i t take s o n th e element s o f
the sam e orbi t ar e alway s equal .
Invariants ma y tak e value s i n arbitrar y sets . I n th e previou s

example, th e se t N consiste d o f unordere d triple s o f integers . Th e
least o f these three numbers (denote d abov e by m) i s also an invarian t
of th e grou p actio n unde r study . However , thi s invarian t doe s no t
possess th e secon d property : fo r example , th e tw o necklace s show n
in Figur e 7 are different , bu t th e value s o f m fo r the m ar e th e same .
Such a n invarian t canno t b e use d t o distinguis h differen t orbits .
Definition 26 . A n invarian t ip : X— > N i s sai d t o b e complete, i f i t

takes differen t value s o n differen t orbits .
Exercise 1 20 . Construc t a complet e invarian t fo r necklace s wit h 4
black an d 6 white beads.
Let M b e th e grou p o f al l plan e movements . Thi s grou p act s o n
the plan e i n a natura l way . Thi s actio n i s transitive ; thu s th e plan e
is a homogeneou s spac e o f th e grou p M. Invariant s o f thi s actio n
present n o interest : thes e ar e onl y constan t mappings .
The grou p G als o act s o n th e se t o f al l straight-lin e segment s
in th e plane . Thi s actio n i s no t trivial . Tw o segment s belon g t o on e
orbit, i f and onl y i f their length s are equal. Th e lengt h o f the segmen t
is thus a complet e invarian t o f thi s action .
Exercise 1 21 . Indicat e som e complet e invariant s fo r th e actio n o f
the grou p G o f plan e movement s o n th e se t o f (a ) al l triangles ,
(b) al l quadrangles.
If H C G i s a subgroup , the n i t als o act s o n th e plane . I f H i s
small enough , the n it s orbit s canno t b e big , an d therefor e th e actio n
might hav e nontrivia l invariants . Fo r example , i f H i s th e grou p
of rotation s aroun d a poin t A, the n it s orbit s ar e circle s centre d a t
A. Th e distanc e o f a poin t fro m A i s th e complet e invarian t o f thi s
action. I n th e pola r syste m o f coordinate s wit h centr e A (se e p. 33) ,
the distanc e i s th e pola r coordinat e r , an d ever y invarian t ha s th e
form / ( r ) , i.e. , i s a functio n o f th e complet e invariant .
Consider th e actio n o f th e dihedra l grou p D 3 o n th e plan e (se e
Figure 1 ) . Le t O b e th e pola r centr e an d OM th e pola r axis . Th e
polar distanc e r i s an invarian t o f this action. Bu t i t i s not a complet e
invariant. T o mak e i t complete , w e wil l ad d on e mor e functio n t o
r. Not e tha t th e functio n cosip i s a n invarian t o f th e grou p D3 .
7. Crystallographi c group s 151
Indeed, th e grou p i s generate d b y th e reflectio n ip —i > — ip and th e

rotation (p *—> (p + 1 20° , an d th e expressio n cos3</ ? doe s no t chang e
under thes e transformations . Th e pai r o f number s (r , cos 3ip) makes
a complet e invarian t o f th e action . Indeed , i t i s eas y t o chec k tha t
the simultaneou s equation s
cos Sep = 6 ,
r— c
for an y rea l c > 0 an d |6 | < 1 may hav e 3 or 6 solution s tha t corre -
spond t o th e point s o f on e orbit .
7. Crystallographi c group s
We now have all the techniques necessary to revisit the question abou t
the symmetr y o f ornament s pose d i n th e introductio n (se e pag e 4) .
The symmetr y o f ornament s — plan e pattern s infinitel y repeate d i n
two or mor e differen t direction s — is described b y the so-calle d plane
crystallographic groups. A n exampl e o f suc h a grou p i s th e rollin g
group o f the equilateral triangl e studie d i n Problem 2 7 — it describe s
the symmetry of the ornament show n in Figure 4b in the Introduction .
Crystallographic group s ar e als o referre d t o a s wallpaper groups.
The exac t definitio n read s a s follows .
Definition 27 . A crystallographic group is a discret e grou p o f plan e

movements tha t ha s a bounde d fundamenta l domain .
We will explai n th e tw o term s tha t appea r i n thi s definition .
Definition 28 . A group of plane movement s G is said t o b e discrete,

if every orbit i s a discrete set i n the plane, i.e., fo r ever y point A ther e
is a disk centred a t A an d containin g n o other poin t o f the same orbit .
A simpl e exampl e o f a discret e grou p i s provide d b y th e cycli c

group generate d b y on e translation . O n th e contrary , th e grou p con -
taining tw o translation s wit h collinea r vector s o f incommensurabl e
lengths i s no t discrete , becaus e th e orbi t o f ever y poin t A i s a dens e
subset o f th e lin e passin g throug h A i n th e directio n o f th e transla -
tions.
Exercise 1 22 . Prov e that the group generated by a rotation through

a degree s is discrete if and onl y if the number a i s rational.
Exercise 1 23 . Prov e that th e stabilizer o f any point wit h respect t o
a discret e group of plane movements is finite.
The second notion that need s explanation i s that o f a fundamental
domain.
Definition 29 . A domain 1 F i s said t o b e fundamental for the group
G, i f
• an y poin t i n th e plan e belong s t o th e orbi t o f som e poin t
x G F (whic h ca n als o b e a boundar y point) , an d
• n o tw o differen t inne r point s o f F belon g t o th e sam e orbit .
These tw o propertie s mea n tha t th e image s o f th e domai n F un -

der th e grou p transformation s ar e al l distinc t (wit h th e exceptio n o f
boundary points ) an d fil l th e plan e withou t overlapping . Anothe r
wording i s that w e have a tiling, o r tessellation, o f the plan e b y copie s
of th e figure F. Fo r example , th e rollin g grou p o f th e equilatera l tri -
angle (Proble m 27 ) i s crystallographic, an d th e initia l triangl e ca n b e
chosen a s its fundamental domain . Th e assertion claimed i n the state -
ment o f this proble m i s exactl y th e secon d propert y i n th e definitio n
of a fundamenta l domain .
Exercise 1 24 . Fin d the fundamental domain s for the groups C n an d
The term "crystallographic " ha s its origin in the fact tha t discret e
groups o f space movement s ar e use d t o describ e th e symmetr y o f
natural crystals . Ther e exist s a specia l universa l syste m o f notatio n
for th e crystallographi c groups , bot h plan e an d spatial . Fo r example ,
the rolling group of the equilateral triangl e i s traditionally denote d b y
p3ml. W e will giv e som e mor e example s o f crystallographi c group s
and correspondin g ornaments .
The simples t suc h group , denote d b y pi , i s th e grou p generate d
by two translations b y non-collinear vector s a and b . Figur e 8 a shows
the generator s o f th e grou p an d th e orbi t o f a point .
It woul d t a k e som e effor t t o giv e a n exac t meanin g o f t h e notio n o f domain .

However, i t i s saf e t o replac e t h e wor d "domain " b y "polygon " everywher e i n thi s
section.
7. Crystallographi c group s
f
153
-» •
M
(§> / <$ > : (§ )
/ D / D •
D • D • J) •
(h • ( $ > ; (s>;
F i g u r e 8 . Th e simples t crystallographi c grou p
As a fundamenta l domain , on e ca n tak e th e parallelogra m wit h

sides a an d b . Th e fundamenta l domai n ca n b e chose n i n a variety of
ways. Figur e 8 b show s severa l differen t fundamenta l domain s fo r th e
same grou p p i . Tw o o f thes e domain s ar e parallelograms , whil e th e
third on e i s a hexagon . Th e side s o f th e parallelogra m ar e b — a , b
and b , 2 b — a, respectively .
Exercise 1 25 . Prov e that th e parallelogra m wit h side s ks. -f lb an d
raa + nb, where k, I, ra, n ar e integers, i s a fundamental domai n
if and onl y if \kn — lm\ = 1 .
The grou p pi , simples t amon g al l ornamenta l groups , describe s
the purel y translationa l symmetr y o f an ornament . A n ornamen t ha s
symmetry pi , i f i t ha s n o symmetrie s othe r tha n translations . I t i s
very eas y t o inven t suc h a n ornament . Al l yo u hav e t o d o i s dra w
an arbitrar y figure wit h a trivia l symmetr y group , lyin g insid e th e
fundamental parallelogram , an d conside r th e unio n o f al l th e copie s

of thi s figure obtaine d b y paralle l translation s o f th e give n grou p
(Figure 8c) .
If the chose n figure lies strictly insid e the parallelogram , the n th e
ornament obtaine d ha s symmetry group exactly equal to pi. However ,
if yo u allo w th e figure t o touc h th e border , the n th e ornamen t ma y
have a wide r symmetr y group . A n exampl e o f thi s phenomeno n i s
shown i n Figur e 8d .
The grou p p i i s importan t no t onl y becaus e i t i s th e simples t
crystallographic grou p i n th e plane , bu t als o becaus e o f the followin g
fact.
Lemma 1 . Every plane crystallographic group contains a sub-

group of type pi, i.e., generated by two non-collinear translations.
Proof. A crystallographi c grou p mus t contai n a t leas t on e transla -

tion, a s otherwis e i t wil l reduc e t o a finite group .
Suppose tha t G is a discret e grou p o f plane movement s suc h tha t
all the translations belongin g to G have the same direction (say , paral-
lel to th e lin e I). W e are goin g to prov e that th e fundamenta l domai n
of th e grou p G canno t b e bounde d i n thi s case .
Let u s first conside r wha t movements , othe r tha n rotations , ca n
belong to the group G. Not e first that th e axe s of all glide symmetrie s
that belon g t o G mus t als o b e paralle l t o th e lin e I — becaus e th e
square of a glide symmetry i s a translation. Th e rotations that belon g
to th e grou p ca n onl y b e rotation s throug h 1 80° , because, i f R^ G G
and y> -=f=- 1 80° , then, togethe r wit h every translation T a e G the grou p
also contains th e translatio n R?oT aoR~v, non-collinea r wit h T a (se e
the answe r t o Exercis e 58) . Th e compositio n o f two rotations b y 1 80 °
is a translatio n alon g th e lin e tha t connect s thei r centres . Therefore ,
the centre s o f al l rotation s tha t belon g t o G mus t li e o n on e lin e
parallel t o I. Withou t los s o f generality , w e ca n assum e tha t I is th e
line that passe s throug h th e centre s o f al l rotations. Finally , w e leave
it t o th e reade r a s a n exercis e t o find ou t wha t kin d o f reflections ou r
group ma y contain .
Exercise 1 26 . Prov e that th e grou p G may contain onl y reflection s

whose axes are perpendicular t o I or coincide with I.
Prom al l the observation s made , w e can conclud e tha t tw o point s

can belong to one orbit onl y if their distance s from th e line I are equal.
The fundamenta l domai n mus t contai n on e point o f each orbit; there -
fore i t canno t b e bounded , an d th e grou p G i s non-crystallographic .
•
We have thus proved that ever y ornamental grou p G contains tw o
non-collinear translations , an d henc e a subgroup o f type pi tha t the y
generate. I n fact , a stronge r assertio n holds .
Lemma 2 . The set of all translations belonging to G is a group

of type pi.
This lemma is an immediate consequence of the following exercise .

Exercise 1 27 . Prov e that ever y ornamental group that consist s only
of translations i s generated b y two non-collinear translations .
In a certain sense , an y ornamenta l grou p i s reduced t o th e grou p

pi an d a finite grou p o f plan e movements . W e wil l explai n how . Le t
G b e a n arbitrar y ornamenta l group . Denot e b y H it s subgrou p o f
translations (w e alread y kno w tha t i t i s o f typ e pi) . Th e subgrou p
H i s norma l i n G , becaus e th e conjugat e o f a translatio n b y an y
movement i s alway s a translatio n (se e page 234) .
Lemma 3 . For any crystallographic group G, the quotient group

G/H is finite and may only belong to one of the ten types C n, D n,
where n = 1 ,2,3,4,6 .
We are no t goin g to prov e thi s fac t now . Th e reade r wil l verify i t

later, usin g th e tabl e o f plane crystallographi c group s (Exercis e 1 30) .
The typ e o f th e grou p G/H i s referred t o a s th e ornamental class o f
G.
Let u s now discus s th e relatio n betwee n th e group s G an d H an d
their fundamenta l domain s i n mor e detail .
If $ i s a fundamenta l domai n o f G an d g\ , ... , g^ i s the complet e

set o f representatives o f the coset decompositio n G/H, the n th e unio n
n = Ua*
forms a fundamental domai n fo r H. Th e domai n i s called th e motif
of the ornament , an d th e domai n II , it s elementary cell. I n principle ,
$ an d gi ca n alway s b e chose n i n suc h a wa y tha t I I become s a
parallelogram, bu t sometime s i t i s mor e convenien t t o us e polygon s
of anothe r shape , notably , regula r hexagons , a s th e elementar y cell .
The rati o o f th e are a o f I I t o th e are a o f $ i s equa l t o th e inde x
of H i n G , i.e. , th e numbe r o f coset s i n G/H. Th e bigge r G/H, th e
smaller th e fundamenta l domai n $ i n II .
Problem 48 . Find the motif and the elementary cell of the ornament
that has symmetry group of type p3ml. Describe the cosets of G with
respect to its subgroup of translations H.
Solution.
F i g u r e 9 . A n ornamen t wit h symmetr y grou p p3ra l

Look a t Figur e 9a , whic h show s a n ornamen t wit h

symmetry grou p p3ral, i.e. , the group generated b y thre e
reflections i n the sides of an equilateral triangle. (Instea d
of Figure 9a , w e could a s well use Figur e 4 b from th e In -
troduction.) Th e ornamen t i s obtaine d b y rollin g ove r
the triangl e MNB (th e motif) o n th e plane . Yo u ca n
see fro m th e pictur e tha t th e ornamen t admit s transla -
tions b y vector s tha t connec t th e point s A y B an d D .
As th e elementar y cel l w e ca n choose , fo r example , th e
parallelogram ABCD, whos e side s AB an d AD gener -
ate th e subgrou p o f translations H. I t i s easy to se e tha t
SABCD '• SMNB = 6 : 1 . I t i s impossibl e t o divid e th e
parallelogram ABCD int o six equilateral triangle s whic h
are th e fundamenta l domain s o f the ornamen t (althoug h
it i s possibl e t o divid e i t int o si x fundamenta l domain s
of anothe r typ e — tr y t o d o this!) . I n thi s case , i t i s
more convenient to choose the fundamental domai n for H
having th e shap e o f a hexagon , fo r exampl e BQCLDM,
which is naturally divide d int o six fundamental triangles .
To find th e numbe r o f coset s i n G/H, le t u s not e
that th e element s o f th e subgrou p H d o no t chang e th e
relative positio n o f th e moti f i n th e plan e (th e directio n
and th e orientatio n o f th e "leg") . Differen t movement s
that belon g t o th e sam e coset , sa y g o h\ an d g o /i2 ,
hi, h,2 G H, chang e th e relativ e positio n o f th e moti f i n
the sam e way , becaus e hi an d h^ d o not chang e i t a t all .
In th e picture , yo u ca n see si x differen t position s o f th e
motif; henc e th e numbe r o f the coset s i n G/H i s six. W e
can numbe r the m fro m 1 to 6 , a s in Figure 9b . Then , fo r
example, al l th e movement s belongin g t o th e cose t gH,
where g G G is the reflectio n i n a vertical line , induce th e
following permutatio n o f thes e numbers : 1 <-» 6, 2 <- » 5,
3^4.
We have thus arrived a t the following conclusion : th e
quotient grou p G/H act s o n th e se t o f motif s containe d
in th e elementar y cel l i n th e sam e wa y a s th e grou p D%
acts on the set of vertices of an equilateral triangle : ther e
are thre e rotations , includin g th e identity , an d thre e re -

flections. W e can write the resul t a s follows: G/H = D3 .
In th e terminolog y introduce d above , thi s fac t ca n als o
be state d a s follows : th e ornamenta l clas s o f G i s D3.
Note, finally, tha t writin g pSml/pl = D 3 i s not cor -
rect, becaus e ther e ar e man y subgroup s o f typ e pi con -
tained i n th e grou p o f typ e p3ml.
Exercise 1 28 . Wha t i s the order o f the quotient grou p G/K, wher e

G is the group just studied , an d K C H i s the subgroup of trans-
lations generate d b y AC an d AR (se e Figure 9a) . Describ e th e
structure of this group; in particular, find whether it is isomorphic
to one of the groups C n, D n •
How many differen t subgroup s o f types pi ar e ther e i n th e grou p
of plane movements MP. O f course, an infinite number : th e choice of a
specific grou p i s determine d b y tw o basi c vector s a an d b . However ,
any tw o suc h group s ar e isomorphic . I n fact , a stronge r assertio n
holds: an y tw o such group s ar e conjugat e t o eac h othe r vi a a suitabl e
linear transformatio n (se e pag e 1 68 ) o f th e plane : i f H i s generate d
by translation s T a an d T b an d K b y translation s T c an d Td , the n
LHL~l = K, wher e L i s a linea r transformatio n suc h tha t L(a ) = c
and L(b ) = d .
Definition 30 . Tw o groups of plane movement s ar e said t o b e equiv-

alent, i f the y ar e conjugat e t o eac h othe r b y a suitabl e linea r trans -
formation.
We ca n no w stat e th e theore m tha t give s th e exac t meanin g t o

the statemen t tha t ther e ar e 1 7 types o f wallpaper symmetry .
Theorem 1 2 (Fedorov—Schoenflies) . Up to the equivalence for-

mulated above, any plane crystallographic group is equivalent to one
of the 1 7 groups given in the table below. These 1 7 groups are not
isomorphic to each other.
Outline o f th e proof . Th e proo f i s no t ver y difficult , bu t rathe r

lengthy. W e ar e onl y givin g a n outline , leavin g th e detail s t o th e
industrious reader .
(1) Prov e tha t th e onl y possibl e rotation s i n a crystallographi c

group ar e o f orde r 2 , 3 , 4 or 6 .
(2) Le t G + C G b e th e subgrou p o f al l prope r (orientatio n pre -
serving) movement s i n th e grou p G . The n G + i s a norma l
subgroup o f inde x 1 or 2 .
(3) I f G + = G , i.e . th e grou p consist s onl y o f translation s an d
rotations, the n i t i s equivalen t t o on e o f th e group s pi, p2,
p3, p4 , p6 , dependin g o n th e bigges t orde r o f a rotatio n i t
contains.
(4) I f G + 7 ^ G, the n G is generated b y th e subgrou p G + , whic h
belongs t o on e o f the five types liste d above , an d on e move -
ment / fro m G \ G + . Considerin g th e variou s possibilitie s
that ma y aris e ( / i s eithe r a reflectio n o r a glid e reflection ,
its axis may pass or not pas s through the centres of rotation s
etc.), w e establish tha t
(a) I f G + = pi, the n G = pm , pg o r era .
(b) I f G + = p2 , the n G = prara , pmg, pgg o r crara .
(c) I f G + = p3 , then G = p31 m o r p3ml.
(d) I f G + = p4 , the n G = p4r a o r p4g.
(e) I f G + = p6 , then G = p6ra .
•
Now, th e table . Fo r ever y group , th e tabl e o f plan e crystallo -
graphic group s include s (lef t t o right) :
• Smbl: Th e canonica l crystallographi c notatio n o f the group .

