323 f09 Pracprobs Sol
323 f09 Pracprobs Sol
323 f09 Pracprobs Sol
Practice Problems-1
Gallian
(p.148)
7. Find all of the left cosets of H = {1, 11} in U (30).
8. Suppose that a has order 15. Find all of the left cosets of a5 in a.
SOLUTION:
5 5 10
a = a , a , e ; the left cosets are a5 , a a5 , a2 a5 , a3 a5 , and a4 a5 .
9. Let |a| = 30. How many cosets of a4 in a are there?
10. Let a and b be nonidentity elements of different orders in a group G of order 155. Prove
that the only subgroup of G that contains a and b is G itself.
SOLUTION:
By Lagranges Theorem, the orders of a and b are divisors of |G| = 155 = 5 31, and neither
is 1, since a, b = e. Let H G contain both a and b. If either of a, b has order 155, then
obviously |H| 155, so |H| = 155, so H = G. Otherwise the orders of a and b are 5 and 31 in
some order; but then 5 and 31 both divide |H|, so again |H| 155, so again H = G.
13. Let |G| = 60. What are the possible orders for the subgroups of G?
14. Suppose that K is a proper subgroup of H and H is a proper subgroup of G. If |K| = 42
and |G| = 420, what are the possible orders of H?
SOLUTION:
We need 42 dividing |H| and |H| dividing 420, with |H| =
42, 420. The multiples of 42 that
are divisors of 420 are 42 2 = 84 and 42 5 = 210.
15. Suppose that |G| = pq, where p and q are prime. Prove that every proper subgroup of
G is cyclic.
19. Suppose |G| = n and gcd (m, n) = 1. If g G and g m = e, prove that g = e
20. Suppose H, K are subgroups of G. If |H| = 12 and |K| = 35, find |H K|.
SOLUTION:
Note that there are many groups G for which the condition holds. We dont need to investigate G. Instead, we notice that H K K and H K H. It tells us that the order
of H K divides the order of H, and the order of K. Since, gcd (|H| , |K|) = 1, we get that
|H K| = 1.
21. Suppose G is abelian of odd order. Show that the product of all of the elements of G is
the identity.
SOLUTION:
The order of any element must divide the order G, by Lagranges theorem; since G has odd
order then no element can have order 2. Hence for every x = e in G, x = x1 . Thus we can
1
1
write the product of all the elements of G as ea1 a1
1 a2 a2 ak ak = e for some appropriate
choice of a1 , a2 , , ak .
22. Suppose |G| > 1 and G has no proper nontrivial subgroups. Prove that |G| is prime.
SOLUTION:
Choose an x G \ {e}. Then x G; since G has
no proper, nontrivial subgroups, then,
x = G. Thus G is cyclic. If G is infinite, then x2 is a nontrivial proper subgroup of G, so
G must not be infinite. Thus G is a finite cyclic group. By the Fundamental Theorem of finite
cyclic groups, for each divisor k of |G| there is a subgroup of order k of |G|; thus G must not
have any divisors other than itself and 1. Hence |G| is prime, as desired.
23. Let |G| = 15. If G has only one subgroup of order 3 and only one of order 5, prove that
G is cyclic.
24. Let |G| = 25. Prove that G is cyclic or g 5 = e for all g G.
SOLUTION:
Suppose G is not cyclic. Choose any nonidentity g G. By Lagranges Theorem,
|g| | |G| = 25, so the order of g must 5 or 25. But we have assumed that G is not cyclic, so
not element can have order 25, so |g| = 5. This holds for any nonidentity g G, and certainly
e5 = e, so g 5 = e for all g G.
25. Let |G| = 33. What are the possible orders for the elements of G? Show that G must
have an element of order 3.
26. Let |G| = 8. Show that G must have an element of order 2.
27. Let G be finite and H K G be subgroups of G. Prove that |G : H| = |G : K| |K : H|.
28. Show that (Q, +) has no proper subgroup of finite index.
SOLUTION:
Suppose that H < Q has finite index |Q : H| = n. Choose any nonzero x Q. Consider the
left cosets H, x + H, 2x + H, , nx + H. These are n + 1 cosets, so since H has index n in
Q, they cannot be distinct; two of them must be equal. Say, then, that ix + H = jx + H with
i < j, so (j i) x + H = H so (j i) x H. Now, 1 j i n, so n! is a multiple of j i,
so n!x H.
x
H, so we have
Hence for every nonzero x Q, n!x H. Now for any nonzero x Q,
n!
1
(n!x) H; but this just says that every nonzero x Q is in H. Thus H = Q, contradicting
n!
the fact that H < Q. We have reached a contradiction, so there can be no H < Q with finite
index.