• Symmetries: A n elementar y cel l (eithe r a squar e o r a regu -
lar hexagon ) wit h th e symbol s fo r th e movement s containe d
in th e grou p ( a soli d lin e mean s th e axi s o f a reflection , a
dashed lin e mean s th e axi s o f a glid e symmetry , an d th e
symbols O , A , O , # designat e th e centre s o f rotatio n o f
orders 2 , 3 , 4 and 6) .
• Sample: A sample ornament wit h this symmetry group. Th e
sample show s onl y on e cel l o f th e ornament . Th e ornamen t
is obtaine d fro m th e elementar y cel l b y translation s i n tw o
non-collinear directions . Insid e th e elementar y cell , a fun -

damental domai n i s hatched .
• Generators and relations: A se t o f generator s an d definin g

relations o f th e group .
In th e table , fo r group s numbe r 1 -1 2 , w e giv e a representativ e

with a n elementar y cel l i n th e for m o f a square , an d fo r group s 1 3 -
17, i n th e for m o f a regula r hexagon . Not e tha t th e cel l ca n b e a n
arbitrary parallelogra m fo r group s pi an d p2 , a n arbitrar y rectangl e
for group s pm, pg, pram, pmg, an d pgg, an d a n arbitrar y rhombu s
for group s cm, cmm.
Using thi s table , th e reade r ca n determin e th e symmetr y typ e of
any ornament , startin g fro m th e wallpape r desig n o n th e wall s o f hi s
or he r room .
Table o f plan e crystallographi c group s
Smbl Symmetries Sample Generators and relations
Non-collinear
Pi translations I \ , T 2
TiT2 = T 2T1
6
Half turn s Ri , R 2, # 3
p2 R^ = R 2 = i? 3 := id,
!.M>>, 1
o
(RiR2R3)2 = i d
6
Reflections Si, S2 an d
!|
translation T
pm
SiT = TSuS 2T = TS 2,
iililj 5? = Si = id
Parallel glid e
reflections C/i , C/2
Pg
Reflection 5
and glid e reflectio n U
cm
S2 = id , SU 2 = U 2S
Reflections i n th e side s
of a rectangle Si , 52 , S3,
pmm <p rt rt
Q2 _ o 2_ Q 2_ _ <? 2 _
D D
l— ° 2— 3~ ° 4—
6 (!) 6
id, (SiS 2 ) = (S 2 S 3 ) 2 =
2
(S3S4)2 = (S 4 Si) 2 = id
Reflection S an d centra l
O O O
1
1
1
1
1
1
symmetries R\, R2
1 ! 1
pmg 1i 1
=
11 1
0Q 0 D= R^ — -LI2 id )
R1SR1 = R2SR2
Perpendicular glid e
reflections C/i , C/2
(CW2) 2 (C/f1^)2 =
id
Reflections Si , S 2 an d
central symmetr y R
cmm S2 = S | = i? 2 = id ,
(S1S2)2 = (S 1 RS2R)2 =
id
162 5. Orbits an d Ornament s

<> o < >
-1
|lr I ! Central symmetry # an d
•
p4 <j> O <j >
90° rotatio n R 1
R? = R\ = (RiR) 4 = i d
 o - 6
i_ Reflections Si , 52 , S 3
^- < KX T in th e side s o f a n isosce -
les righ t triangl e
p4m ri / C 7f
[^j! ui —
(SlS2) =
-O o
2
=
(S2S3)
(SsSi)4 = i d
=
^ 3 -~ - 1 0- ,
4
=
P4g
pFj j i Reflection 5
and 90 ° rotatio n R
2
S2=R*=(R-iSRS) = id
<^::.J..!\<1.!...::^>
Three rotation s i?i , #2 ,

\/
i? 3 throug h 1 20 °
p3
i?l = R1 R2R3 = i d
..A..^
Reflection 5 an d rota -
tion R throug h 1 20 °
p31m i\/ A \/ i RS = S 2 =
A. /X / \ A (R^SRS)3^ i d
Reflections Si , 52 , S 3
„' i V in the sides of an equilat -
eral triangl e S 2 = S f =
IN-* î
4^
p3ml
S | = id ,
(Si5 2 ) 3 = (S 2 S 3 ) 3 =
(S 3 Si) 3 = i d
7. C r y s t a l l o g r a p h i c g r o u p s 163
p6
yy r<h Half tur n i ? an d
120° rotatio n R 1
og
AA
j6
^ 1 E2 = Rl = (flii*) 6 = i d
V
Reflections Si , S2 ,
S3 i n th e side s o f a
YXJ\': AJ^Y (30°,60°,90°) triangl e
p6m ^
*^l : = : *^ 2 == 3 ~ =
'
; =;\!%i
^4r' ^ (5!5 2 ) 2 = (S 2S3f =
(5 3 5x) 6 = i d
E x e r c i s e 1 2 9 . Fin d th e symmetr y group s o f th e ornament s show n
in Figur e 4 (pag e 4 ) and Figure 8d.
E x e r c i s e 1 3 0 . Determin e th e ornamental clas s of every grou p i n the
table, an d thus prov e Lemm a 3 (p. 1 55) .
E x e r c i s e 1 3 1 . Tr y to guess th e meaning o f the letters an d number s
used i n the notation o f crystallographic groups .
Chapter 6
Other Type s o f
Transfer mat ions
The mai n protagonist s o f the boo k — transformation group s — hav e

so fa r appeare d i n th e particula r cas e o f group s o f plan e movements .
In th e presen t chapter , w e ar e goin g t o discus s othe r type s o f plan e
transformations: affin e an d projectiv e transformations , similitude s
and inversions . Al l thes e transformation s ca n b e describe d b y frac -
tional linea r function s o f eithe r tw o rea l o r on e comple x argument .
1. Affin e transformation s
Affine transformation s constitut e a n importan t clas s o f plan e trans -
formations whic h i s a natura l generalizatio n o f movements . I n fact ,
the grou p o f plan e movement s A4 i s a subgrou p o f th e affin e grou p
Aff(2,R). Th e transitio n fro m movement s t o affin e transformation s
is easil y achieve d i n coordinates .
Problem 49 . Find a description of plane movements in Cartesian

coordinates.
Solution. I f Oxy i s a syste m o f Cartesia n coordinate s

in th e plan e (whic h w e wil l cal l th e 'old ' system) , the n
its imag e unde r a movemen t / i s anothe r Cartesia n co -
ordinate syste m 0\Xiyi, calle d th e 'new ' system .
166 6. Othe r Type s o f Transformation s
If a poin t A ha s coordinate s (p , q) i n th e ol d coordi -

nate system , the n it s imag e A! ha s the sam e coordinate s
(p, q) i n th e ne w system .
Figure 1 . Movemen t i n coordinate s
For an orientation-preserving movemen t (translatio n

or rotation) , i t i s eas y t o find th e coordinate s (p r,q') o f
point A' i n th e ol d system , usin g Figur e 1 :
J p' = p cos a — q sin a - f po,

(38)
\q f= p sin a + q cos a - h<?o»
where a i s the angl e betwee n th e ray s Ox an d 0\X\.
For a n orientation-reversin g transformatio n w e ge t
similar formula s wit h q changed t o —q:
J P f — pcosa + qsina +po >

(39)
\ q' = p sin a — q cos a + g o •
Affine transformation s ar e give n by a formul a simila r t o (38 ) an d

(39), where the coefficient s ma y be arbitrary numbers , no t necessaril y
sines an d cosines .
1. Affin e t r a n s f o r m a t i o n s 167
D e f i n i t i o n 3 1 . A n affine transformation of the plane i s a transfor -

mation t h a t take s a poin t (x,y) t o th e poin t (x'.y') accordin g t o th e
equations
ax + by + XQ ,
(40)
cx + dy + y 0.
Formulas (40 ) mak e sens e fo r an y value s o f th e coefficient s a , 6 , c ,

d. However , i f w e w rant t o obtai n a genuin e (one-to-one ) transforma -
tion o f th e plane , w e mus t suppos e t h a t ad — be (th e determinan t o f
the matri x ( achd)) i s differen t fro m zero . Indeed , basi c vector s (1 ,0 )
and (0,1 ) ar e taken , b y th e transformatio n (40) . int o vector s (a,c)
and (b.d), an d th e are a o f th e parallelogra m constructe d o n thes e
two vector s i s equa l t o ad — be.
Under a n affin e transformation , paralle l line s g o int o paralle l lines ,
but th e angle s ar e no t preserved : a squar e ma y becom e a n arbitrar y
parallelogram. Figur e 2 shows a n exampl e o f a n affin e transformatio n
11 / 2
with th e matri x equal t o
0 1
Figure 2 . A n affin e transformatio n
D e f i n i t i o n 3 2 . Th e grou p o f affin e transformation s o f th e plane ,

denoted b y Aff(2,E) , consist s o f al l affin e transformation s (40 ) wit h
ad - be ^ 0 .
The grou p o f affin e transformation s act s transitivel y o n th e plane .

The stabl e subgrou p o f th e origi n O i s th e group of linear transfor-
mations G L ( 2 , R ) .
Definition 33 . A linear transformation of the plane i s a transfor -

mation give n i n Cartesia n coordinate s b y th e equation s
f x' = ax + by,
\ y' — ex + dy.
The group GL(2, R) consists of all such transformations wit h ad—be /
0.
Exercise 1 32 . Le t R 2 denot e the group of plane translations. Prov e
the isomorphis m Aff(2,R)/IR 2 2* GL(2,R) .
The fundamenta l propert y o f affin e transformation s i s tha t the y
preserve the ratio of points on straight line s (see page 1 6) : i f a point C
divides a segmen t AB i n th e rati o k : Z, then it s imag e C wil l divid e
the correspondin g segmen t A'B' i n th e sam e rati o k : l. I n fact , on e
can prov e tha t Aff(2,R ) coincides wit h th e se t o f al l transformation s
of th e plan e tha t tak e straigh t line s int o straigh t line s an d preserv e
the rati o o f point s o n ever y line .
Affine transformation s ar e usefu l fo r th e solutio n o f geometri c
problems wher e th e statemen t i s invarian t unde r affin e transforma -
tions, bu t th e solutio n i s easier fo r som e specia l case o f th e construc -
tion.
To tak e a simpl e example , conside r th e propert y o f th e median s
that w e talked abou t i n Chapte r 1 : the three medians in any triangle
meet in one point and this point divides each of them in the ratio 2 : 1
(see Exercise 1 0) . B y a n affin e transformation , th e give n triangle ca n
be reduce d t o a n equilatera l one , fo r whic h th e assertio n i s evident .
Exercise 1 33 . Fin d anothe r solutio n o f Problem 4 (page 1 9) , using
affine transformations .
The notion s of linear an d affin e transformation s mak e sense in the
one-dimensional case , too. Linea r transformations o f the line have the
form n - > o x , whil e affin e transformation s ar e describe d b y th e for -
mula x— i > ax + b. Th e correspondin g group s ar e distinguishe d b y th e
condition o ^ O an d denote d b y GL(1 ,R ) an d AfT(l,R) , respectively .
Instead o f real numbers R we can conside r residue s over a prime num-
ber, thu s arrivin g a t finite groups . On e can als o use complex number s
instead o f real : th e correspondin g group s wil l com e u p late r i n thi s
chapter (se e sectio n 3) .
2. P r o j e c t i v e t r a n s f o r m a t i o n s 169
E x e r c i s e 1 3 4 . (a ) Ho w man y element s ar e ther e i n th e grou p G =

GL(2,Z2)? Amon g th e group s tha t w e considere d earlier , fin d a
group isomorphi c t o G. (b ) Th e sam e question s fo r th e grou p
Aff(l,Z 3 ).
2. Projectiv e transformation s
T h e notio n o f projectiv e transformation s come s fro m dail y life .
I
H ->- ® I ^.-- \ aoaao/
Figure 3 . Photograph y a s projectio n
Prom th e mathematica l poin t o f view, photography , a s wel l a s stil l

life drawing , i s a perspective transformation, o r a central projection.
Photography take s ever y poin t A o f the give n objec t int o th e poin t A'
where th e lin e AO (O bein g th e optica l centr e o f th e camera ) meet s
the plan e o f th e fil m (Figur e 3) . T o mak e a drawin g o f nature , th e
artist doe s essentiall y th e sam e thing , wit h th e differenc e t h a t th e
plane o f th e canvas s i s place d between th e objec t an d th e eye .
It i s clea r t h a t unde r suc h transformation s straigh t line s g o int o
straight lines . Henc e i t i s possibl e t o stud y th e perspectiv e transfor -
mations o f a line , too .
D e f i n i t i o n 3 4 . Suppos e t h a t I an d V ar e tw o line s i n th e plane ,

and 5 i s a fixe d poin t i n th e sam e plan e (Figur e 4a) . A perspective
transformation i s a mappin g p : I — • V t h a t take s ever y poin t A G I
into th e intersectio n poin t A! o f th e line s SA an d V. Give n tw o
planes H i , II 2 an d a poin t S i n space , on e ca n defin e a perspectiv e
transformation p : IIi— > II 2 i n a simila r wa y (se e Figur e 4b) .
D e f i n i t i o n 3 5 . A projective transformation of a line or a plane into

itself i s a compositio n o f severa l perspectiv e transformation s wher e
auxiliary line s o r plane s ar e used .
Figure 4 . Perspectiv e transformatio n o f line s (a ) an d plane s (b )
We begi n th e discussio n o f th e propertie s o f projectiv e transfor -

mations wit h th e followin g question . Suppos e tha t w e take a pictur e
of a serie s o f equidistan t object s arrange d alon g a lin e (e.g. , tree s
along a road) . I t i s clea r tha t th e image s o f thes e point s o n th e pic -
ture d o not nee d to be equidistant. I t i s also clear that the y canno t b e
arbitrary an d ther e mus t exis t a certai n invarian t whic h i s preserve d
by the projectiv e transformations . Suc h a n invarian t mus t depen d o n
more tha n thre e points , becaus e th e distanc e betwee n tw o point s ca n
change an d als o th e mutua l relatio n o f th e thre e point s ca n chang e
arbitrarily: fo r example , i n Figur e 4 a the poin t B lie s between A an d
C, bu t it s imag e B' i s n o longe r betwee n th e respectiv e image s A'
and C
The remarkabl e fac t i s tha t a certai n functio n o f fou r points ,
called thei r cross ratio, o r anharmonic ratio, doe s no t chang e unde r
projective transformation s o f th e line .
Definition 36 . Th e cross ratio o f point s A, B, C an d D i s define d

as follows :
where th e length s o f th e lines , AC, BC, AD, BD ar e considere d a s

signed numbers , positiv e o r negativ e dependin g o n th e orientatio n o f
the give n pai r o f points .
2. Projectiv e transformation s 171
Theorem 1 3 . The cross ratio of four points is preserved under pro-

jective transformations.
Proof. I t i s enough t o prov e th e invarianc e unde r perspectiv e trans -

formations.
To d o so , le t u s expres s th e area s o f th e triangle s tha t yo u ca n
see i n Figur e 5 , usin g tw o differen t formulas :
SASAC = ^h-AC=^SA' SC si n Z AS C,
SASBC = ^h-BC=^SB- SC si n ZBSC,
SASAD = ^h-AD= ]-SA • SD si n ZASD,
SASBD = ]-h-BD = ]-SB • SD si n ZBSD.
F i g u r e 5 . Derivatio n o f th e cros s rati o
Now,
AC AD _ S^SAC . SASAD
BC BD S^SBC S&SBD
SA • SC si n ZASC • SB • SD si n ZBSD
SB-SCsinZBSC -SA-SD sinZASD
sin ZASC si n ZASD
Licensed to Penn St Univ, University Park. Prepared onsin ZBSC
Tue Mar ' EDTsi2019for
12 18:16:11 n ZBSD'
download from IP 132.174.254.159.
We se e tha t th e cros s rati o o f fou r point s i s expresse d throug h

the angle s a t whic h th e correspondin g segment s ar e viewe d fro m th e
centre S. Thes e angle s d o not chang e unde r th e perspectiv e transfor -
mation, an d th e theore m i s proved. •
Problem 50 . Let A, B, C, D be four sequential equidistant trees

along a straight-line road, and let A', B f, C, D' be their images in a
photo. Suppose that the distance A'B' is 6 cm and the distance B'C
is 2 cm. What is the distance CD' ?
Solution. W e hav e
AC AD _ 2 3 _ 4
~BC : ~BD ~ 1 : 2 ~ 3 '
Denoting CD' b y x, w e hav e
A'C ^ A'D' _ 8 m x +8
Wc* : B'D' ~~ 2 :
x + 2'
and fro m th e equatio n
8Q + 2 ) _ 4
2(x + 8 ) " 3
we find tha t x = 1 .
A simila r argumen t ca n b e use d t o deriv e th e genera l formul a

that expresse s projectiv e transformation s i n coordinates . Suppos e
that x i s th e coordinat e o f a variabl e poin t M o n th e lin e / an d x'
the coordinat e o f it s imag e M' G /; . Fi x thre e differen t point s A , B,
C o n Z , denote b y a, b, c thei r coordinate s an d le t a r, &' , c' b e th e
coordinates o f thei r image s A', B' ', C. The n th e relatio n
(A,B;C,M) = (A\B';C',M')
can b e rewritte n a s
c—a x — a d — a' x' — a'
c — b'x — b d — b' x' — br
Prom thi s equatio n w e ca n expres s x' i n term s o f x. Th e resul t
looks lik e
2. P r o j e c t i v e t r a n s f o r m a t i o n s 173
where m , n , p , q ar e certai n constant s dependin g o n a , 6 , c , a' , 6' , c' .

Functions o f thi s kin d ar e calle d fractional linear functions.
Note t h a t i n thi s formul a th e expressio n mq — np (determinan t
of th e matri x ( ™ g) ) mus t b e differen t fro m 0 . Otherwis e th e pai r
( r a , n ) woul d b e proportiona l t o th e pai r (p , g), an d th e fractio n wil l
give on e an d th e sam e valu e fo r al l x.
Now suppos e t h a t ra, n , p , q ar e fou r rea l number s suc h t h a t
mq — np ^ 0. I s i t tru e t h a t th e formul a
. mx - fn
px + q
defines a one-to-on e mappin g o f th e rea l lin e R ont o itself ? T h e

answer i s negative : i n fact , th e poin t x = —q/p (i f p ^ 0 ) ha s n o
image unde r / .
E x e r c i s e 1 3 5 . Indicat e th e rea l numbe r tha t ha s n o inverse imag e

under thi s mapping .
We hav e alread y encountere d thes e difficultie s i n on e particula r

case (se e pag e 1 40) . Th e wa y ou t i s t o introduc e on e mor e poin t o o
(infinity) an d exten d th e actio n o f t h e projectiv e transformatio n t o
the se t1 R = R U oo usin g th e rules :
• - = c o fo r an y a / 0 ,
^ m-o o + n = { j , ifpÔ ,
p • o o -f q ) oo , i f p = 0 .
Fractional linea r function s wit h mq — np ^ 0 defin e one-to-on e

transformations o f th e extende d lin e R .
E x e r c i s e 1 36 . Chec k tha t th e se t o f al l transformation s give n b y

formula (42 ) wit h mq — np ^ 0 forms a group .
This grou p i s calle d the group of projective transformations of the

(extended) real line an d denote d b y P G L ( 1 , R ) .
E x e r c i s e 1 3 7 . Chec k directly , usin g formul a (42) , tha t th e cros s ra -

tio i s a n invarian t o f th e projectiv e grou p actin g o n th e se t o f
quadruples o f points .
The argumen t tha t le d u s t o formul a (42 ) show s tha t an y tripl e

of distinc t point s ca n b e take n int o an y othe r suc h tripl e b y a suit -
able projectiv e transformation . Thi s mean s tha t th e actio n o f th e
projective grou p o n th e se t o f triple s ha s n o nontrivia l invariants .
We have already severa l times considere d th e grou p generate d b y
two projectiv e transformation s x —t > 1 /x an d x > - > l - x (se e Exercis e
74, Proble m 43 , etc.) . Thi s i s no t th e onl y finite subgrou p i n th e
group o f projectiv e transformations .
Exercise 1 38 . Prov e that the two transformations x—
f > 1 /x an d x H->
(x — l)/(x + 1 ) generate a group of eight elements , isomorphic t o
D4.
Exercise 1 39 . Fin d al l projectiv e transformation s o f th e lin e tha t
have finite order.
That's al l abou t projectiv e transformation s o f th e line . No w a

few word s abou t th e plane .
The se t o f al l projectiv e transformation s o f th e plan e i s a grou p
denoted by PGL(2, M). I t contain s the set of all affine transformation s
Aff (2, M.) a s a subgroup .
Exercise 1 40 . I s Aff(2, R) a normal subgrou p of PGL(2, E)?
Arguments simila r t o thos e tha t w e use d fo r projectiv e transfor -
mations o f th e lin e impl y th e followin g theorems .
Theorem 1 4 . Projective transformations of the plane are those and

only those transformations which are described by formulas
aix + biy + c i
(43) a 0x + b 0y + c 0 '
a2x + b 2y + c 2
a0x + b 0y + c 0
in a Cartesian (or affine) coordinate system.
Theorem 1 5 . The group PGL(2,R) acts transitively on the set con-
sisting of all quadruples of points no three of which are collinear.
The last theore m ma y turn ou t t o be quite useful fo r solvin g some

problems i n elementar y geometry . I f a proble m involve s nothin g bu t
the incidenc e between point s an d lines , then we can make a projectiv e
3. Similitude s 175
transformation tha t take s an y give n quadrangl e int o anothe r quad -

rangle, fo r whic h th e solutio n migh t b e easier . Remember , however ,
that projectiv e transformation s chang e no t onl y th e angles , distance s
and areas , bu t als o ratios o f segments o n a lin e an d ratio s o f area s of
different figures. The y onl y preserv e straigh t line s an d cros s ratio .
Here i s a n exampl e o f suc h a n application .
Figure 6 . Pappus' s theorem
Exercise 1 41 . Prov e th e theore m o f Pappu s (se e Figur e 6) : if the

three points A, B, C are collinear and the three points D, E, F
are collinear, then the intersection points AE D BD, AF D CD
and BF n CE are also collinear.
3. Similitude s
Definition 37 . A similitude i s a plan e transformatio n tha t change s
all distance s b y on e an d th e sam e positiv e factor .
Like affme transformations , similitude s constitut e a class of plan e

transformations whic h is wider than the class of movements. I t i s clear
from Definitio n 3 7 that th e se t o f al l similitude s i s a transformatio n
group.
The simples t typ e o f similitudes , differen t fro m movements , i s
provided b y homotheties.
Definition 38 . A homothety H\ with centre A and coefficient k ^ 0

is the transformation tha t take s every point M int o the point M' suc h
that AM' = k • AM (se e Figure 7) .
F i g u r e 7 . Homothet y
The se t o f al l homothetie s o f th e plan e doe s no t constitut e a

group, bu t th e se t o f al l homothetie s wit h a fixed centr e does . Usin g
a comple x coordinat e z , suc h transformation s ca n b e describe d b y
the formul a z —
t > kz, wher e k i s a non-zer o rea l number . Thi s grou p
is thu s isomorphi c t o M* , th e multiplicativ e grou p o f non-zer o rea l
numbers.
Exercise 1 42 . Prov e the isomorphis m GL(2,M)/R * ^ PGL(1 ,R) .
Below ar e som e example s showin g th e us e o f homothetie s i n ele -
mentary geometry .
Problem 51 . In a given triangle ABC, inscribe a square in such a
way that two of its vertices belong to one side of the triangle and the
remaining two vertices lie on the other two sides.
Solution. I t i s ver y eas y t o construc t a squar e wit h
three vertice s satisfyin g th e requirement s o f the proble m
(square KLMN i n Figur e 8) .
Any homothet y centre d a t verte x A preserve s thes e
properties, an d i t remains to find the coefficient k so that
H\ map s th e poin t N int o a poin t E belongin g t o th e
side BC. Th e whole construction i s clear from th e figure.
3. Similitude s 7 7
A L M C
Figure 8 . Inscribin g a square int o a triangl e
E x e r c i s e 1 4 3 . Int o a give n triangle , inscrib e a triangl e whos e side s

are paralle l t o th e thre e give n lines .
One mor e usefu l propert y o f homothetie s i s t h a t the y preserv e
the directio n o f straigh t lines : th e imag e o f a lin e / i s alway s a lin e
parallel t o I. W e wil l us e thi s fac t i n th e followin g problem .
P r o b l e m 5 2 . Several circles are inscribed into one circular segment

(Figure 9).
F i g u r e 9 . Circle s inscribe d int o a circula r segmen t
Let Ai and Bi be the tangency points of the i-th inscribed circle

with the arc and the chord, respectively. Prove that all lines AiBi
pass through a common point.
178 6. O t h e r T y p e s o f T r a n s f o r m a t i o n s
S o l u t i o n . W e wil l prov e t h a t ever y lin e AiBi passe s

through th e poin t C whic h i s th e tangenc y poin t o f th e
line I paralle l t o th e chor d MN an d tangen t t o th e bi g
circle. Conside r th e homothet y h\ wit h centr e A\ an d
coefficient k — OA\ : 0\A\. I t transform s th e smal l
circle 5 i (centre d a t 0\) int o th e bi g circl e S. Therefore ,
the imag e o f th e lin e MN tangen t t o 5 i i s th e lin e / ,
tangent t o S an d paralle l t o MN. Th e poin t f?i , whic h
is th e commo n poin t o f MN an d S± , goe s unde r hi int o
the poin t C , commo n t o I an d S. Th e sam e argumen t
can b e repeate d fo r eac h smal l circl e Si. Thi s complete s
the proof .
E x e r c i s e 1 4 4 . Give n tw o concentri c circles , construc t a lin e whic h

intersects the m i n th e fou r consecutiv e point s A, B, C , D s o tha t
the followin g relatio n hold s fo r th e length s o f th e segment s tha t
are cu t b y th e circles : AB = 2BC = CD.
E x e r c i s e 1 4 5 . Give n a triangle , prov e tha t th e thre e lines , eac h o f

which passes through th e midpoint o f a side parallel to the bisecto r
of th e opposit e angle , mee t i n on e point .
E x e r c i s e 1 46 . Prov e tha t fo r an y triangl e ABC ther e exist s a circl e

that contain s th e midpoint s o f th e sides , th e fee t o f altitude s an d
the midpoint s o f th e segment s KA, KB, KC, wher e K i s th e
intersection poin t o f th e altitudes . (Thi s circl e i s calle d th e circle
of 9 points, o r Euler's circle.)
T h e grou p o f plan e similitude s i s no t exhauste d b y t h e se t o f al l

homotheties.
D e f i n i t i o n 3 9 . A spiral similarity i s define d a s th e compositio n o f a

homothety an d a rotatio n wit h th e sam e centr e (se e Figur e 1 0) .
In thi s book , w e hav e alread y encountere d suc h transformation s

when studyin g comple x numbers : w e sa w t h a t multiplicatio n b y a
number a i s equivalen t t o a homothet y wit h coefficien t \a\ an d sub -
sequent rotatio n throug h th e angl e a r g a aroun d th e origi n (se e (4)) .
Let u s conside r som e geometri c application s o f spira l similarities .
3. Similitude s 179
Figure 1 0 . Spira l similarit y
Problem 53 . Given an arbitrary triangle ABC, draw two triangles

ABP and BQC, lying outside of AABC, having right angles at ver-
tices P, Q and equal angles j3 at the vertex B (see Figure 1 1 ).
Figure 1 1 . Righ t triangle s buil t o n th e side s o f triangl e ABC
Find the angles of APQK, where K is the middle point of the

side AC.
k
Solution. Conside r tw o spira l similarities : Fp i P>
k d
Rdp^ F Q = H^ oR Q, wher e d = 90 ° an d k = PB:PA =
QB : QC. I t i s clea r tha t F P{A) = B an d F Q(B) = C ;
hence (FQ O Fp)(A) = C. Whe n tw o spira l similari -
ties are performed on e after anothe r thei r coefficient s ge t
multiplied, an d thei r rotatio n angle s ar e adde d (w e wil l
explain thi s a littl e later) . Therefore , th e compositio n
F = FQ O Fp mus t b e a rotatio n throug h 1 80° . Sinc e

F(A) = C , th e centr e o f rotatio n i s th e poin t K an d
thus F{K) = K. Le t F P(K) = # i ; the n F Q ( # i ) = # .
Both righ t triangle s KPK\ an d QKK\ hav e th e sam e
angle / ? at th e verte x K\\ henc e the y ar e equa l (Figur e
12).
F i g u r e 1 2 . Produc t o f tw o spira l similaritie s
It follow s tha t PQ 1 KK X an d ZKPQ = ZKQP ••

(1-
In th e previou s argument , w e have use d th e fac t tha t th e compo -

sition o f tw o spira l similaritie s i s a spira l similarit y whos e coefficien t
is th e produc t o f th e tw o coefficients , whil e th e angl e o f rotatio n i s
the su m o f th e tw o rotatio n angles . Thi s fac t i s eviden t i f th e tw o
transformations hav e a common centre . T o prove i t i n full generality ,
we will use calculations with complex numbers, based o n the followin g
theorem.
Theorem 1 6 . A plane transformation is a similitude if and only if,

in the complex coordinate z, it can be written as either
(44) z ^- > pz + a
3. Similitude s 181
or
(45) z >— » pz + a,
where p and a are arbitrary complex numbers, p ^ 0 . T/i e fauo cases

(44) and (45 ) correspond to proper (i.e., orientation-preserving) and
improper (i.e., orientation-reversing) transformations.
Proof. Indeed , recal l tha t w e have alread y prove d i n (8 ) tha t prope r

movements o f the plan e correspon d t o the linea r function s w = pz + a
with \p\ = 1 . Now , suppose tha t F i s a proper similitud e o f the plane ,
i.e., a transformatio n tha t stretche s al l distance s b y a certai n facto r
k an d preserve s th e orientation . Le t H b e the homothet y wit h coeffi -
cient k and centre 0. Th e composition H~ 1 oF preserve s the distance s
and th e orientation ; henc e i t i s a prope r movemen t an d correspond s
to a functio n w = pz + a wit h \p\ = 1 . The n th e transformatio n
F = H o ( i J _ 1 o F) ca n b e writte n a s w = k(pz + a) , whic h i s a n
arbitrary linea r function .
Conversely, give n a comple x functio n pz + a with arbitrar y coef -
ficients, w e ca n verif y tha t i t stretche s th e distance s betwee n point s
by th e facto r k = \p\:
\(pzi + a) - (pz 2 + a)| = |p | • \zi - z 2\.
The cas e o f imprope r transformation s i s reduce d t o th e cas e o f

proper transformation s b y th e simpl e observatio n tha t th e functio n
(45) i s the compositio n o f (44 ) an d th e standar d reflectio n z\-^z. •
In th e terminolog y an d notatio n o f section 1 , Theorem 1 6 mean s

that th e group of proper similitude s i s Aff (1, C), where C is the grou p
of comple x numbers .
Using the descriptio n o f similitudes i n terms of complex numbers ,
we ca n easil y prov e tw o importan t facts , relate d t o eac h other :
• An y transformation o f similitude pz-\-a whic h is not a trans-
lation (i.e. , p ^ l) ha s a uniqu e fixe d point .
• An y transformation o f similitude pz + a which is not a trans-
lation i s a spira l similarit y (i n particular , a homothety) .
182 6. O t h e r T y p e s o f T r a n s f o r m a t i o n s
Indeed, a fixed poin t i s a numbe r ZQ such t h a t pz$ + a = z$. I f

p ^ l , thi s equatio n ha s a uniqu e solutio n ZQ = a/(l—p). T h e formul a
/a \ a
pz + a = p[z- +
shows t h a t thi s transformatio n i s in fac t a spira l similarit y wit h centr e

a / ( l — p), stretchin g coefficien t |p | an d rotatio n angl e a r g p .
Now w e can prov e th e fac t use d i n Proble m 5 3 above. T h e compo -
sition o f tw o spira l similaritie s w = pz-\-a an d u = qw + b correspond s
to th e functio n
u = q(pz + a) + b = pqz - f (aq + b).
This i s a spira l similarit y wit h coefficien t \pq\ = \p\\q\ an d angl e o f

rotation arg(pg ) = a r g p + a r g g .
E x e r c i s e 1 47 . Tw o map s o f th e sam e country , draw n t o differen t
scales o n transparen t paper , ar e pu t o n th e tabl e i n suc h a wa y
that on e o f the map s completel y cover s th e other . Prov e tha t on e
can pierc e bot h map s wit h a pi n i n a poin t tha t correspond s t o
the sam e plac e o n bot h maps .
E x e r c i s e 1 4 8 . Give n fou r point s A, B, C , D i n th e plane , suc h tha t
AB T ^ CD, prov e tha t ther e exist s a poin t E fo r whic h th e tw o
triangles ABE an d CDE ar e similar .
E x e r c i s e 1 4 9 . Point s M , N an d P ar e centre s o f th e square s con -
structed o n th e side s AB, BC, CA o f a n arbitrar y triangl e ABC
outside o f it . Prov e tha t th e segment s NP an d CM ar e perpen -
dicular an d hav e equa l lengths .
4. Inversion s
P r o b l e m 5 4 . A circle S touches two circles S\ and S2 at the points A
and B. Prove that the line AB passes through the centre of similitude
of the circles S\ and S2 .
S o l u t i o n . Le t K b e th e intersectio n poin t o f th e line s

AB an d O1 O2 (se e Figur e 1 3) . W e wan t t o prov e t h a t K
is th e centr e o f similitud e o f th e circle s S\ an d S2 . W e
will construc t th e require d similitud e i n a rathe r indirec t
manner.
4. Inversion s 183
Figure 1 3 . Thre e tangen t circle s
Let / b e the transformation o f the plan e which, wit h

every poin t M , associate s th e poin t M' belongin g t o th e
half-line KM a t a distanc e fro m K tha t satisfie s KM •
KM' = KA-KB = const . Obviously , f(A) = B an d
f(B) = A.
We clai m tha t th e circl e S goe s int o itsel f unde r / .
This follow s fro m a well-know n theore m o f elementar y
geometry (i f you don' t kno w it , tr y t o prov e i t yourself) :
for a fixed circle S, a fixed point K and an arbitrary line
I that passes through K and meets S at the two points L
and V', the product of lengths of the two segments KL
and KL' does not depend on the choice of the line I.
Now tak e a poin t M G S 2 . It s imag e M' lie s o n th e
half-line KM an d satisfie s
KAKB
KM'
KM
Let M i b e the secon d intersectio n poin t o f KM wit h 52 .
By the theorem that w e quoted, KMKM\ — C — const.
Therefore,
KAKB
KM' KMU
C
which mean s tha t M' i s obtaine d fro m M i b y a homo -
thety centre d a t K\ Therefore , th e imag e o f $ 2 unde r
/ i s a circle , sa y S 2. Sinc e S2 passe s throug h B an d i s

tangent t o th e circl e S , th e circl e S 2 passe s throug h th e
point A an d i s tangent i n tha t poin t t o S ; thu s S 2 = Si -
We hav e prove d tha t th e tw o circle s S i an d 5 2 ca n
be transforme d int o on e anothe r b y a homothet y wit h
centre K.
The transformatio n / tha t w e used i n the previou s proble m i s a n

example o f inversion .
Definition 40 . Th e inversion with respect to the circle T with centre
O and radius r is the transformation tha t map s every point M int o the
point M' tha t belong s t o th e half-lin e OM an d satisfie s th e equalit y
OM OM' = r 2.
The inversio n ma p take s th e insid e o f th e circl e outsid e an d th e

outside inside . I t preserve s th e circl e itself . A well-know n jok e o f
H. Petar d (" A contributio n t o th e mathematica l theor y o f bi g gam e
hunting") suggest s th e followin g metho d t o catc h a lion . Th e hunte r
gets int o a cag e an d waits . Whe n th e lio n appears , h e perform s a n
inversion. No w th e lio n i s inside th e cage .
Inversion i s a n almost one-to-on e transformatio n o f th e plane :
it i s define d an d one-to-on e everywher e excep t a t th e centr e o f th e
circle O. Whe n th e poin t M move s toward s O , it s imag e M' move s
infinitely fa r fro m O. Thi s i s why i t i s natura l t o ad d th e poin t
00 ("infinity" ) t o th e plane , similarl y t o wha t w e di d fo r projectiv e
transformations o f th e lin e o n pag e 1 73 , an d conside r inversio n a s a
one-to-one transformatio n o f th e extende d plane .
Suppose tha t ou r plan e i s th e plan e o f comple x numbers . W e
know tha t ever y comple x numbe r z an d it s conjugat e z satisf y th e
relation zz = \z\ 2. Therefore , th e algebrai c formul a fo r a n inversio n
t > r 2/z.
of radiu s r wit h centr e 0 is z—
Exercise 1 50 . Wha t transformation grou p is generated b y the set of
all inversions with a fixed centre 0?
During th e discussio n o f Proble m 54 , we found tha t a n arbitrar y
circle tha t doe s no t pas s throug h th e centr e o f th e inversio n / map s
under / int o a certai n circle .
4. I n v e r s i o n s 185
))
Figure 1 4 . Us e of inversion t o catc h a lion (afte r H . Petard )
E x e r c i s e 1 5 1 . Wha t i s the image , under a n inversion , o f a circle tha t

passes throug h th e centr e o f inversion ?
E x e r c i s e 1 5 2 . Wha t i s th e image , unde r a n inversion , o f a straigh t
line?
All thes e facts , pu t together , mea n t h a t the inversion preserves

the set of all lines and circles. Viewin g a straigh t lin e a s a circl e
passing throug h infinity , w e ca n sa y t h a t inversion s ar e circular trans-
formations, i.e . transformation s t h a t preserv e th e clas s o f al l (gen -
eralized) circles . A littl e late r w e wil l se e t h a t th e se t o f circula r
transformations i s no t exhauste d b y inversions .
Now w e giv e som e mor e application s o f inversion s i n elementar y
geometry.
P r o b l e m 5 5 . Each of four circles touches two of its neighbours (Fig-

ure 1 5a). Prove that the points of contact lie on one circle.
S o l u t i o n . Le t u s appl y a n inversio n wit h centr e A (on e

of th e contac t points ) an d a n arbitrar y radius . W e wil l
Figure 1 5 . Fou r circle s touchin g eac h othe r
see that th e problem become s simpler afte r thi s transfor -

mation.
Denote b y S[ th e imag e o f Si. I t follow s fro m ou r
previous considerations tha t S[ an d 5 ^ ar e straight lines ,
and $ 3 an d S f4 ar e circles . Th e relation s o f tangenc y
between S ^ are the same as between Si, i.e. , S ( i s tangent
to S 2 , S 2 to S' 3, S' s to 54, and S 4 to S[. Not e also that th e
lines S[ an d S f2 must b e parallel, becaus e S i an d S 2 have
only on e commo n poin t A, whic h goe s t o infinit y unde r
the inversion . W e arriv e a t th e configuratio n show n i n
Figure 1 5b . Th e problem is to prove that th e three point s
of contac t B\ C" , D' belon g t o on e straigh t lin e — thi s
will impl y tha t th e invers e image s o f thes e point s B, C ,
D belon g t o a circl e tha t passe s throug h th e centr e o f
the inversio n A.
To prov e tha t B'C'D' i s a straigh t line , le t u s dra w
the commo n tangen t o f th e circle s S 3 an d S' 4 unti l th e
intersection wit h th e line s S [ an d S 2 i n th e point s M
and N. Conside r th e tw o triangle s MC'D' an d NC'B'.
They ar e equilatera l an d hav e equa l angle s /-M — Z.N.
Therefore thei r angle s a t th e verte x C ar e als o equa l t o
5. Circula r transformation s 187
each other . Henc e B'C'D' i s a straigh t line . Th e proo f

is complete .
In problem s 5 4 an d 5 5 we use d th e eviden t propert y tha t i f tw o

lines ar e tangen t t o eac h other , the n thei r image s unde r th e inversio n
are als o tangent . Th e nex t exercis e i s a generalizatio n o f thi s fact .
Exercise 1 53 . Defin e th e angl e betwee n th e tw o circle s a t a poin t
of intersection t o be the angl e made by their tangen t line s drawn
through that point. Prov e that inversion with respect to any circle
preserves angle s between circles.
5. Circula r transformation s
The set o f all inversions in the plane is not a transformation group . I n
this section, we shall study the group generated b y all inversions. Thi s
group is called the group of circular transformations. I t consist s of two
halves: th e subgroup of orientation-preserving transformations , whic h
coincides with the complex projective group PGL(1, C) (se e page 173),
and a cose t consistin g o f orientation-reversin g transformations .
We start wit h a n illustrativ e problem .
Problem 56 . Fix a circle C with centre O. Let A he the midpoint of
its radius OB. Suppose that we are allowed to perform two transfor-
mations: inversion with respect to the circle C and half turn around
point A. What is the maximal number of different points that can be
obtained from a given point by successive applications of these trans-
formations ?
Solution. Le t u s writ e bot h transformation s a s func -
tions o f a comple x variable , assumin g tha t th e poin t O
has comple x coordinat e 0 an d poin t B comple x coordi -
nate 1 .
The point A correspond s t o the numbe r 1 /2 , an d th e
symmetry a t thi s poin t i s described b y th e functio n
h(z) = l-z.
To find th e formul a fo r th e secon d allowe d transfor -
mation, not e tha t point s z an d w tha t correspon d t o
each othe r unde r th e inversio n satisf y th e tw o relation s
\z\\w\ = 1 and arg z = argw . I t follow s tha t w = 1 /z ,

and thu s
Mz) = i/f .
Each o f th e tw o allowe d transformation s i s involu -
tive, and so the only way to obtain differen t composition s
of the tw o function s i s to appl y the m b y turns . Startin g
from z an d applyin g firs t / i , the n /2 , then agai n / i , etc. ,
we obtain th e followin g list : z , 1 — 2, 1 /( 1 — 2), z/(z — 1),
1 - 1 /z , I/2 , z , 1 - z , 1 /( 1 - z) , z/( z - 1 ) , 1 - 1 /z ,
1/z. Afte r thi s we obtain z onc e again , an d th e sequenc e
begins looping . Therefore , th e inversio n an d th e centra l
symmetry generat e a grou p G o f 1 2 elements , an d it s
orbit canno t contai n mor e tha n 1 2 points . A n exampl e
where the orbit contain s exactly 1 2 points is given below.
Exercise 1 54 . Fin d al l the possibilitie s fo r th e numbe r o f points i n

the orbit s o f th e grou p G. Dra w picture s o f differen t type s o f
orbits.
Now le t u s fin d th e fundamental domain o f th e grou p G , i.e. ,
the par t o f th e plan e whos e image s unde r th e grou p actio n cove r
the plan e withou t overlapping . Th e imag e o f th e circl e C unde r th e
transformation f\ i s the circl e C (se e Figur e 1 6) .
The imag e o f C unde r fa i s the lin e MM'. On e mor e importan t
line i s th e straigh t lin e OB whic h separate s th e tw o region s corre -
sponding t o eac h othe r unde r th e comple x conjugatio n z1— > z ( a
mapping belongin g t o ou r group) . Thes e line s divid e th e plan e int o
12 domain s tha t g o int o on e anothe r unde r th e grou p action . Eac h
of thes e domain s ha s th e propert y tha t i t doe s no t contai n interio r
points equivalen t unde r G.
Any of the 1 2 domains (e.g. , domain 1 ) can be taken a s the funda -
mental domai n o f the grou p actio n unde r study . Th e reade r i s invited
to check wha t ar e the image s of domain 1 under differen t transforma -
tions o f th e group .
The unio n o f th e fou r line s draw n i n Figur e 1 6 i s th e se t o f al l
points o n th e plan e lef t invarian t b y som e nontrivia l elemen t o f th e
group. Th e orbi t o f an y interio r poin t consist s o f exactl y 1 2 points .
5. Circula r transformation s 189
F i g u r e 1 6 . Grou p o f comple x function s generate d b y z