30. Prove that every subgroup of Dn of odd order is cyclic.
31. Let G = {(1) (12) (34) , (1234) (56) , (13) (24) , (1432) (56) , (56) (13) , (14) (23) , (24) (56)}.
a. Find the stabilizer of 1 and the orbit of 1.
b. Find the stabilizer of 3 and the orbit of 3.
c. Find the stabilizer of 5 and the orbit of 5.
33. Let |G| = pn where p is a prime. Prove that |Z (G)| =
pn1 .
SOLUTION:
Suppose |Z (G)| = pn1 and take any a G \ Z (G). Then CG (a) contains both a and Z (G),
so CG (a) is strictly larger than Z (G). That means that |Z (G)| = pn1 divides |CG (a)| = pn ,
so CG (a) = G. This works for any a G (since also CG (a) = G for every a Z (G)), so G is
abelian. But then Z (G) = G, so |Z (G)| = pn , contradicting the fact that |Z (G)| = pn1 .
There is a shorter proof.
pn
|G|
= n1 = p. But then G/Z (G) has prime
If |Z (G)| = pn1 then |G/Z (G)| =
|Z (G)|
p
order and is therefore cyclic, whence by the G/Z (G) theorem, G is abelian. Then Z (G) = G,
contradicting the fact that |Z (G)| < |G|.
60
= 15.
4
19. What is the order of the factor group (Z10 U (10)) / (2, 9)?
21. Prove that an abelian group of order 33 is cyclic.
22. Determine the order of Z Z/ (2, 2). Is the group cyclic?
SOLUTION:
Notice that the set {(n, 0) | n Z} is an infinite subset of Z Z/ (2, 2)", since (m, 0) +
(2, 2) = (n, 0) + (2, 2) iff n = m. Hence this factor group is of infinite order. We now show
that this group cannot be cyclic.
Claim. An infinite cyclic group cannot contain an element of finite order.
Proof. Suppose G is an infinite cyclic group, say G = a, and b G is an element of
finite order, say |b| = n. Then since a generates G b = a! + a +"# + a$k = ka for some
k Z, so that 0 = bn = nb = n (ka) = ank or in other words, a has finite order. But a
generates an infinite cyclic group, so this is a contradiction. This proves the claim. To show
that Z Z/ (2, 2) is not cyclic, then, we need only to produce an element of finite order. Note
that (1, 1) + (2, 2) + (1, 1) + (2, 2) = (2, 2) + (2, 2) = (0, 0) + (2, 2). This completes the
proof.
23. Determine the order of Z Z/ (4, 2). Is the group cyclic?
24. The group Z4 Z12 / (2, 2) is isomorphic to one of Z8 , Z4 Z2 , or Z2 Z2 Z2 .
Determine which one by elimination.
25. Let G = U (32) and H = {1, 31}. The group G/H is isomorphic to one ofZ8 , Z4 Z2 ,
or Z2 Z2 Z2 . Determine which one by elimination.
27. Let G = U (16), H = {1, 15}, and K = {1, 9}. Are H and K isomorphic? Are G/H and
G/K isomorphic?
28. Let G = Z4 Z4 , H = {(0, 0) , (2, 0) , (0, 2) , (2, 2)}, and K = (1, 2). Is G/H isomorphic
to Z4 or Z2 Z2 ? Is G/K isomorphic
to Z4 or Z2 Z2 ?
29. Let G = GL (2, R) H = A G | det (A) = 3k , k Z . Prove that H G.
30. Express U (165) as an internal direct product of proper subgroups in four different ways.
34. In Z, let H= 5 and K = 7.Prove that Z = HK. Does Z = H K?
35. Let G = 3a 6b 10c | a, b, c Z under multiplication and H = 3a 6b 12c | a, b, c Z
under multiplication. Prove that G = 3 6 10 whereas G = 3 6 12.
37. Let G be finite and H G. Prove that the order of gH must divide |g|.
38. Let H G, a G. If aH has order 3 in G/H and |H| = 10. What are possibilities for
|a|
SOLUTION:
First, aH has order 3 in G/H. Hence G = H aH a2 H has 30 elements and is a subgroup
of G: if we take x, y G then x = ai h1 , y = aj h2 for some i, j and h1 , h2 H. Then
1
xy 1 = ai h1 aj h2
=
=
=
=
j
ai h1 h1
2 a
ai haj [for h = h1 h1
2 H]
i j
a a h [for some h H, since H G]
aij h
and this last is an element of G . Hence a is an element of a group of order 30, so |a| divides
30.
Now |a| must be a nultiple of 3. If not, say |a| = 3k + i where i = 1 or 2. Then H = eH =
k
3k+i
H = (aH)3k+i = (aH)3 (aH)i = H k (aH)i = H (aH)i = (aH)i = ai H; so H = ai H for
a
i = 1, 2, so aH does not have order 3 in G/H, a contradiction.