\ — z an d z — i > 1 /z.
The orbit s o f th e point s o f th e line s differen t fro m thei r intersectio n

points hav e cardinalit y 6 . Th e intersectio n point s spli t int o thre e
orbits: tw o o f lengt h 3 (1 /2 , 2 , -1 an d 0,1 , co) an d on e o f lengt h 2
(points M an d M' , correspondin g t o th e number s 1 / 2 ± i>/3/2).
The fou r line s show n i n Figur e 1 6 ar e divide d int o 1 8 segment s
by the intersectio n points . Thes e segments go into one another b y th e
group actio n an d spli t int o 3 orbit s o f lengt h 6 , show n i n th e figure
as normal , bol d an d dotte d lines .
Now w e shal l stud y th e geometri c meanin g o f plan e transforma -

tions describe d b y fractiona l linea r function s wit h arbitrar y comple x
coefficients, i.e. , element s o f th e grou p PGL(1 ,C) , a s wel l a s simila r
functions wit h z replace d b y th e conjugat e variabl e z.
Theorem 1 7 . Let a, b, c, d be any complex numbers such that ad —
be ^ 0. Then:
(1) The transformation defined by the function
az + b
w cz + d
(an improper fractional linear transformation) is a compo-
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(2) The transformation defined by the function

az + b
w=
cz + a
(a proper fractional linear transformation) is a composition
of an inversion, a spiral similarity and a reflection.
Proof. I f c = 0 , then the second formula give s a linear functio n a\z+
&i, which , a s w e know , correspond s t o a similitud e transformation .
The firs t formul a give s a±z + &i , a linea r functio n i n z, whic h i s th e
composition o f th e reflectio n z —
i > z an d a similitude .
Suppose no w that c ^ 0 . The n the fractio n — : can be writte n
cz + a
as
az + b ( 1 \
- , , = P ~ — + z Q]+r,
cz + a \ z — Zo J
where ZQ = — d/c, p = (bc — ad)/c2 an d r = a/c — pzQ. The expressio n
inside th e parenthese s i s th e conjugatio n o f th e standar d inversio n
(with centr e 0 and radiu s 1 ) by the translatio n z \— > z -f ZQ ; therefore,
it represent s th e inversio n o f radius 1 centred a t th e poin t zo . T o th e
result o f th e inversion , th e similitud e transformatio n z \— > pz + r i s
applied, an d w e get th e require d composition .
The prope r fractio n (az -f - b)/(cz + d ) i s reduced t o th e imprope r
one b y th e chang e z »- » f, an d w e obtai n th e secon d par t o f th e the -
orem. I t i s funn y tha t i n thi s cas e imprope r transformations , thos e
that chang e orientation , ar e easie r t o handl e tha n th e prope r trans -
formations. Thi s observatio n i s accounte d fo r b y th e importanc e o f
inversions i n thi s contex t — and inversion s ar e imprope r transforma -
tions. •
Exercise 1 55 . Chec k that all (proper and improper) fractional linea r
transformations for m a group , an d th e se t o f proper transforma -
tions is a normal subgrou p o f it. Fin d th e quotien t group .
Theorem 1 7 implies that bot h classe s of fractional linea r transfor -
mations ar e circular, i.e. , the y preserv e th e se t o f generalize d circle s
(circles an d straigh t lines ) i n th e plane . I t i s als o tru e tha t an y cir -
cular transformatio n i s describe d b y eithe r a prope r o r a n imprope r
fractional linea r function . Thi s i s why th e grou p o f al l suc h transfor -
mations i s called the circular group. Anothe r noteworth y propert y o f
6. Hyperboli c geometr y 191
these transformation s i s tha t the y ar e conformal, i.e. , the y preserv e

angles betwee n curves . However , th e clas s o f al l conforma l mapping s
is much wider tha n tha t o f circular transformation s — for example , i t
includes comple x function s P(z)/Q(z), wher e P an d Q ar e arbitrar y
polynomials.
Exercise 1 56 . Prov e tha t al l transformation s o f the comple x plan e
given by formula s
(46) w = — ; , a,6,c,deM , ad - bc> 0 ,
cz + a
and
(47) w =^ ; , a,b,c,d€R, ad-bc<0,
cz + d
form a group.
6. Hyperboli c geometr y
Let u s chec k tha t transformation s (46 ) an d (47 ) ma p th e uppe r half -
plane y > 0 int o itself . I f z = x + iy, w = u + iv, the n a simpl e
calculation show s tha t th e comple x formul a (46 ) i s equivalent t o th e
pair o f rea l formula s
(ax + b) (ex + d) + acy 2
(ex + d) 2 + y 2
(ad — bc)y
(ex + d) 2 + y 2 '
and w e se e tha t v ha s th e sam e sig n a s y. Th e formul a (47 ) i s con -
sidered i n a simila r way .
Let L b e the grou p o f all transformations (46 ) an d (47 ) actin g o n
the uppe r half-plan e H = {(x,y)\y > 0} . Th e half-plan e H i s calle d
the hyperbolic plane, o r th e Lobachevsky plane, an d th e grou p L i s
called th e grou p o f hyperbolic movements o f H. Thi s terminolog y ha s
the followin g meaning .
As we know, the transformation s i n L tak e an y circl e into a circl e
(or a line , whic h w e vie w a s a particula r cas e o f th e circle) . I n th e
plane H, ther e i s a distinguishe d se t o f circle s whic h i s preserve d b y
the grou p L . Thes e ar e th e (half)-circle s an d (half)-line s perpendic -
ular t o th e lin e Ox (se e Figur e 1 7) . W e wil l cal l thes e circle s th e
L-lines, becaus e throug h an y tw o point s o f H ther e passe s on e an d
only on e L-lin e — a propert y owne d als o b y th e se t o f al l usua l line s

in th e usua l plane .
M
\M
F i g u r e 1 7 . L-line s i n hyperboli c plan e
The grou p L o f hyperboli c movements , o r L-movements, ha s

properties simila r t o thos e o f th e grou p o f plan e movement s actin g
on th e usua l plane . I n particular , an y poin t ca n b e take n int o an y
other b y a hyperboli c movement , bu t th e actio n o f L o n th e se t o f
L-segments (arc s o f L-lines ) i s no t transitive . Th e mai n geometri c
difference betwee n th e hyperboli c an d th e usua l plane s appear s whe n
we think abou t paralle l lines .
In ordinar y Euclidea n geometry , tw o line s ar e calle d paralle l i f
they d o no t hav e commo n points , an d th e mai n propert y o f paralle l
lines is that fo r an y line a and an y point A outsid e of a there is exactly
one lin e passin g throug h A an d paralle l t o a . No w loo k a t Figur e 1 8 ,
which show s a n L-lin e / an d a n L-poin t A. Amon g th e fou r line s
drawn throug h A, ther e i s one (I) that intersect s th e lin e a , an d ther e
are thre e (k, n , m) tha t hav e n o commo n point s wit h a (w e recal l
that th e point s o f th e boundar y horizonta l lin e d o no t belon g t o H) .
We thu s se e tha t i n Lobachevsk y geometr y on e ca n dra w man y line s
passing throug h th e give n poin t an d no t intersectin g th e give n line .
Let u s d o a computationa l exercis e i n Lobachevsk y geometry .
The angl e betwee n tw o L-line s i s by definitio n measure d a s the usua l
Figure 1 8 . Mutua l positio n o f tw o L-line s
Euclidean angl e betwee n th e tangen t line s (not e tha t wit h thi s defi -
nition o f th e angl e w e hav e th e propert y tha t L-movement s preserv e
L- angles).
Problem 57 . Find the sum of angles of the Lobachevsky triangle with

vertices A(0J), £(4,3) , if(0,5) .
Solution. Th e given triangle ABK i s the hatched regio n

in Figur e 1 9 .
Figure 1 9 . A triangl e i n th e hyperboli c plan e
The sid e AK lie s on the axi s Oy. Th e sid e KB i s a n

arc o f a circl e wit h centr e a t O . Th e sid e AB i s a n ar c
of th e circl e wit h centr e a t th e poin t M .
Exercise
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coordinate s o f the poin t M.
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The angl e K o f our triangl e i s a right angle , becaus e

it i s formed b y a circl e an d it s radius :
ZK = 90°.
The angl e B betwee n th e tw o circle s i s equa l t o th e
angle betwee n th e tangent s an d henc e t o th e angl e be -
tween th e radii :
ZB = ZOBM.
Similarly,
ZA = ZOMA.
If yo u hav e foun d th e coordinate s o f M , the n yo u
can find tha t
7
tan ZOMA = - ,
tanZOBM = tan(ZBO£ i - ZBMB X) = |- ,

1
+1-§ - 4 3
t a n ( Z ^ + ZB ) = ^ ± 1 ^= ^ .
1 Z 4
" 3 ' 37
Since th e tangen t i s positive , w e infe r tha t ZA +
ZB < 90° . Therefore , th e su m o f the thre e angle s o f th e
triangle ABK i s les s tha n 1 80° .
It i s interesting t o not e tha t th e bigge r th e Lobachevsk y triangl e

(in a certai n sense) , th e smalle r it s su m o f angles . Fo r example , yo u
can chec k that th e isoscele s triangle ABC, whic h i s twice the triangl e
ABK, ha s a smalle r su m o f angles .
Finally, w e wil l giv e on e exampl e o f a crystallographi c grou p i n
the hyperbolic plane — the so-called modular group U that consist s of
all prope r fractiona l linea r transformation s wit h intege r coefficients :
[ az + b | T T 1
U = < \a,b,c,deZ>.
I cz -h a )
This grou p i s generate d b y th e tw o element s
S : z i- > —l/z an dT : zH- V 1 + z.
Exercise 1 58 . Chec k the relations S 2 = (ST) 3 = id.

Figure 2 0 show s th e fundamenta l domai n o f th e grou p U
$ = {z = x + yi\\z\>l, \x\ < \}
and it s image s unde r T , S , T ~ \ TS, ST, etc .
F i g u r e 2 0 . Fundamenta l domai n o f th e modula r grou p
The domai n $ i s in fact a n L-triangle , an d i t ha s a finite are a (w e

have no t define d are a i n Lobachevsk y geometry , s o you canno t chec k
that!). Copie s of<J> cover all the upper half-plan e withou t overlapping .
Thus U i s reall y a crystallographi c group . Th e reade r i s invite d t o
draw a moti f an d repea t i t throughou t th e Lobachevsk y plan e usin g
the actio n o f th e grou p U, an d thu s obtai n a hyperbolic ornament
Chapter 7
Symmetries o f
Differential Equation s
In thi s chapte r w e will apply th e machiner y o f transformation group s

to th e solutio n o f differential equations . W e assume tha t th e reade r i s
acquainted with the notions of derivative and of definite an d indefinit e
integral.
1. Ordinar y differentia l equation s

In thi s boo k w e will only stud y th e simples t clas s o f differential equa -
tions: ordinary differential equations of first order resolved with re-
spect to the derivative.
Definition 4 1 . A differential equation i s a n equatio n o f th e for m

(48) y' = f(x,y),
where y i s a variable dependin g o n x , th e prim e mean s th e derivativ e
with respec t t o x , an d f(x,y) i s a give n functio n o f tw o variables ,
x an d y , whic h i s suppose d t o b e "goo d enough " (continuou s an d
different iable).
Definition 42 . A solution to equation (48 ) i s a functio n y = </>(# )
which, upo n substitutio n int o th e equation , make s i t a tru e identity ,
so tha t
</>'(x) = f{x,4>(x))
198 7. Symmetrie s o f Differentia l Equation s
for an y valu e o f x.
Since w e ar e intereste d i n th e wa y y depend s o n x , w e call x th e
independent an d y th e dependent variable . Equatio n (48 ) ca n als o
be writte n a s dy/dx = / ( x , y ) , wher e dy an d dx ar e differentials, i.e .
infinitesimal ("infinitel y small" ) increment s o f y an d x whos e rati o i s
by definitio n equa l t o th e derivativ e y f. 1
Here i s an exampl e o f a differentia l equation :
(49) y' = y-x.
As yo u ca n chec k b y a direc t substitution , eithe r o f th e function s
y = x + 1 and y = e x + x + lis& solutio n o f this equation .
The mai n theorem o f the theory o f ordinary differentia l equation s
implies that ever y differential equatio n ha s a one-parameter famil y o f
solutions tha t ca n b e describe d b y a formul a y — ^(x,c) containin g
a constan t c whos e valu e ma y b e arbitrary . Suc h a functio n ?/?(# , c)
is calle d th e general solution o f th e give n equation . Fo r example ,
equation (49 ) ha s the genera l solutio n y = ce x - f x +1 whic h give s th e
two particula r solution s quote d above , whe n c = 0 and c = 1 .
Note that th e famil y o f solutions o f a differential equatio n ^(x,c)
cannot b e a n arbitrar y one-paramete r famil y o f functions .
Exercise 1 59 . I s ther e a (firs t order ) differentia l equatio n tha t ha s
the followin g pai r o f particula r solutions : (a ) y = 0 an d y = 1
(constant functions) ? (b ) y — 1 and y — x!
The function f(x,y) i n the right-hand side of the differential equa -
tion (48 ) ca n b e fre e o f x, o f ?/ , o r o f both . Fo r example , w e ca n
consider th e followin g equations :
(50) y' = 2,
(51) y' = cosx ,
(52) y' = y 2.
Exercise 1 60 . Fin d th e genera l solutio n o f (50 ) an d (51 ) . Tr y t o
guess a particular solutio n o f (52) .
1
Prom th e moder n viewpoint , th e notio n o f differentia l i s formalize d usin g dif-
ferential forms, but , a s w e canno t touc h upo n tha t i n thi s book , w e ar e goin g t o
treat th e differential s i n th e abov e intuitiv e sense , followin g th e mathematician s o f th e
seventeenth
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University h centuries
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1. Ordinar y differentia l equation s 199
Equations (50 ) an d (51 ) belon g t o th e clas s o f equation s whos e

right-hand side s depen d onl y o n x:
(53) y' = / ( * ) .
The reade r know s tha t th e genera l solutio n t o suc h a n equatio n i s
obtained b y indefinit e integration :
(54) y = / f{x)dx,
" /
where th e right-han d sid e i s define d "u p t o a n additiv e constant" .
More exactly , i f F{x) i s a certai n primitive o f / ( # ) , i.e. , a functio n
such tha t F'(x) = f(x), the n th e genera l solutio n t o (53 ) ca n b e
written a s
(55) y = F(x) + C.
This formula , fo r arbitrar y value s o f th e constan t C , give s al l th e
solutions o f (53) . Th e graph s o f al l function s (55 ) d o no t intersec t
and fill all the plane (x, y). Fo r example, for (51 ) we obtain the pictur e
shown i n Figur e la .
\- \
\- \
-\
\- x -
\- Jr -
Figure 1 . Graph s o f solution s an d fiel d o f direction s o f a

differential equatio n
Not onl y th e se t o f solutions , bu t th e differentia l equatio n itsel f

can b e represente d a s a geometri c object . Th e equalit y y f = f(x,y)
means tha t th e slop e (mor e exactly, th e tangen t o f the slop e angle) of
the grap h o f the unknow n solutio n a t th e poin t (x , y) shoul d b e equa l
to th e know n numbe r f(x,y). Therefore , a t ever y poin t o f th e plan e
(x,y) w e kno w th e directio n i n whic h th e integra l curv e (th e grap h
of a solution ) shoul d pass . W e arriv e a t th e followin g conclusion :

the geometric object associated with the differential equation (48 ) is a
field of directions in the plane. A field o f directions i s fixed whenever,
for ever y poin t o f th e plane , on e define s a lin e passin g throug h tha t
point.
Geometrically, th e proble m o f integratin g a differentia l equatio n
is formulated a s follows: given a field of directions in the plane, find all
the curves that are everywhere tangent to the given field. Suc h curve s
are referre d t o a s integral curves o f th e field o f directions . Figure s
1 an d 2 sho w th e directio n fields an d th e familie s o f integra l curve s
corresponding t o equation s (51 ) an d (49) , respectively .
/ / /
/ / —
/
/
/ X -\ \
/ / SJ ^ \ \
// - | \ \ \
Figure 2 . Fiel d o f direction s an d solution s o f anothe r differ -

ential equatio n
Exercise 1 61 . Dra w the direction field s an d th e familie s o f integral

curves for equation s (50 ) an d (52) .
Exercise 1 62 . Wha t ar e the integral curves of the field o f directions
shown i n Figur e 3 ? Doe s this field correspond t o an y differentia l
equation?
Using only indefinite integration , on e can solve not only equation s
of clas s (53 ) (independen t o f y) , bu t als o equation s o f the for m
(56) y' = f(x)g(y),
This ca n b e done b y the followin g classica l trick. Writ e y' a s the rati o
of tw o differential s dy/dx an d the n rewrit e (56 ) a s
dy
(57) f(x) dx.
9(y)
As yo u see , th e variable s ar e separated : o n th e left , w e hav e onl y y ,
on th e right , onl y x. Thi s i s wh y equation s o f typ e (56 ) ar e calle d
1. Ordinar y differentia l equation s 201
l
1— • * -t-4— • t I •• * *•
Figure 3 . A field of directions i n the plan e
equations with separating variables. Integratin g bot h side s of (57) , we

obtain
(58) f(x) dx
J g(y) J
(it i s understoo d tha t on e sid e o f thi s equalit y contain s a n arbitrar y
additive constan t C) . Thi s i s a n implici t formul a fo r th e genera l
solution o f (41 ) . I f y i s expressed i n terms o f x, w e will get a n explici t
general solution .
Problem 58 . Find the general solution of the differential equation

2
(59) y' = (2x + l)/(Zy ).
Solution. Rewrit e the equation i n terms o f differentials :

Sy2 dy — (2x + l)dx. Findin g the indefinite integra l give s
y3 = x 2 + x - f C , whenc e y = yjx 2 + x + C. Thi s i s th e
general solutio n o f Equatio n 59 .
Special case s o f equation s wit h separatin g variable s consis t o f

equations whos e right-han d side s depen d o n onl y on e variable , x o r
Exercise 1 63 . Fin d th e genera l solution of equation (52) .

2. Chang e o f variable s
We now discuss o f the relatio n betwee n th e differentia l equation s an d
the mai n them e o f th e boo k — transformations o f th e plane .
It turn s ou t tha t i n variou s method s o f findin g solution s o f dif -
ferential equation s a crucia l rol e i s played b y changes of variables. I f
we pas s fro m variable s x, y t o ne w variable s it , v accordin g t o som e
formulas
U V
(60) { = ^ \
then th e equatio n y' = f(x,y) transform s t o anothe r equatio n

(61) v' = g(u,v)
where th e prim e mean s th e derivativ e wit h respec t t o u, no t x. I f i t
turns ou t tha t i n thi s equatio n variable s separate , the n w e ca n solv e
it an d then , usin g (60) , return t o th e initia l variable s x , y an d obtai n
the solutio n o f th e initia l equation .
What w e have described i s a very simple, bu t ver y effective meth -
od of integration: mak e a change of variables that lead s to an equatio n
with separatin g variables .
Problem 59 . Solve the equation y' = y — x (49) by reducing it to an
equation with separating variables.
Solution. Le t u s make the following chang e of variables:
J u — x,
\v = y — x — 1.
Since u = x, th e derivativ e wit h respec t t o u i s the sam e
thing a s th e derivativ e wit h respec t t o x\ therefor e n o
confusion arise s i f w e denot e bot h o f the m b y a prime .
Next w e hav e y' — v' + 1 and, substitutin g thi s int o th e
given equation , w e obtai n th e equatio n v' = v. Thi s i s
an equatio n wit h separatin g variables , i t ha s th e genera l
solution v = Ce u. Comin g bac k t o th e variable s x , y,
we get th e genera l solutio n o f th e initia l equatio n i n th e
form y = Ce x+x+l.
3. Th e Bernoull i equatio n 203
Exercise 1 64 . Fin d the change of variables that transforms the equa-

tion y = y 2+2xy-\-x2 — 1 to an equation with separating variables.
The formula s o f th e chang e o f variable s (60 ) hav e a doubl e geo -
metric meaning .
First, regardin g x , y a s Cartesia n coordinate s o f a poin t i n th e
plane, we can view u, v as the coordinates of the same point i n anothe r
curvilinear coordinat e system . Fo r example , th e formula s
2
fu = \A + ?/2>
\y
v= arcta n -
vX
or th e equivalen t formula s
x = wcosv ,
2/ = i x sin v
introduce th e syste m o f pola r coordinate s (u, v).
Second, w e ca n thin k tha t w e dea l wit h a transformatio n o f th e
plane, whic h take s a point wit h coordinate s (x, y) int o th e poin t wit h
coordinates (u,v) wher e u — <p(x, y), v — il>(u, v). Al l the coordinate s
are i n thi s cas e calculate d i n on e an d th e sam e coordinat e system .
To visualiz e th e transformatio n i n thi s case , i t i s usefu l t o dra w th e
images o f th e coordinat e line s x = const an d y = const .
3. Th e Bernoull i equatio n
Historically, th e first perso n wh o successfully applie d transformation s
of variable s t o differentia l equation s wa s probabl y Johan n Bernoulli ,
who solve d th e equatio n (no w bearin g hi s name )
(62) y' = Ay + By n,
where A an d B ar e give n function s o f x. H e manage d t o reduc e thi s
equation t o a simple r (linear) equatio n
(63) y' = Py + Q,
where P an d Q ar e agai n function s o f x.
Let u s firs t explai n ho w th e linea r equatio n i s solved . Writ e th e
unknown functio n y a s a product y = uv, wher e u an d v ar e unknow n
functions o f x. Substitutin g thi s int o (63) , we ge t

u'v + uv' = Puv - f Q.
This equation i s satisfied, i f the following tw o relations hold : u' = Pit ,
v' — Q/u. Th e firs t i s a n equatio n wit h separatin g variables , fro m
which w e can fin d th e functio n u(x). Feedin g i t int o th e secon d one ,
we ca n fin d v(x) b y simpl e integration . W e thu s ge t th e solutio n o f
the initia l equatio n y = u(x)v(x).
Exercise 1 65 . Fin d th e genera l solution o f the equatio n
y — 2— — x + x.
x
Problem 60 . Find a transformation that reduces Bernoulli's equa-
tion (62 ) to the linear equation (63) .
Solution. Equation s (62 ) an d (63 ) diffe r onl y i n th e
exponents o f th e dependen t variabl e y. Therefore , i t i s
natural t o tr y a transformatio n o f the for m y — v k (pre -
serving th e independen t variabl e x). Le t u s substitut e
this expressio n int o th e equatio n an d se e what happens :
kvk-xv' = Av k + Bv kn
,
or
kk
If k = 1 /( 1 — n) , th e secon d exponen t kn — k + 1 be -
comes 0 an d w e arriv e a t a linea r equation ! Therefore ,
the require d transformatio n i s y = -u 1 /( 1 -n ).
Exercise 1 66 . Fin d th e genera l solution of the equatio n

, xy 2 + 1
y=
-2y—
A reade r wh o ha s solve d thi s exercis e (o r looke d a t th e answe r
in th e bac k o f th e book ) migh t b e perplexe d b y th e fac t tha t th e
solution i s not give n by a conventional formula , a s a closed expressio n
in elementar y functions . W e mus t therefor e sa y a fe w word s abou t
integration in closed form. Th e functio n
r^da
/-
3. Th e Bernoull i equatio n 205
although i t i s no t a n elementar y functio n (i t canno t b e writte n a s a

combination o f polynomials, trigonometri c functions , logarithm s an d
exponents), i s in fac t almos t a s goo d a s an y elementar y function . It s
numeric value s ca n b e foun d b y compute r t o an y degre e o f precisio n
and it s propertie s ar e wel l known , becaus e thi s functio n i s widel y
used i n probabilit y theor y an d statistics . Th e sam e refer s als o t o th e
integral o f any elementary function . Thi s lays the groundwor k fo r th e
following importan t notion .
A differentia l equatio n i s sai d t o b e integrable in closed form, i f
its genera l solutio n ca n b e writte n b y a formul a involvin g elementar y
functions, indefinit e integral s an d invers e functions .
The simples t exampl e o f a n equatio n whic h i s no t integrabl e i n
closed for m i s the equatio n
y' = y 2 + x
(this fac t wa s prove d b y J . Liouvill e i n 1 841 ) . Thi s equatio n i s a
particular cas e o f th e so-calle d Riccati equation
(64) y' = a(y 2 + x n).
Exercise 1 67 . Fin d the general solution of the Riccati equation (64 )
for n = 0 .
In the year 1 742 , D. Bernoulli an d J . Riccat i discovere d a discret e
series o f value s o f th e paramete r n fo r whic h th e equatio n (64 ) ca n
be integrate d i n close d form . Thi s wa s don e b y a ver y elegan t trick ,
actually b y mean s o f a cycli c group o f transformations. Th e ide a wa s
to fin d a chang e o f variables whic h take s (64 ) int o a n equatio n o f th e
same form , bu t wit h a different valu e of the exponen t n , an d the n tr y
to reduc e th e equatio n t o th e cas e n — 0 (whic h i s integrable , a s yo u
know fro m Exercis e 1 67) .
Let u s firs t mak e th e chang e o f the dependen t variabl e accordin g
to th e formul a
1 1
x2v ax
Feeding thi s int o (64) , afte r som e simplification s w e obtai n a n equa -
tion fo r v:
(65) xn+2v2
where th e prime , a s before , mean s th e derivativ e wit h respec t t o x.

This i s no t ye t a Riccat i equation , bu t w e wil l ge t on e i f w e
change th e independen t variabl e accordin g t o th e rule 2 u = x n+s (o r
x — ii 1 /( n + 3 ). Indeed , b y th e chai n rul e w e hav e
, dv dv du dv, . _ ,9 . _ n+2 dv
v' = — = — — = — (n + 3 ) z n + 2 = ( n + 3 ^ - .
arc a w ax a w a w
A simpl e calculatio n show s tha t afte r thi s chang e equatio n (65 ) be -
comes
dv (o _S±4 \
dw n +3
This is again a Riccati equation, bu t wit h the exponent n change d
to - ( n + 4)/( n + 3) .
If, fo r example , w e ha d a n equatio n wit h n = —4 , afte r thi s
transformation w e would obtai n th e equatio n wit h n = 0 — whic h i s
integrable. Therefore , Riccati' s equatio n wit h n — — 4 i s integrable ,
too.
Exercise 1 68 . Fin d th e genera l solutio n o f th e equatio n y = y 2 +
x~\
Exercise 1 69 . Fin d on e more value of n fo r which the Riccat i equa-
tion i s integrable i n closed form .
Now let u s make our observation s int o a general theory. W e know
that, i f the Riccat i equatio n i s integrable fo r a certai n exponen t m , i t
is also integrabl e fo r th e exponen t n suc h tha t — (n + 4)/(n + 3 ) = m ,
i.e. n = - ( 3 m + 4)/( m + 1 ) .
Consider th e fractiona l linea r functio n
3m + 4
By th e previou s argument , i f th e Riccat i equatio n (64 ) i s integrabl e

for som e exponen t m , the n i t i s als o integrabl e fo r th e valu e q(m).
Repeating th e transformation , w e deduc e tha t i t i s als o integrabl e
for th e exponent s q(q(m)), q(q{q(m))) an d i n genera l fo r an y q k(m),
where q k mean s th e fc-th powe r o f th e transformatio n q.
Exercise 1 70 . Fin d a n explicit formul a fo r q k(m).
We onl y quot e th e transformation s invente d b y Bernoull i an d Riccati . Nobod y

knows how the y foun d them !
4. Poin t transformation s 207
Note tha t th e invers e transformatio n q" 1 ha s th e sam e property :

it take s a n "integrable " exponen t int o a n "integrable " exponent . W e
obtain a n infinite cycli c group generated b y the fractional linea r trans -
formation q. Thi s grou p act s o n th e se t o f al l rea l number s (expo -
nents o f th e Riccat i equation) . Th e propert y o f th e equatio n t o b e
integrable i n close d for m i s a n invarian t o f thi s action . Therefore ,
each orbi t eithe r consist s entirely o f exponents fo r whic h the equatio n
is integrable , o r contain s onl y suc h exponent s fo r whic h th e equatio n
is not integrable .
In particular , th e orbi t o f th e numbe r 0 furnishe s a n infinit e se -
ries o f Riccat i equation s tha t ar e integrabl e i n close d form . Thei r
exponents ar e
where k G Z i s a n arbitrar y integer . Not e tha t q k(0) tend s t o th e

value —2 , when k goe s t o infinity .
Exercise 1 71 . Prov e that the Riccati equation (64 ) is also integrable
for th e exponent n = — 2.
We must, however , war n th e reade r that , startin g fro m th e "inte -
grable" valu e n = — 2 and usin g th e transformatio n g , it i s impossibl e
to find an y ne w integrabl e cases , becaus e th e numbe r — 2 is a fixed
point o f q and it s orbi t consist s o f only on e point .
We have thus foun d tw o integrable orbit s i n the se t o f Riccati ex -
ponents. J . Liouvill e proved that fo r al l the remaining values Riccati' s
equation canno t b e solve d i n close d form .
4. Poin t transformation s
So far , w e have onl y encountere d change s o f variable s o f th e for m
X u)
(66) { = * ',
i.e., wher e th e independen t variabl e x i s expresse d i n term s o f th e
new independen t variabl e only . I n thi s cas e i t i s eas y t o expres s th e
derivative dy/dx i n term s o f u, v an d dv/du usin g th e chai n rul e
dy/dx — dy/du • du/dx.
One can , however , us e arbitrar y transformation s o f th e indepen -

dent an d dependen t variable s x = ip(u,v), y = ijj{u,v) {point trans-
formations). T o deriv e th e transformatio n formul a fo r th e derivativ e
dy/dx i n thi s case , w e will nee d th e notio n o f partial derivatives.
Definition 43 . Le t z = h(x, y) b e a functio n o f tw o variables . I f

the valu e o f y i s fixed, y = yo ? w e obtai n a functio n o f on e variabl e
z = h(x,yo). Th e derivativ e o f thi s functio n a t th e poin t XQ is calle d
the partial derivative of the function h(x,y) with respect to the variable
x at the point (#o , 2/o)- Th e partial derivative is denoted b y §^(#o 5 Vo)-
Symbolically,
dh . dh(x,y 0)l v h(x 0 + e,y 0) - h(x 0,yo)
lx
Ox dx - x ° e^ oe
Problem 61 . Compute the partial derivative over x of the function
z = y/9 — x 2 — y 2 at the point (2,1 ) .
Solution. Assignin g y = 1 , w e ge t a functio n o f on e

variable z = y/S — x 2. It s derivativ e i s —x/\/ 8 — x 2. Fo r
x = 2 we obtai n
When th e poin t (xo,2/o ) varies , th e valu e f f (^0^2/0 ) become s a

function o f the variable s XQ an d y$. Usin g the norma l notation s (x , y)
instead of (a?o, 2/o)> o ne g e^s a function o f the variables x an d y denote d
by dz/dx o r simpl y z x. Thus , fo r th e functio n z = y/9 — x 2 — y 2 w e
obtain
dz x
dx ~ y/ g _ x 2 _ y 2 '
Once again , t o comput e th e partia l derivativ e z x, on e ha s t o dif -

ferentiate z(x,y) wit h respec t t o x , treatin g th e variabl e y a s a n
arbitrary constant . Th e partia l derivativ e wit h respec t t o y i s de -
fined i n th e simila r way , treatin g x a s a parameter . Fo r th e functio n
z = . y/9 — x 2 — y 2 w e hav e
dz y
We als o explai n th e geometri c meanin g o f partia l derivatives .

Consider th e surfac e i n 3-spac e consistin g o f al l point s (x,y,z(x,y))
— th e graph o f th e give n functio n o f tw o variables . Figur e 4 depict s
the grap h o f ou r favourit e functio n ^/ 9 — x 2 — y2.
Figure 4 . Partia l derivative s
Given a poin t (xo,yo) > a ^ whic h the - functio n z(x,y) i s defined ,

draw the plane y = yo. I t cut s the surface alon g a certain plan e curve .
The slop e o f the tangen t lin e t o thi s curv e a t th e poin t (xo,yo) give s
the valu e o f th e partia l derivativ e z x(xo, yo).
Another partia l derivativ e z y(xo,yo) i s th e slop e o f th e tangen t
line t o th e sectio n o f the surfac e b y the plan e x = xo . Figur e 4 shows
both section s an d thei r tangen t line s fo r th e functio n y^ 9 — x 2 — y 2
at th e poin t (2,1 ) .
The plan e passin g throug h th e tw o tangen t line s i s th e tangen t
plane t o th e grap h o f th e functio n a t th e give n point . It s equatio n i s
(67) z - z 0 = p(x - x 0) + q(y - y 0 ),

where z 0 = h(x 0,y0), p = f|(ô,2/o ) an d q = ^(x 0,y0). Indeed ,
substituting x = xo int o (67) , we get th e equatio n o f th e tangen t lin e
to th e grap h o f h(xo,y) viewe d a s a functio n o f y , an d substitutin g
y = yo , the simila r equatio n fo r h{x,yo).
The poin t o f the surfac e an d th e poin t o f the tangen t plane , bot h

corresponding to one and the same point (x , y) of the horizontal plane,
are ver y clos e t o eac h other , i f th e poin t (x,y ) i s clos e enoug h t o
( ^ y o ) - Therefore , th e differenc e z — z$, compute d b y formul a (67) ,
can b e viewe d a s a n incremen t o f th e functio n h(x,y), whe n it s ar -
gument move s fro m (xo,2/o ) t o (x,y) , provide d tha t thi s shif t i s "in -
finitely small" . Denotin g th e infinitesima l increment s o f th e thre e
variables b y dx, dy an d dz, w e can writ e
(68) dz = z xdx + Zydy
(the formul a o f th e differential of a function of two variables).
With th e help of this formula, w e will now obtain th e transforma -
tion rul e fo r th e derivativ e dy/dx whe n the variable s x an d y underg o
an arbitrar y poin t transformatio n
X = M
(69) { ^ '
Using th e notatio n y' — dy/dx, v' = dv/du (not e tha t th e prim e ha s

different meanin g i n eithe r case) , b y formul a (68 ) w e ca n writ e
dv
Vudu + Vvdv yu + Vv
(70^ ' =^ = = d% = y* + yv v'
dx x udu + x vdv x u + xv ^ x u + x vv''
We se e tha t th e derivativ e y' i s expresse d a s a fractiona l linea r

function o f v' wit h coefficient s dependin g o n u and v, i.e. , a s a projec -
tive transformatio n o f v' (se e (2)) . Thi s remarkabl e fact , b y th e way ,
leads to a deep connectio n betwee n projectiv e geometr y an d ordinar y
differential equations , bu t w e will no t discus s tha t i n thi s book .
As w e notice d before , (69 ) ca n b e viewe d eithe r a s th e passag e
from on e coordinat e syste m i n the plan e t o another , o r a s a mappin g
of th e plan e int o itsel f accordin g t o th e rul e (u,v) *- » (x,y). I n th e
latter cas e (70 ) describe s ho w th e slop e o f th e plan e curve s change s
under thi s mappin g (se e Figur e 5) : v' i s th e tangen t o f th e angl e t o
the horizonta l lin e fo r a give n curve , whil e y f i s th e tangen t o f th e
similar angl e fo r th e imag e o f this curve .
Let u s cal l a poin t o f th e plan e togethe r wit h a directio n ( a
straight line ) attache d t o thi s poin t a contact element. A contac t
element i s describe d b y thre e number s (x,y,p), wher e (x,y) ar e th e
\vn
y
UK
/KU/
bf '
M
i i i
1 \1
F i g u r e 5 . Poin t transformatio n
coordinates o f the give n poin t an d p i s the slop e o f the lin e (th e tan -
gent o f th e angl e i t make s wit h th e horizonta l axis) . Th e se t o f al l
contact element s thu s form s a three-dimensiona l spac e — th e space
of contact elements. Som e o f it s element s ar e show n i n Figur e 6 .
y
v.
x'
7
F i g u r e 6 . Contac t element s
Formula (70) , take n togethe r wit h (69) , define s a certai n trans -

formation o f the spac e o f contac t elements , whic h correspond s t o th e
plane transformatio n give n b y formul a (69 ) alone .
Problem 62 . Find the transformation of the space of contact el-

ements that corresponds to the inversion with respect to the circle
x2 +y 2 = 1 .
Solution. Unde r th e inversion , th e coordinate s o f a

point transfor m a s follows :
x
x2 + y 2'
y
x2 -f - 7/ 2
The partia l derivative s o f thes e function s ar e
y2 — x 2 —%xy
Ux =
(x 2+y2)2' Uy= {x 2
+ y 2)21
2
_ —2xy _ x - y 2
Vx = 2 212 Vy= 2
(x +y ) {x +y2)2'
By (70 ) w e obtai n
,= v + v yy' = (x 2 - y 2)y' - 2x y
x
(71)
ux + %2/ ; ~2xyy f + y 2 -x 2'
As a corollar y o f this result , w e can prov e th e followin g theorem ,

which generalize s th e assertio n o f Exercis e 1 5 3 (se e page 1 87) .
Theorem 1 8 . A plane inversion preserves the angles between curves.
Proof. Th e angl e betwee n tw o curve s is , b y definition , th e angl e

between thei r tangen t lines . I f w e have tw o curve s i n the (x , y)-plane
such tha t th e tangent s o f th e slop e angle s ar e pi an d p2 , the n th e
angle a betwee n th e curve s satisfie s
Pi - V2
tan a — .
1 + P1P2
Let q\ an d q<i b e th e correspondin g tangent s fo r th e image s o f
the tw o curve s afte r inversion , an d le t (3 b e th e angl e betwee n them .
Then
(72) tsm p = SLZSL m
K
+ 41 4 2
According t o (71 ) , we hav e
api + b
Qi =
bpi — a'
ap2 + b
42 =
bp2019for
Licensed to Penn St Univ, University Park. Prepared on Tue Mar 12 18:16:11 EDT 2- a'download from IP 132.174.254.159.
where a an d b are certai n constant s dependin g o n th e poin t (x,y).

Feeding thi s int o (72) , we get
api + 6 ap 2 + b
a bpi - a bp 2 -a Pi - P2 ,
tan/? = — ^ , , ^—— r = = -tana ,
1 _ L a Pl+b a P21 +b + PlP2
bpi — a bp2 — a
which mean s tha t th e inversio n preserve s th e angles , bu t change s th e
orientation o f th e plane . •
We now giv e an exampl e wher e a generi c poin t transformatio n i s
used t o solv e a differentia l equation .
Problem 63 . Solve the differential equation
2
(y2 -x 2) (x + (x 2 +</ 2 ) 2 ) + 2 x y (y 2 + ( x 2 + V ) 2 )
2xy (x 2 + {x 2 + y 2f) + (x 2 - y 2) (y 2 + (x 2 + y 2 2
))
Solution. Le t u s mak e th e chang e o f variable s
u2 - f v 2'
u2 + u 2 '
2m ; (w 2 - i> 2)i>/ - -
—2uvv' + v 2—u2'
Substituting thes e expression s i n th e give n equation , w e
obtain, afte r simplifications ,
2
,u + l
Here the variable s separate: (v 2 — l)dv = (u 2 + l)dtx ,

and th e general solution i s given by the following implici t
function:
v3 u 3 _ ,
Going bac k t o th e variable s (x , ?/), we obtain th e an -

swer:
yS _ x 3= 3(a .+ y){x 2+ y2) 2+ c(a; 2+ y2)3 j
C bein g a n arbitrar y constant .

Exercise 1 72 . (a ) Fin d th e expressio n o f dr/d(p in term s o f dy/dx,

if (a;, y) ar e Cartesian coordinates and (r, ip) ar e polar coordinate s
in th e plane , (b ) Usin g th e formula s obtaine d i n (a) , solv e th e
differential equatio n yy + x = (x 2 + y 2)(xy' — y) b y passin g t o
polar coordinates .
5. One-paramete r group s
Definition 44 . A one-parameter group of plane transformations i s
an actio n o f th e additiv e grou p R o n th e plane .
This mean s tha t fo r ever y rea l numbe r t a transformatio n gt i s

defined i n suc h a wa y tha t th e equalit y gt°g s — 9s+t hold s fo r ever y
pair s , t G l . I n other words , we deal with a homomorphism fro m th e
group R int o th e grou p o f plane transformations . I n thi s cas e w e say
that {gt} i s a one-parameter group of transformations o f th e plane .
Let u s stres s tha t a one-paramete r grou p i s no t simpl y th e se t {gt},
but thi s se t togethe r wit h th e parametrizatio n t ^ gt-
The simples t exampl e o f a non-trivial one-paramete r grou p i s the
group o f paralle l translations , say , i n th e directio n o f th e axi s Ox: g t
is the translatio n b y te±, wher e e i i s the horizonta l uni t vector .
Exercise 1 73 . Prov e tha t an y tw o transformation s belongin g t o a
one-parameter grou p commute.
A one-parameter grou p ca n b e written i n coordinate s a s a pair of
functions o f three variables :
/ ?3 x ( x t= <f(x,y,t),
Here (x , y) ar e the coordinate s o f an arbitrar y poin t i n the plane , an d

(xt,yt) ar e th e coordinate s o f it s imag e unde r th e transformatio n g t.
For an y fixed valu e o f t w e obtain a pai r o f functions o f two variable s
that defin e a concret e transformation .
The grou p law , i.e. , th e relatio n g t o gs = g s+t, ca n b e writte n i n
terms o f function s </? , ip as follows :
\ ip((f{x, y, s),il){x, y , s),t) = ^(x , y, s + t).

This is the definition o f a one-parameter grou p written i n coordinates .
5. O n e - p a r a m e t e r g r o u p s 215
For example , th e grou p o f horizonta l translation s i s represente d

by th e function s
xt = x + t,
yt = y,
for whic h th e relation s (74 ) ar e obviousl y fulfilled .
E x e r c i s e 1 74 . Conside r th e se t o f al l homothetie s wit h a commo n

centre an d positiv e coefficients . I s thi s a one-paramete r group ?
T h e observan t reade r wil l notic e t h a t th e questio n o f thi s exercis e

is no t correctl y posed , becaus e a one-paramete r grou p presuppose s a
fixed parametrizatio n o f the give n se t o f transformations. I f we assign ,
to ever y numbe r t G M , th e homothet y wit h stretchin g coefficien t £ ,
we won' t obtai n a one-paramete r group , becaus e th e compositio n o f
homotheties wit h coefficient s s an d t i s the homothet y wit h coefficien t
st, no t s + t. Fortunately , w e kno w th e tric k t h a t t u r n s additio n
into multiplication : thi s i s th e exponentia l function . Assignin g t h e
homothety wit h coefficien t e * t o th e numbe r t , w e ge t a genuin e one -
parameter group . Placin g th e centr e o f homothetie s a t t h e origin , w e
can describ e th e grou p b y th e function s
Xt — e t x,
Vt = e*y .
Relations (74 ) obviousl y hold .

A one-paramete r grou p ca n b e visualize d throug h th e se t o f it s
orbits. Figur e 7 show s th e orbit s o f th e tw o group s t h a t w e hav e
mentioned: translation s an d homotheties .
o X
Figure 7 . Orbit s o f one-paramete r group s o f translation s an d homothetie s

216 7. S y m m e t r i e s o f Differentia l E q u a t i o n s
We mus t not e t h a t th e se t o f orbit s doe s no t uniquel y defin e th e

one-parameter group . A simpl e exampl e o f thi s i s provide d b y th e
group o f translation s wit h a doubl e spee d
xt = x + 2t ,
Vt = y ,
which ha s th e sam e orbit s a s th e grou p o f translation s previousl y
discussed (show n i n Figur e 7a) .
E x e r c i s e 1 7 5 . Defin e a one-parameter grou p of rotations with a com-
mon centre , writ e it s coordinat e representation , an d dra w it s or -
bits.
E x e r c i s e 1 7 6 . Chec k tha t th e relation s
at
Xt = e (x cos bt — y sin bt),
at
yt = e (x sin bt -f y cos bt)
define a one-parameter group . Explai n it s geometric meanin g an d
draw it s orbits .
E x e r c i s e 1 7 7 . Le t xt an d yt b e th e root s o f th e quadrati c equatio n
for th e unknow n w
(w - x)(w - y) + t = 0 ,
chosen i n suc h a wa y tha t xt, yt continuousl y depen d o n t an d
x0 = x , y 0 = y.
The number s x t an d yt ar e function s o f thre e variable s x , y
and t. Prov e tha t thes e function s defin e a one-paramete r grou p
of transformations , an d dra w it s orbits .
6. Symmetrie s o f differentia l equation s

A differentia l equation , viewe d a s a field o f direction s i n t h e plane ,
may posses s som e symmetry . On e glanc e a t Figur e l b i s sufficien t t o
understand t h a t thi s fiel d o f direction s i s preserved b y an y translatio n
along Oy a s wel l a s translation s alon g Ox b y whol e multiple s o f 2n.
Transformations o f th e firs t kin d for m a one-paramete r grou p xt = x,
yt = y + t. Transformation s o f th e secon d kin d for m a n infinit e cycli c
group (se e p . 87) .
It turn s ou t t h a t the knowledge of a one-parameter group of sym-
metries enables one to find the general solution of the equation under
study in closed form.
6. Symmetrie s o f differentia l equation s 217
Passing t o exac t definitions , suppos e tha t w e ar e give n a differ -

ential equatio n
(75) y' = f(x,y)
and a transformatio n o f th e plan e
(x = (p(u,v),
(76)
\y = il)(u,v).
By (70) , w e ca n fin d th e correspondin g expressio n o f y' — dy/dx
through u, v an d v' — dv/du.
Definition 45 . Transformatio n (76 ) i s called a symmetry of the dif-
ferential equation (75) , i f th e equatio n fo r v(u) obtaine d afte r th e
expressions fo r it , v an d v' ar e substitute d int o (75 ) ha s th e sam e
function / i n it s right-han d side :
v' = f{u,v).
In geometri c languag e thi s mean s tha t th e transformatio n o f th e

space of contact element s given by (69 ) an d (70 ) preserve s th e surfac e
in thi s spac e tha t consist s o f al l contac t element s belongin g t o th e
given field o f directions .
F i g u r e 8 . Differentia l equatio n a s a surfac e i n 3-spac e
Figure 8 show s suc h a surfac e fo r th e differentia l equatio n y'

cosx, an d yo u ca n se e tha t th e transformation s
xt = x ,
yt = y + t,
Pt = p
and
Xt = X + 27T& ,
Vt = 2/ ,
Pt = P
map thi s surfac e int o itself .
There ar e tw o mai n problem s abou t th e interrelatio n o f differen -
tial equation s an d one-paramete r groups :
(1) Give n a differential equation , find al l (o r some ) group s of its
symmetries.
(2) Give n a one-parameter grou p o f plane transformations , find
all (o r some ) differentia l equation s preserve d b y thi s group ,
i.e., suc h tha t th e grou p consist s o f thei r symmetries .
For practica l need s (solvin g differentia l equations ) th e first ques -
tion i s mor e important . Bu t i t i s als o mor e difficult . Therefore , le t
us first discus s th e secon d question .
We start wit h tw o simple examples , wher e th e answe r i s obvious:
• Th e genera l equatio n preserve d b y th e grou p o f x-transla -
tions i s
(77) y' = f(y).
We have earlier considered a particular cas e of this, equatio n
(52).
• Th e genera l equatio n preserve d b y th e grou p o f ?/-transla -
tions i s
y' = fix).
We hav e encountere d equatio n (51 ) , belonging t o thi s class .
Let u s no w conside r mor e interestin g groups .
Problem 64 . Find the general form of a differential equation pre-
served by the group of rotations of the (x,y) plane centred at (0,0) .
Solution. Unde r th e rotatio n throug h a n angl e a ever y

contact elemen t move s together wit h th e poin t o f attach -
ment an d turn s b y th e sam e angl e a (Figur e 9) .
6. Symmetrie s o f differentia l equation s 219
Figure 9 . Actio n o f rotation s o n contac t element s
Hence, th e angl e i t make s wit h th e radius-vecto r o f

the poin t (x,y) doe s no t change . Therefore , a fiel d o f
directions ( = a differentia l equation ) i s invarian t unde r
the grou p o f rotations , i f an d onl y i f th e angl e betwee n
the directio n o f th e fiel d an d th e radius-vecto r depend s
only o n th e distanc e fro m th e origin . W e ca n tak e th e
tangent o f th e angl e instea d o f th e angl e itself , an d th e
square o f the distanc e instea d o f the distance . Usin g th e
formula fo r th e tangen t o f th e differenc e o f tw o angles ,
we obtain th e genera l form o f the equation admittin g th e
group o f rotations aroun d th e origin :
xy' -y 2
= f(x +y2),
yy' + x
f bein g a n arbitrar y functio n o f its argument . Resolvin g
this with respect to y\ w e can write the answer as follows:
xf(x2 + y 2) + y
(78) y
x-yf(x2 + y 2)'
Exercise 1 78 . Fin d al l differentia l equation s tha t admi t th e one -

parameter grou p of homotheties.
Exercise 1 79 . Fin d al l differentia l equation s tha t admi t th e one -
parameter grou p of spiral homotheties describe d i n Exercise 176.
7. Solvin g equation s b y symmetrie s

In thi s section , w e wil l prov e th e followin g fact : if a one-parameter
group of symmetries of a differential equation is known, then it can
be reduced, by a change of variables, to an equation with separating
variables — and hence its general solution can be found in closed
form.
To find the new coordinate system in which the variables separate ,
we will use invariants o f one-parameter groups , so let u s first mentio n
some o f thei r propertie s an d conside r som e examples .
We recal l (se e Sectio n 6 ) tha t a n invarian t o f a grou p actio n i s a
function whic h i s constant o n the orbits. I n other words , a function i s
an invariant, i f it has equal values at any two points that ma p into each
other by a transformation o f the group. T o give a simple example, an y
function o f y is an invariant o f the one-parameter grou p of translation s
along Ox. Th e functio n y itsel f i s th e universa l (complete ) invarian t
of thi s grou p action , becaus e it s value s o n al l orbit s ar e different .
In th e sam e way , th e functio n x i s th e universa l invarian t o f th e
group o f translation s i n th e directio n o f the axi s Oy.
What i s the universa l invarian t o f the grou p o f homotheties wit h
centre 0 actin g o n th e plan e withou t th e origin ? On e i s tempte d t o
think tha t i t i s th e pola r angl e (p. Indeed , th e pola r angl e <p take s
equal value s a t al l point s o f ever y ra y (half-line ) tha t issue s fro m
the origin , an d differen t ray s correspon d t o differen t value s o f th e
function. However , th e pola r angl e (p is no t a norma l single-value d
function o n the plane , fo r example , t o th e poin t (—1 ,0 ) one can, wit h
equal truth, assig n the values 1 80 ° and —1 80 ° (an d an infinite numbe r
of others). O f course , on e ca n mak e ip a single-valued functio n using ,
for example , th e conventio n tha t (p mus t alway s tak e value s betwee n
0° (including ) an d 360 ° (excluding ) — bu t the n i t wil l becom e a
discontinuous function . I n fact , th e grou p o f homothetie s doe s no t
have any continuous universal invariant wit h values in M. O f course, it
does have continuous invariant s whic h ar e not universal , fo r example ,
the functio n simp.
7. Solvin g equation s b y symmetrie s 221
Exercise 1 80 . Fin d an invariant of the group of rotations around the

origin. Doe s thi s grou p hav e a universa l continuou s real-value d
invariant?
There ar e two approaches to the problem o f finding th e invariant s
of a give n one-paramete r group :
(1) The formal approach. Assumin g tha t th e grou p i s give n
in coordinate s b y a pai r o f function s (73) , th e proble m i s
to mak e u p a combinatio n o f th e expression s (p(x,y,t) an d
i/j(x,y,t) tha t doe s no t contai n th e variabl e £ , i.e., t o fin d a
function h(<p, ip) that doe s no t depen d o n t.
(2) The geometric approach. I n th e plane , w e dra w a curv e K
that meet s ever y orbi t o f th e grou p i n exactl y on e poin t
(see Figur e 1 0) . W e choose a n arbitrar y functio n o n thi s
curve that take s different value s at differen t points , and the n
prolong i t t o th e whol e plane , followin g th e rul e tha t th e
value of the functio n a t an y poin t A i s set equa l t o it s valu e
at th e poin t B wher e th e orbi t meet s th e chose n curv e K.
Figure 1 0 . Invarian t o f a one-paramete r grou p
Problem 65 . Find an invariant of the one-parameter group

xt = x + t,
Vt = e V
Solution. Formal method. Lookin g a t th e abov e for -

mulas fo r a while , on e ca n gues s tha t th e combinatio n
ety. e x+t — y e-x ^ oes n 0 £ d e p e n ( i o n f a n c j thu s provide s
an invarian t o f the group .

Geometric method. Th e orbit s o f th e give n grou p ar e
shown i n Figur e 1 1 .
Figure 1 1 . Orbit s o f th e grou p studie d i n Proble m 6 5
The coordinat e axi s Oy meet s ever y orbi t i n on e

point, s o we can choos e i t a s th e curv e K. Th e functio n
v(0, y) = y takes different value s at different point s of this
line, s o let u s fin d it s prolongatio n t o al l th e plan e alon g
the orbits of the group. Le t A(x,y) b e an arbitrary poin t
of th e plan e an d le t B(Q,v) b e th e intersectio n poin t o f
the correspondin g orbi t wit h th e y-axis . Pro m th e equa -
tions o f th e grou p w e se e tha t ther e i s a numbe r t suc h
that x = t, y = e tv. Pro m thes e equation s w e fin d tha t
v = ye~ x. Thus , th e invarian t i s ye~ x.
Exercise 1 81 . Fin d a non-trivial invarian t o f the group of spiral ho-

motheties (Exercis e 1 76 ) by the abov e two methods.
We finally ge t t o the questio n o f integrating differentia l equation s
with a known symmetry group. Th e first principl e is that i t is useful t o
pass to new coordinates such that on e of the new coordinate function s
is a n invarian t o f th e group . Afte r suc h transformatio n i t ofte n (bu t

not always ) happen s tha t th e variable s separate , an d th e equatio n
can b e solve d i n close d form . Le t u s consider a n exampl e tha t come s
up quit e often .
Problem 66 . Find the general solution of the differential equation

(79) y' = f(y/x).
Solution. Suc h equation s ar e calle d homogeneous. A s
we kno w fro m Exercis e 1 78 , homogeneous equation s ar e
invariant wit h respec t t o th e one-paramete r grou p o f
homotheties wit h th e centr e (0,0) . Sinc e th e functio n
y/x i s a n invarian t o f thi s group , le t u s tak e i t fo r th e
new dependen t variable , leavin g th e independen t vari -
able x unchanged . W e thu s se t v = y/x, o r y = xv;
hence y' — v + xv f. Substitutin g thi s int o (79) , w e ge t
v + xv' = f(v), o r
This i s indee d a n equatio n wit h separatin g variables .
Exercise 1 82 . Continuin g th e previou s argument , fin d a n explici t

answer i n the particula r cas e of the equation y' = 1 -j - 2y/x.
Exercise 1 83 . D o the variables in (79) separate in polar coordinates?
Exercise 1 84 . Adap t th e argumen t o f Problem 6 6 to th e equation s
of the for m
, _ / ax + by + c \
\aix + biy + ci) '
We must stres s tha t takin g th e ne w dependent variable s t o b e a n
invariant o f the grou p doe s not guarante e th e separatio n o f variables .
The choice of the independent variabl e is also very important. Fo r ex -
ample, if , i n the equation (78 ) tha t admit s the grou p of rotations, on e
goes ove r t o pola r coordinates , the n th e ne w equatio n ha s th e for m
udv/du = f(v), wher e th e variable s separate . However , th e transfor -
mation u — x, v — x 2 + y 2 (th e secon d variabl e i s a grou p invariant )
does not lea d t o an equatio n wit h separatin g variable s (pleas e check) .
To ge t a universa l rul e fo r th e integratio n o f equation s wit h a

known one-paramete r grou p o f symmetries, w e will try t o find a coor -
dinate syste m (u,v) suc h that i n these coordinate s th e grou p look s a s
simple a s possible, fo r example , consist s o f parallel translation s alon g
the axi s u. Th e genera l for m o f a n equatio n tha t admi t thi s grou p
is, a s w e know , dv/du = ip{v). Her e th e variable s separate , an d th e
equation i s solvable i n close d form . I t remain s t o understan d ho w we
can reduc e th e initia l one-paramete r grou p t o thi s simpl e for m b y a
change o f coordinates .
In the coordinates (u , v) w e want the transformations o f the grou p
to hav e th e for m u t = u + t, v t = v . Thes e transformation s ma p th e
coordinate lin e u = 0 int o th e paralle l line s u = t. Choos e a real -
valued functio n v(x,y) whic h i s an invarian t o f the group . Th e orbit s
are give n b y th e equatio n v — const . Choos e a curv e K whic h meet s
every orbi t i n on e point , an d assum e tha t K i s the i>-axi s o f th e ne w
coordinate system , i.e. , i s describe d b y th e equatio n u = 0 . Then ,
to fulfil l ou r plane , w e mus t assum e tha t th e imag e o f K unde r th e
group transformatio n gt shoul d b e describe d b y th e equatio n u — t.
If, instea d o f this function u, w e take for th e independent variabl e
another functio n w tha t ha s th e sam e leve l line s w = cons t (th e lin e
w = t shoul d b e th e imag e o f w — 0 unde r som e transformatio n o f
the group , bu t no t necessaril y g t), the n w e will also hav e a n equatio n
with separatin g variable s i n th e ne w coordinates. Indeed , i n this cas e
it; i s a functio n o f u: w — h(u), which , upo n substitutio n int o th e
equation dv/du = <p(v), gives a n equatio n dv/dw = (p(v)ip(w).
We ca n stat e thi s resul t a s th e followin g variable separation the-
orem.
Theorem 1 9 . Suppose that we know a one-parameter group G =
{gt} of symmetries of a differential equation E. Then the equation
E becomes an equation with separating variables in any coordinate
system (u,v) such that the coordinate lines v = cons t are the orbits
of the group G, and the lines u = cons t go into each other under the
transformations gt.
y' = l(y 2 + x-2).

7. Solvin g equation s b y symmetrie s 22 5
Solution. I f x i s multiplie d b y a constan t k an d y b y

its invers e A; -1 , the n y' get s multiplie d b y A: -2 (yo u ca n
also chec k tha t usin g formul a (70)) . Al l the term s o f th e
given equatio n increas e b y th e sam e factor , s o tha t th e
equation actuall y doe s no t change . Thi s mean s tha t th e
group
Xt = e~ fx,
yt = e ly
is a grou p o f symmetrie s o f th e give n equation .
This group is called the group of hyperbolic rotations.
Its orbit s ar e branche s (connecte d components ) o f th e
hyperbolas xy = cons t (se e Figur e 1 2) . Th e produc t
xy i s a n invarian t o f thi s group . W e tak e i t a s th e ne w
dependent variable : v — xy.
y
J \
\ v. — *•—
1
•—-
ft
-**- . — - •
- \
Figure 1 2 . Orbit s o f hyperboli c rotation s
Note tha t th e vertica l line s x = cons t ar e mappe d

one int o anothe r b y hyperboli c rotations : th e imag e o f
the lin e x = a i s th e lin e x = e~ la. Therefore , w e ca n
set x t o b e th e ne w coordinat e u. Th e require d chang e
of variables i s thu s
x= u ,
v
y = -.
u
After thi s chang e o f variables th e give n equation be -

comes
, _ 2v 2 + 5v + 2
The variable s separate , an d th e genera l integra l i s
^v +42= Cu*l\
Expressing u an d v throug h x an d y , w e obtai n th e an -
swer:
32
y = 2(x-Cx8/5) x
Remark. Th e sam e procedur e solve s th e equatio n
y' = a(y 2 + x~ 2) fo r a n arbitrar y valu e o f a. Th e reaso n
why we only considere d a special cas e is that th e genera l
answer i s rather cumbersome .
2
Exercise 1 85 . Fin d th e genera l solution o f y' = (x + y )/(xy).
As th e las t example , w e conside r a n equatio n wher e i t i s impos -
sible t o separat e th e variable s b y a chang e o f th e for m x = <p(u),
y = ip(x,y).

(y 2-x2)2-5(y-x) +
/ = 4
y( y2 _ x 2)2 + 5 ( y _ x ) + 4
using the one-parameter group
it = ^ - x + ^ f - y ,
Solution. Thes e transformation s ar e hyperboli c rota -

tions (se e 7) , bu t considere d i n a coordinat e syste m ro -
tated throug h 45 ° wit h respec t t o th e initia l one . Th e
function v — y 2 — x 2 i s a grou p invariant . Not e tha t th e
families o f vertica l o r horizonta l line s ar e no t preserve d
by th e transformation s o f the group ; henc e i t i s no goo d
to tak e u = x o r u = y. However , th e line s y — x — const
do hav e thi s property . Se t u = y — x. The n afte r th e

transformation th e equatio n become s
dv 5v — v2 — 4
du u
The variable s separate !
Solving this equatio n an d makin g th e invers e chang e
of variables, w e obtain th e implici t solutio n t o th e initia l
equation:
y1 - x 2
- 4
Exercise 1 86 . Solv e th e differentia l equatio n y' = e~ xy2 - y + e x

with th e help of the symmetry grou p xt = x + t, yt = e ty.
We hav e thu s learne d ho w t o solv e a differentia l equation , i f a
one-parameter grou p o f it s symmetrie s i s known. "This , however , b y
no mean s implie s tha t an y differentia l equatio n Xdy — Ydx = 0 ca n
be solve d i n close d form . Th e difficult y consist s i n finding th e one -
parameter grou p tha t leave s i t invariant" . Thes e word s wer e writte n
by Sophu s Lie , th e Norwegia n mathematicia n wh o create d th e the -
ory o f continuou s group s an d foun d it s application s t o differentia l
equations, th e simples t cas e o f whic h wa s describe d i n th e previou s
pages. A s a las t remark , le t u s mentio n tha t durin g th e las t 3 0 year s
or s o som e algorithmi c method s fo r findin g th e symmetrie s o f differ -
ential equation s hav e bee n elaborate d an d implemente d i n numerou s
software system s o f compute r algebra .
Answers, Hint s an d
Solutions t o Exercise s
1. Example : triangl e wit h vertice s (0,0) , (1 2,9) , (24,-7) .

2. Th e proble m ca n b e solve d eithe r b y a direc t constructio n
or usin g th e resul t o f exercis e 6 . Answer : th e polygo n 3>
may hav e 3 , 4, 5 or 6 vertices.
3. A necessar y an d sufficien t conditio n i s k — I = 1 . T o prov e
this, rewrit e th e give n expressio n a s A\ + A^ — B\ + As —
B2 + --.
4. Proble m 2 show s tha t th e give n assertio n i s equivalen t t o
Euler's theorem, whic h say s tha t MH = 2 0 M , wher e M i s
the median intersectio n point , O is the outcentre (th e centr e
of th e circumscribe d circle ) an d H i s th e orthocentr e (th e
intersection poin t o f the altitudes ) o f the triangle . T o prov e
this theorem , construc t th e triangl e A\B\C\ whic h i s twic e
as larg e a s ABC an d wit h side s paralle l t o thos e o f ABC
(see Figur e 1 ) an d observ e tha t th e perpendicula r bisector s
of th e side s o f triangl e ABC coincid e wit h th e altitude s o f
triangle A\B\C\.
5. Th e necessar y an d sufficien t conditio n i s a + /3 + ---+o ; = l .
The proo f follow s fro m th e vecto r equalit y
aPA + (3PB + • • • + uPZ
= aQA + PQB + • • • + 00QZ + ( Q + / 3 + . . . + u)PQ.
230 Answers , Hint s an d Solution s t o Exercise s
Figure 1 . Euler' s theore m
6. The se t o f al l point s a\A\ + CL2A2 - f • • • + a nAn wher e al l

coefficients ai ar e nonnegativ e an d a\ H - 0,2 + • • • - h an = 1 .
7. Express al l th e point s unde r consideratio n i n term s o f th e
four vertice s o f th e quadrangle .
8. Express thes e points i n terms o f the vertice s of the hexagon .
9. Express al l th e point s unde r consideratio n i n term s o f th e
four vertice s o f th e quadrangle .
10. Take th e intersectio n poin t o f th e tw o median s fo r th e pol e
and us e th e resul t o f Proble m 2 .
11. Answer: th e diagona l i s divided i n th e rati o 1 :6 . Hint : tak e
one verte x o f th e parallelogra m fo r th e pole , tw o other s fo r
the basi c points , an d expres s th e intersectio n poin t o f th e
given lin e an d th e diagona l i n tw o differen t ways .
12. (a) y = 6 , (b ) x = a, (c ) ay = bx.
13. The unio n o f th e thre e medians .
14. Let K b e th e pol e an d A, B, th e basi c points . Expres s th e
coordinates of the points D, E, F i n terms of the coordinate s
(a, b) of th e poin t C.
15. See th e solutio n t o th e previou s exercise .
Answers, Hint s an d Solution s t o Exercise s 231
16. T o both question s th e answer i s negative.

17. Expan d th e poin t K(0, 1 ) i n term s o f th e basi s E, A an d
prove tha t K 2 = — E. Therefore , thi s multiplicatio n coin -
cides with the multiplication of complex numbers (se e p. 30).
Using the trigonometric representatio n o f complex numbers ,
we obtain th e followin g answer :
E A B C D
E E A B C D
A A B C D E
B B C D E A
C C D E A B
D D E A B C
18. (a ) 0 , (b ) db( 2 — i) , (c ) 1 (not e tha t th e cub e o f th e give n
number i s —1 ) .
19. (a ) Circle of radius 5 centred a t point —3 . (b ) Perpendicula r
bisector o f the segment [—4, 2i\. (c ) Circl e (us e the identit y
\p\2 + \q\ 2 — (\p + Q\ 2 + \p — Q\ 2)/% f° r comple x number s p
and q).
20. Th e su m i n th e left-han d sid e o f th e equalit y i s equa l t o
the lengt h o f a certai n broke n lin e connectin g th e point s 0
and 5 + 5i o f the comple x plane . T o see this, observ e tha t
the su m o f th e number s x\ - f ( 1 — #2) ^ ( 1 — #3) + x 2h ••• ?
XQ + ( 1 - xio)i, ( 1 - xi) + xioi i s 5 + 5 i
21. 0° , 90°, 180°, 270°, 45°, 330°.
22. (a ) Se e Figure 2 .
(b) For example, ( r —2)(r —2 —| sin30|) = 0 or, in Cartesia n
coordinates,
(V^Ts - 2)(v^T 7 - 2 - ffi ~jCl ) = o-

23. Prov e tha t z = cos a ± i sin a.
24. Answer : 2n . Proof : simila r t o Proble m 9 .
25. I f the point z lies outside of the polygon, the n the argument s
of all differences z — ai belong to one segment o f width 180°.
Therefore, th e arguments of all numbers \/{z — a,i) als o lie in
Figure 2 . Th e curv e o f Exercis e 22(a )
one segmen t o f th e sam e width , an d thes e number s canno t

sum u p t o 0 .
26. Us e tw o translations .
27. On e possible solution is to draw the lines parallel to the sides
of th e give n triangle , throug h th e endpoint s o f th e give n
vector, thu s circumscribing th e triangl e aroun d th e vector ,
and the n mak e a parallel translatio n t o mov e the triangl e t o
its initia l position .
28. Le t th e trapezoi d b e ABCD wit h paralle l side s AB an d
CD. Translat e th e poin t B b y vector DC t o a poin t B' an d
consider th e triangl e ACB'.
29. Us e tw o reflection s i n th e side s o f th e give n angle .
30. Th e poin t symmetri c t o a vertex of the triangl e wit h respec t
to a bisecto r lie s o n th e opposit e sid e o f th e triangl e (o r it s
prolongation).
31. Rol l the angle by successive reflections fou r time s and unfol d
the reflection s o f th e ray . Th e phenomeno n occur s fo r al l
angles 90°/ n wit h intege r n.
>
32. Afte r th e rotatio n throug h 90 ° th e vectors MAi becom e th e
Licensed to Penn St Univ, consecutive side
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33. Us e rotation s throug h 60° .

34. Le t K b e th e resul t o f rotatio n o f th e poin t M aroun d th e
centre of the square which takes vertex A int o B. Sho w tha t
AK_ L M , BK _ L CM, CK 1_ _ DM, DK _ L AM.
35. Us e th e poin t symmetr y wit h respec t t o th e intersectio n
point o f th e circles .
36. Th e firs t playe r shoul d pu t th e firs t coi n i n the centr e o f th e
table, the n us e poin t symmetry .
37. Simila r t o th e proo f o f formul a (7 ) (p . 52) . Us e a rotatio n
around th e intersectio n poin t o f th e give n lin e wit h th e x-
axis.
38. Introduc e a suitabl e comple x structur e i n th e plane .
39. Answer : th e identity . Proof : us e formul a (1 2 ) (p . 56) .
40. Give n thre e number s tha t correspon d t o th e point s A, B,
C, b y formul a (1 7 ) fin d th e number s tha t correspon d t o M ,
N an d P .
41. I t i s eas y t o chec k tha t R dM o R^ o Rdp o R^ = id . Rewrit e
this conditio n i n term s o f comple x numbers .
42. (a ) RB O R.A = ^2AB ' (k ) Si o RA i s a glid e reflectio n wit h
axis AK an d vecto r 2AK, wher e K i s th e bas e o f th e per -
pendicular draw n fro m th e poin t A t o th e lin e I.
43. Conside r th e compositio n o f fiv e poin t symmetrie s i n th e
centres o f the consecutiv e side s o f the pentagon . Prov e tha t
it i s a point symmetr y wit h respec t t o on e o f the vertice s of
the pentagon .
44. Fin d the axis of the glide symmetry whic h is the compositio n
of th e thre e give n reflections . Conside r th e lin e tha t passe s
through th e give n poin t paralle l t o thi s axis .
45. Us e th e resul t o f Exercis e 37 .
46. No , becaus e eac h sho t change s th e orientatio n o f th e se t o f
three pucks .
47. Us e th e definitio n o f the argument .
48. (a ) R o Si = Si o R , (b ) R oR =
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234 Answers, Hint s an d Solution s t o Exercise s
49. (a ) Th e lin e Z i s th e perpendicula r bisecto r o f th e segmen t

AB, (b ) Th e line s Z , m an d n mee t i n on e point .
50. Th e set of all points (5-6AH-1 2Z , 3+12Jfe-6Z), (7-6&+1 2Z , 2+
12fc-6Z), (3-6Jfc+1 2Z,l+1 2fc-6Z) , (4-6*H-1 2Z,-l+1 2fc-6Z) ,
(2 - 6/ c + 1 2Z, 9 + 1 2k - 6Z) , ( 3 - 6/ c + 1 2Z , 1 0 + 1 2k - 6Z) ,
where k an d Z ar e arbitrar y integers .
51. (a ) yes , (b ) yes , (c ) no .
52. Tak e an arbitrary elemen t g G G. B y property (2) , g~ x G G.
By propert y (1 ) , g o g~l = i d G G.
53. Us e Figure 3.1 b an d th e fac t tha t a movement i s completel y
determined b y th e image s o f thre e non-collinea r points .
54. Al l thes e group s ar e different .
55. Us e th e resul t o f Proble m 21 .
56. (a ) Yes , for example , a rectangle , (b ) no . I f RA an d RB ar e
symmetries o f a certai n figure , the n Re wit h C = RA{B) i s
also a symmetry .
57. A circle , i n particula r a point , o r th e unio n o f a se t o f con -
centric circles .
58. Al l possibl e conjugation s i n th e grou p o f plan e movement s
are displaye d i n th e followin g table , which , o n th e intersec -
tion o f a ro w labelle d g an d a colum n labelle d / , contain s
the elemen t / o g o f" 1 .
Th RB &m uh
Ta Ta T
R%{*)
T
Sm(a) T
Sm(a)
pa pa n
U U
Tb(A) SM{A) U*(A)
Si S S S
Th(l) S
R%(1) Sm(l) u*(D
u? U
Th(l) U
K(D
U
Sm(l)
59. Th e multiplicatio n tabl e fo r th e grou p D 3 i s shown o n pag e

92.
60. Th e identit y transformatio n commute s wit h everything . Al l

rotations commut e betwee n themselves . Apar t fro m these ,
in th e grou p D n wit h od d n ther e ar e n o mor e commutin g
pairs. I n the grou p D n wit h eve n n w e also have commutin g
pairs o f reflection s i n mutuall y perpendicula r lines .
61. Verif y th e equalit y {f~ x)k o fk = id .
62. Th e orde r o f f k i s equal t o n/ GCD(n , k).
63. (a )
2 3 4 5 6 7 8 9 10 11 12 13 14 15
1 2 2 4 2 6 4 6 4 10 4 12 6 8
(b) (p(m) = ra(l - 1 /pi ) • • • (1 - 1 /pfc) , wher e pi, ... , p fc
are al l prim e divisor s o f m.
64. No . Som e straight line s go into themselves unde r thi s move -
ment.
65. Th e compositio n o f n rotation s R\ yield s a n identit y i f an d
only i f the tota l angl e o f rotation, na , i s equal t o a multipl e
of 360° .
66. Th e grou p generate d b y A an d B i s C5 , a cycli c grou p o f
order 5 . A i s a rotatio n b y a n angl e a = ±72 ° o r ±1 44° , B
is the rotatio n aroun d th e sam e centr e b y th e angl e —2a .
k
67. No . An y word is equivalent t o a word of the form EI 'AlUm,
where h, I, m ar e integer s betwee n 0 and 6 .
67. On e ca n tak e eithe r a suitabl e pai r o f reflection s o r a pai r
consisting o f a reflectio n an d a suitabl e rotation . Fo r ex -
ample, tw o reflection s S\ an d S 2 whos e axe s ar e adjacen t
satisfy th e definin g relation s Sf — S\ = (Si o S2)11, = id .
69. Star t b y provin g tha t non e o f th e movement s Fi ca n b e a
rotation.
70. Unde r th e assumption s o f th e exercise , prov e th e followin g
two facts : (1 ) fo r an y give n element a there ar e tw o integer s
m an d n suc h tha t a m = a n , (2 ) th e la w o f cancellatio n
holds: i f xy = xz, the n y = z.
71. (1 ) no : th e number s \/ 2 an d —\[2 ar e irrational , bu t thei r
sum i s rational. (2 ) yes, (3 ) yes, (4 ) no : th e invers e o f 3/4 i s
not binary-rational , (5 ) fo r example , th e se t o f al l number s

{ t a n n | n £ Z}.
72. x = ba~ l.
73. Chec k tha t thi s operatio n i s associative .
74. G = = {x, 1 /( 1 - x), {x - l)/x , 1 - x , 1/x, x/( x - 1 )} . Compo -
sitions o f thes e function s ar e give n i n th e followin g table :
1 x-1 1 x
X
1-x X
1-X X x-1
1 x-1 1 X
X X
1-x X
1-x X x-1
1 1 x-1 1 X
X x-1 1-X
1-x 1-x X X
x-1 x-1 1 X 1
X X
X
1-x x-1
1-x X
X 1 x-1 1
1-x 1-x x-1 X
X X 1-x
1 1 X 1 x-1
X X
1-x x-1 1-x
X X
X X 1 x-1 1
x-1 x-1 x 1-x X 1-x X
75. Fo r example , th e su m o f square s o f al l expression s foun d i n

Exercise 74 .
76. Yes .
77. Element s 5 a , £& , Sc ca n b e arbitraril y permuted . Th e tota l
number o f isomorphisms , includin g th e identity , i s 6 .
78. A direc t verificatio n o f th e definitio n o f th e isomorphism .
79. Poin t E correspond s t o th e identit y transformation , whil e
the point s A, B, C , D, K correspon d t o rotation s b y 60° ,
120°, 1 80° , 240° , 300° , respectively .
80. Th e circl e corresponds t o th e identit y transformation , whil e
the triangle an d th e square correspond t o the remainin g tw o
elements o f th e grou p C3 . Ther e ar e tw o differen t isomor -
phisms.
81. Writ e ou t th e tw o multiplicatio n tables . B y permutin g th e
rows an d column s i n on e table , mak e i t loo k lik e th e othe r
one, u p t o notatio n o f th e elements .
82. Yes . Th e correspondenc e k <- • 2k i s a n isomorphism .

83. Th e onl y pai r o f isomorphi c group s i s D\ an d C2 .
84. Considerin g th e expressio n <^(<?o<7 -1 ), prove tha t v?(# -1 ) =
h~\ I t follow s tha t ip{g- n) = ip{{g~ l)n) = (h~ l)n = h~ n.
85. (1 ) Numbe r th e vertices of the equilatera l triangle . (2 ) Con -
sider th e actio n o f thi s grou p o n th e extende d rea l lin e an d
the permutation s o f th e se t {0,1 ,00 } unde r thi s action .
86. Denotin g th e Napie r logarith m b y iV , we will have N(x\X2)
= N{ Xl) + N(x 2)-B.
87. (a ) A direc t chec k o f grou p axioms , (b ) Th e invers e imag e

of thi s grou p unde r th e mappin g cp(x) = tan x woul d b e
an additiv e grou p o f rea l number s containin g som e ope n
interval. Prov e tha t i t coincide s wit h th e whol e o f R .
88. (a ) x • y = yjx 3 - f y 3, (b ) thi s operatio n i s th e pullbac k o f
multiplication alon g th e mappin g x ^ x - 1 .
89. Th e grou p D3 ha s fou r prope r subgroups : on e o f orde r 3
and thre e o f orde r 2 .
90. Yes, th e numbe r —1 .
91. Prove tha t an y subgrou p o f Z i s generate d b y it s smalles t
positive element .
92. Check th e grou p axioms . Th e notio n o f th e quotien t grou p
(section 2 ) provide s a n easie r wa y t o prov e thi s fact .
93. No, because , fo r example , 2 - 3 = 6 = 0 .
94. Yes. T o prov e this , writ e ou t th e multiplicatio n table .
95. Every solution o f the equation x 2 = Sy 2 + 8 is also a solutio n
of th e equatio n x 2 = 3y 2 + 8 (mo d 3), whic h i s equivalen t
to x 2 = 2 (mo d 3). However , thi s las t equatio n ha s n o solu -
tions.
96. Answer : 81 .
The last tw o digits of a positive integer i s the same thin g
as it s residu e modul o 1 00 . Th e numbe r 200 3 i s mutuall y
prime wit h 1 00 , because i t i s not divisibl e b y 2 and 5 . Sinc e
0(100) = 1 0 0 - ( 1 - 1 / 2 ) - ( 1 - 1 / 5 ) = 4 0 and 200 4 mod 40 = 4 ,
2004
by Fermat' s littl e theore m w e have : 2003 = 2003 4 =
3 4 - 81 .
97. A homomorphis m fro m Z m ont o Z n exist s i f an d onl y i f m
is divisibl e b y n . Unde r thi s assumption , on e o f th e possi -
ble homomorphism s i s give n b y th e correspondenc e a »- » a,
where the ba r o n the lef t mean s the residu e clas s modulo m ,
while th e ba r o n th e righ t mean s th e residu e clas s modul o
n.
98. On e ca n ge t on e o f the followin g expressions : x, 1 — x, 1 /x,
1/(1 — #), 1 — 1 /x , x/(x — 1 ) . Compar e thi s wit h th e resul t
of Exercis e 74 .
99. Us e th e classificatio n o f plan e movement s (Theore m 4 i n
section 7) .
100. Prov e tha t ZOAC = ZBEF, denot e thi s angl e b y a an d
consider th e rotatio n o f BE aroun d E an d th e rotatio n o f
AC aroun d A throug h a .
101. Bot h assertion s follo w fro m th e fac t tha t th e determinan t
of th e produc t o f tw o matrice s equal s th e produc t o f th e
determinants o f th e tw o matrices .
102. (a ) Yes . (b ) No .
103. Th e kernel consists of all odd functions, th e image of all even
functions.
104. Conside r th e homomorphis m ip(z) — z n.
105. Conside r th e homomorphis m <p(x) = cos x 4 - isinx.
106. Thi s i s the dihedra l grou p D n.
107. I f aba = bab, the n th e element s x = ab, y = aba satisf y
x2 = y 3.
If x 2 = y 3, the n th e element s a = x _ 1 y, b = y~ 1 x2 satisf y
aba = bab.
108. Th e se t split s int o 5 orbits .

109. {-1 ,2,1 /2} .
110. {l/ 2 + i v ^ / 2 , l / 2 - n / 3 / 2 }.
111. I n bot h case s th e actio n i s transitive . Ever y edg e i s pre -

served b y 2 movements, an d ever y verte x i s preserve d b y 3
movements.
112. (a ) D, S, T , (b ) D.
113. (a ) 2 , (b ) 1 , (c ) 7 .
1 m
114. — y^nGCD(/c,m)^ where G C D stand s f or t h e greates t com -
m ^— '
mon divisor .
115. 60 .
116. 1 6 .
117. 30 .
118. (a ) 23 , (b ) 21 8 .
119. ({ 1 $) + 1 5{l) + 2(*))/3D= 1 85.
120. Her e i s on e constructio n o f a complet e invariant ; i t i s no t
tremendously elegant , bu t w e describe it for want o f a bette r
one. Th e blac k bead s spli t th e se t o f al l whit e bead s int o 4
parts, som e o f whic h ma y b e empty . Le t m b e th e numbe r
of whit e bead s i n th e bigges t part , an d n , k th e number s o f
beads i n the tw o adjacent part s suc h that n> k. Assig n th e
triple (m , n, k) t o th e give n necklace . Fo r a necklac e wit h
several bigges t parts , amon g al l th e triple s (m.n.k) choos e
the lexicographicall y bigges t one . (Actually , ther e i s onl y
one necklac e fo r whic h w e mus t worr y abou t this. ) The n
the tripl e (m , n, k) i s a complet e invarian t o f th e necklace .
121. (a ) Th e well-know n criteri a fo r th e equalit y o f tw o triangle s
(angle an d tw o adjacent sides , side and tw o adjacen t angles ,
three sides ) provid e example s o f complet e invariants .
(b) Th e ordere d se t consistin g o f the lengths o f all sides AB,
BC, CD, DA an d on e angl e ABC i s a n invariant . Thi s
invariant i s complet e o n th e se t o f conve x quadrilaterals .
122. Se e the discussio n o f finit e rotatio n group s o n pag e 79 .
123. Not e tha t a n infinit e se t o f point s o n a circl e canno t b e
discrete.
124. Angl e o f 360°/n fo r C n o r 1 80 ° /n fo r D n wit h verte x a t th e

common centr e o f rotations .
125. Th e conditio n \kn — lm\ — 1 mean s tha t th e are a o f th e
parallelogram i s 1 . Therefore , b y Pick's well-known formula ,
if P i s such a parallelogram wit h one vertex at poin t A , the n
the point s o f th e orbi t o f A tha t belon g t o P ar e onl y th e
vertices o f P.
126. Us e th e tabl e o f conjugation s (pag e 234) .
127. Le t O b e a n arbitrar y poin t o f the plane , an d S th e orbi t o f
A under th e action of the given group. Le t A b e an arbitrar y
point o f S suc h tha t th e segmen t OA doe s not contai n othe r
points o f S. Le t B b e a poin t o f S a t th e minima l distanc e
from th e lin e OA. Prov e tha t th e pai r OA, OB i s a syste m
of generators .
128. Grou p o f order 1 8 with generator s a, 6 , c and relation s a 2 =
b2 = c 2 = {ab) 3 = (be) 3 = (ca) 3 = (abc) 2 = e.
129. p4ra , p4g, p2.
130.
Cx pi
£i pm, pg, cm
c2 P2
D2 pmm, pmg, pgg, emm
c3 P3
D3 p31ra, p3ml
c4 p4
DA p4m, p4g
C6 p6
D6 p6m
131. I t i s eas y t o se e tha t a number , i f present , stand s fo r th e

maximal orde r o f a rotation , th e symbo l m ('mirror') i s fo r
a reflection , g for a glide reflection. Th e distinctio n betwee n
p an d c i s mor e subtle : th e crystallographers ' implicatio n
is primitive o r centred cell , bu t tha t doe s no t see m t o hav e
any mathematica l meanin g othe r tha n th e group s wit h a c
may hav e a cel l i n th e for m o f a n arbitrar y rhombu s (se e
page 1 60) , which leave s a layma n wonderin g wh y n o simila r
notation i s use d fo r rectangle s o r hexagons . Finally , th e
order o f m an d 1 i n th e notation s pSml an d p31 m look s
completely enigmatic .
132. Defin e a homomorphis m o f Aff(2,R) ont o GL(2,R ) an d us e
the first homomorphis m theore m (pag e 1 34) .
133. Ther e i s an affin e transformatio n tha t take s the give n trape -
zoid t o a trapezoi d wit h equa l sides .
134. Bot h groups have order 6 and ar e isomorphic t o the dihedra l
group D 3.
135. Poin t m/p, i f p ^ 0 .
136. Writ e ou t explici t formula s fo r th e compositio n an d th e in -
verse transformation .
137. B y a direc t computation , chec k tha t
XQ X-^ X^ X-^ £ 3 — X\ X 4 X\
f
X 3 — x' 2 X 4— X f2 £ 3 — £2 X 4 — X2 '
if x\ = (mxi + n)/(pxi + q).

138. A s in Exercise 74, the complete list of elements of this grou p
can b e obtained directly , takin g th e composition s o f the tw o
given function s unti l th e element s begi n t o repeat .
To prove the isomorphism , find tw o generating element s
of this grou p tha t satisf y th e definin g relation s o f the grou p
Dn (se e formul a (25 ) o n pag e 92) .
139. Th e transformatio n x »- » (mx - f n)/(px + q) ha s finite orde r
if an d onl y i f
m2 + q 2 + 2np
where a i s a n angl e measure d b y a rationa l numbe r o f de -

grees.
140. No . Usin g formul a (43) , find a n exampl e o f a translatio n
t an d a projectiv e transformatio n p suc h tha t ptp" 1 i s no t
affine.
141. B y a projective transformation , th e quadrangle ACFD ma y
be mad e int o a square . The n w e ca n introduc e a Carte -
sian coordinat e syste m adapte d t o this squar e an d solv e th e
problem b y a direc t computation . Remark. O f course , thi s
is not a very beautiful solution . I t onl y shows how to reduc e
the amoun t o f computations neede d t o solve the problem b y
brute force . Indeed , i n th e origina l settin g w e ha d 6 basi c
points describe d i n coordinate s b y a se t o f 1 2 real number s
related b y two equations. Afte r th e transformation, th e con -
figuration i s described b y only two independent parameters .
142. Defin e a homomorphis m o f GL(2,R ) ont o PGL(1 ,R ) an d
use th e first homomorphis m theore m (pag e 1 34) .
143. I t i s easie r t o construc t a triangle , circumscribe d aroun d a
given triangle , wit h side s paralle l t o th e thre e give n lines .
144. Us e a homothet y wit h centr e o n th e oute r circl e an d coeffi -
cient 3/5 .
145. Us e a homothet y wit h centr e a t th e media n intersectio n
point an d coefficien t —2 .
146. Conside r th e homothet y wit h coefficien t 1 / 2 centre d a t th e
intersection poin t o f th e thre e altitudes . Prov e tha t th e
outcircle o f th e give n triangl e goe s int o th e desire d circl e
under thi s transformation .
147. W e know tha t an y similitud e wit h coefficien t differen t fro m
1 ha s a uniqu e fixed point . Th e proble m i s t o prov e tha t
this poin t lie s insid e th e smalle r map . Suppos e tha t i t lie s
outside, dra w a lin e throug h thi s poin t tha t intersect s th e
smaller map , an d conside r it s intersectio n point s wit h th e
boundaries o f th e maps .
148. E i s the centr e of homothety tha t take s A int o B an d C int o
D.
149. Fin d th e image s o f segments MC an d PN unde r th e actio n

of th e spira l homothetie s F A(y/2,4b°) an d 2 ^ ( ^ , 4 5 ° ) , re -
spectively.
150. Apar t fro m th e inversion s centre d a t 0 , this grou p als o con -
tains al l positiv e homothetie s wit h th e sam e centre , an d i s
isomorphic t o th e multiplicativ e grou p R * o f non-zer o rea l
numbers.
151. 1 52 . Straigh t line s tha t d o no t pas s throug h th e centr e o f
the inversion , g o int o circle s tha t pas s throug h th e centre ,
and vic e versa .
153. Thi s fact i s proved i n Chapter 7 , Theorem 1 8 (see page 212).
154. Th e lengt h o f a n orbi t ca n b e 2 , 3 , 6 o r 1 2 . Se e Figur e 1 6
and th e discussio n below .
155. 1 56 . Deriv e explici t formula s fo r th e compositio n an d th e
inverse transformation .
157. (-3,0) .
159. (a ) Yes , fo r exampl e y' = 0 . (b ) No .
160. y = 2x + C , y = sinx + C , y — —l/x.
161. A famil y o f straigh t line s fo r equatio n (36 ) an d a famil y o f
hyperbolas fo r equatio n (37) .
162. Circle s centre d a t th e origin . Th e correspondin g differentia l
equation i s y' = —x/y. I t i s define d everywher e excep t fo r
the lin e y = 0 . Ther e i s no such differentia l equatio n define d
in al l th e plane .
163. y = l/(C-x).
164. v = x + y , u = x.
165. y = -x 4/2 + x 2 lo g x + Cx 2.
166. y = e x2/4(fe-x2/2dx + Cy /2
.
167. y = tan(a x + C) .
1Ifift 1
y
z 2tan(l/x + C ) x
169. Fin d n suc h tha t - ( n + 3)/( n + 4 ) = - 4 .
170. Not e tha t

1 1
-1,
q(n) + 2 n +2
whence
- *_= ^ - *.
qk(n) + 2 n +2
171. Se e Proble m 67 .
172 ( a) — = (yy' + x)Vx2 + y2
dcj) xy' — y
(b) Th e solutio n i s given implicitl y b y
(x2+y2)(C-2arctan-) = 1 .
x
173. g togs= g t+s = g s +t = g s ° gt-
174. Yes . Assumin g that th e centre is 0, the correc t parametriza -
t
tion i s given b y g t{x,y) = (e 1x ety).
175. xt = x cost — ysint, y t = xsint + y cost.
176. Thes e formula s defin e th e grou p o f spira l homotheties . It s
orbits ar e logarithmi c spirals .
177. Th e orbit s ar e straigh t line s x + y — const . Th e grou p
property follow s fro m th e Viet a theorem .
178. y' = f(y/x)i wher e / i s a n arbitrar y function .
179. y' = ^ „ i ( , wher e ^ - arcta n ^ - - I n y/x2 + y 2.

x - yf(£) x a
180. Th e functio n x 2 - f y 2 i s a universa l invarian t o f thi s grou p
action.
181. Th e functio n £ from th e answe r t o Exercis e 1 79 .
182. y = Cx 2 - x.
183. Yes .
184. Us e the grou p of homotheties wit h centr e a t th e intersectio n
point o f th e line s ax + by + c = 0 and a\x + b\y + c\ = 0 .
185. Us e th e grou p x\ = e 2tx, y\ = e ty. Answer :
y = yCx 2 — 2x.
Index
Action, 1 3 8 homogeneous, 22 3
transitive, 1 4 1 linear, 20 3
Affine
coordinates, 2 0 Euclid, 1
group,167 Euler, 89 , 1 25 , 1 7 8
transformation, 1 6 7 Extended line , 1 7 3
Alhambra, 3
Associativity, 1 1 , 7 0
Axial symmetry , 4 4 Fedorov, 4 , 1 5 8
Fermat, 1 2 5
Field o f directions , 20 0
Basis, 2 0 First homomorphis m theorem , 1 3 4
Bernoulli, 20 3 Fixed point , 1 3 9
Briggs, 1 1 6 Fundamental domain , 1 5 2
Burnside, 1 4 6
GCD, 8 9
Central symmetry , 5 0 General solution , 1 9 8
Centre o f gravity , 1 8 Generalized circle , 1 8 5
Change o f variables , 20 2 Generator, 8 6
Circular group , 1 9 0 Generators, 93 , 1 3 6
Classification Glide reflection , 6 1
of crystallographi c groups , 1 5 8 Group
of finit e plan e groups , 7 9 abstract, 1 0 8
of movements , 6 3 additive, 9 9
Commutativity, 1 1 affine, 1 6 7
Complex numbers , 3 0 arbitrary, 9 8
Complex structure , 5 0 axioms, 9 8
Composition, 5 5 commutative, 8 4
Conjugate, 8 1 cyclic, 78 , 1 0 2
C o n t a c t element , 21 0 dihedral, 7 9
Coordinates, 2 0 discrete, 1 5 1
Coset, 1 1 8 finite, 7 8
Crystallographic group , 1 5 1 finite cyclic , 8 7
free, 1 3 6
Defining relations , 93 , 1 3 6 infinite, 7 8
Determinant, 1 3 1 infinite cyclic , 8 7
Differential, 1 98 , 21 0 linear, 1 6 7
Differential equation , 1 9 7 multiplicative, 9 9
246 Index
one-parameter, 21 4 Quotient group , 1 3 2

trivial, 8 7
Ratio
Half t u r n , 5 0 cross, 1 7 0
Homogeneous space , 1 4 2 simple, 1 6 , 1 6 8
Homomorphism, 1 2 7 Reflection, 4 4
Homothety, 1 7 6 Relation, 9 3
Hyperbolic Residue class , 1 2 2
rotation, 22 5 Riccati, 20 5
Hyperbolic movement , 1 9 1 Rotation, 4 7
Hyperbolic plane , 1 9 1
Schonfliess, 4 , 1 5 8
Identity transformation , 5 8 Separating variables , 20 1
Index, 1 5 6 Similitude, 1 7 5
Infinitesimal, 1 9 8 improper, 1 8 1
Integrable i n close d form , 20 5 proper, 1 8 1
Integral curve , 20 0 Spiral homothety , 22 2
Invariant, 1 4 9 Spiral similarity , 1 7 8
complete, 1 5 0 Square grid , 7
Inversion, 21 2 Stabilizer, 1 4 4
Involution, 6 8 Subgroup, 7 6
Isomorphism, 1 06 , 1 1 5 normal, 1 3 2
Symmetry
of a differentia l equation , 21 7
Kernel, 1 3 4
of a figure , 7 7
S y m m e t r y group , 7 7
Lagrange, 1 1 8
Lie, 22 7 Transformation
Linear combination , 1 5 affine, 1 6 7
Liouville, 20 5 circular, 1 8 7
Lobachevsky, 1 9 1 linear, 1 6 8
Logarithm, 1 1 6 perspective, 1 6 9
projective, 1 6 9
Median, 1 2 , 2 0 Transformation group , 7 6
Modular group , 1 9 4 Transition o f s t r u c t u r e , 1 1 7
Monomial, 9 3 Translation, 4 1
Movement
improper, 66 Variable
proper, 52 , 6 6 dependent, 1 9 8
independent, 1 9 8
Napier, 1 1 6
Napoleon, 6 0 Wallpaper group , 1 5 1
Orbit, 1 3 8
Order
of a group , 7 8
of a n element , 87 , 1 0 2
Orientation, 6 6
O r n a m e n t a l class , 1 5 5
Parallel translation , 4 1
Pa rt i al derivative , 208 , 20 9
Permutation, 1 0 4
Point addition , 1 0
Point multiplication , 2 5
Point transformation , 20 8
Polar coordinates , 3 3
Pole, 1 0
Primitive, 1 9 9

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Licensed to Penn St Univ, University Park. Prepared on Tue Mar 12 18:16:11 EDT 2019for download from IP 132.174.254.159.

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2000 Mathematics Subject Classification. Primar y 22-01 , 54H1 5 .

For additiona l informatio n a n d u p d a t e s o n t h i s b o o k , visi t

Library o f Congres s Cataloging-in-Publicatio n D a t a

Copying an d reprinting . Individua l reader s o f thi s publication , an d nonprofi t

Chapter 1 . Algebr a o f Point s 7

Chapter 2 . Plan e Movement s 4 1

Chapter 3 . Transformatio n Group s 7 3

Chapter 4 . Arbitrar y Group s 9 7

Chapter 6 . Othe r Type s o1

Chapter 7 . Symmetrie s o f Differentia

Answers, Hint s an d Solution s t o Exercise s 22 9

The first , Russian , versio n o f thi s boo k wa s writte n i n 1 983-1 98 6 b y

Probably, th e on e mos t famou s boo k i n th e whol e histor y o f mathe -

Proof. Ever y hig h schoo l studen t know s th e standar d moder n proo f

Figure 1 . A n isoscele s triangl e

STANDARD PROOF . Le t ABC b e the given isosceles triangle (Fig -

It seem s tha t ther e i s nothin g interestin g abou t thi s theorem .

Figure 2 . Euclid' s proo f

In mediaeval England, Propositio n 1 . 5 was known under th e nam e

This exampl e show s tha t th e us e o f movement s ca n elucidat e ge -

Figure 3 . Asses' s Bridg e

important no t onl y when the y ar e studied separately . I t i s very inter -

F i g u r e 4 . Tw o ornament s fro m Alhambr a

each othe r fo r eac h o f th e tw o patterns . Th e exac t meanin g o f thes e

This boo k provide s a n elementar y introductio n int o th e theor y

actions, orbits , invariants , an d som e classificatio n problems , an d fi -

In thi s chapte r w e wil l introduc e algebrai c operations , additio n an d

F i g u r e 1 . Polygon s i n th e checkere d plan e

Problem 1 . Prove that a regular polygon different from a square

Solution. Suppose , o n th e contrary , tha t suc h a poly -

Repeating th e sam e procedur e fo r th e polygo n

Since ther e ar e onl y finitel y man y intege r point s o n

Figure 3 . I s ther e suc h a n equilatera l triangle ?

If n — 3 o r 6 , th e argumen t fail s (why?) , an d w e

Exercise 1 . Suppos e tha t th e side s o f th e square s makin g th e gri d

Figure 4 . Poin t addition

Definition 1 . W e wil l cal l th e poin t L thu s constructe d the sum of

This definitio n hold s for a n arbitrar y tripl e o f points i n the plane .

Now we can give an exact statement fo r the property of the intege r

holds fo r an y arbitrar y point s A , B, C ove r an y pol e P .

F i g u r e 5 . Associativit y o f poin t additio n

Using propert y 3° , w e ca n defin e th e difference o f tw o point s

Problem 2 . Find the sum A + B + C, where M is the intersection

Figure 6 . Su m o f vertice s o f a triangl e

Solution. Recal l tha t eac h media n i s divide d b y thei r

It i s interesting t o observ e tha t th e intersectio n poin t o f th e me -

The firs t equalit y ca n b e rewritte n a s (A + B) + Q = A + Z? , an d

Figure 7 . Chang e o f bas e poin t

Note t h a t th e poin t P doe s no t appea r i n th e left-han d side s o f

Continuing th e discussio n o f Proble m 2 , suppos e t h a t a poin t

3. Multiplyin g point s b y number s

Definition 2 . Th e product of a point A by a real number a over the

F i g u r e 8 . Multiplicatio n o f point s b y number s

In othe r words , thi s operatio n mean s tha t yo u stretc h th e vecto r

are no t integer . Fo r example , th e poin t M = | ^ 4 + \B i s alway s th e

Figure 9 . Poin t o f a segment expresse d i n terms o f endpoint s

Using poin t additio n an d multiplicatio n b y numbers , i t i s possibl e