Thus |a| is a multiple of 3 that divides 30. The possibilities for |a| are therefore 3, 6, 15, and
30. We can give examples of all 4 possibilities: In G = Z30 , let H = 3. Then for a = 10, 5, 2,
and 1, aH has order 3 and a has order 3, 6, 15, and 30, respectively.
39. If H G, then prove that CG (H) G.
40. Suppose G is abelian and H < G. If every element of H is a square and every element
of G/H is a square, then prove that every element of G is a square.
41. Show, by an example, that in a factor group G/H it can happen that aH = bH but
|a| =
|b|.
SOLUTION:
Take G = Z6 , H = {0, 3} , a = 1, and b = 4.
43. Suppose G is nonabelian with |G| = p3 (where p is a prime) and Z (G) = {e}. Prove
that |Z (G)| = p.
SOLUTION:
Use "G/Z Theorem."
44. If |G| = pq, where p, q are primes, prove that |Z (G)| = 1 or pq.
45. Let N G, H G and let N H. Prove that H/N G/N iff H G.
46. Let G be abelian and let H = {x G : |x| < }. Prove that every nonidentity element
of G/H has infinite order.
47. Determine all subgroups of R that have finite index.
50. Show that the intersection of two normal subgroups of G is a normal subgroup of G.
51. Let N G, H G. Prove that N H G. Give an example to show that N H need not
be a normal subgroup of G if neither N nor H is normal.
52. If N, M G prove that NM G.
53. Let N G. If N is cyclic, prove that every subgroup of N is normal in G.
SOLUTION:
r
m
k
Let N = a , H = ak , and x G. Then x ak x1 = xam x1 = (ar )k = ak H.
55. Let G be finite, H G and let x G. If gcd (|x| , |G/H|) = 1, show that x H.
SOLUTION:
gcd (|x| , |G/H|) = 1 implies gcd (|xH| , |G/H|) = 1. But |xH| divides |G/H|. Thus |xH| = 1
and therefore xH = H = x H.
56. Let G = x1 y 1 xy | x, y G .
a) Prove that G is normal in G.
b) Prove that G/G is abelian.
c) If G/N is abelian, prove that G N .
d) Prove that if H G and G N , then H is normal in G.
57. If N G and |G/N | = m, show that xm N .
SOLUTION:
Let G be a group, with N a normal subgroup such that |G/N | = m. For all x G, we have
that xN G/N . Since G/N is itself a group of order m, we know that (xN)m = e. That is
e = xm N = N . In general, for a normal subgroup N, aN = N a if and only if a N. Therefore,
we can conclude that xm N .
58. Suppose G has a subgroup of order n. Prove that the intersection of all subgroups of G
of order n is a normal subgroup of G.
59. If G is nonabelian, show that Aut (G) is not cyclic.
SOLUTION:
We know that for any group G, we have G/Z (G)
= Inn (G) Aut (G). So if Aut (G) were
cyclic, then Inn (G) would be cyclic. Thus G/Z (G) would have to be cyclic. But in this case G
would have to be abelian. This is contradiction to the given information that G is nonabelian.
60. Let |G| = pn m, where p is prime and gcd (p, m) = 1. Suppose H G with |G| = pk . If
K < G, |K| = pk , show that K H.
61. Suppose G is finite and H G. If G/H has an element of order n, show that G has an
element of order n.
SOLUTION:
Say |gH| = n. Then |g| = nt by Exercise 37, and g t = n. For the second part, consider
Z/ k.
62. Recall that a subgroup N of a group G is called characteristic if (N) = N (we denote
this by N char G)for all Aut (G). If N char G, show that N G.
64. Show that S4 has a unique subgroup of order 12.
65. If |G| = 30 and |Z (G)| = 5, what is the structure of G/Z (G)?
SOLUTION:
30
We know that |G/Z (G)| =
= 6 = 2 p, where p = 3 is a prime greater than 2 We also
5
know that any group of order2p, where p ia prime greater than 2 is isomorphic to Z2p or the
dihedral group Dp of order 2p. In the former case, since Z2p is cyclic, then G/Z (G) is cyclic,
and so G must be abelian by a theorem we proved in class, but this cannot be the case, because
in this case by the hypothesis, we have G = Z (G). In the latter case, we have G/Z (G)
= Dp .
66. If H G and |H| = 2, prove that H Z (G).
SOLUTION:
First, H contains the identity, e, and one other element, call it g. Since H is normal, for any
given a G, aH = Ha. Since ae = ea = a for all a, it must therefore be true that ag = ga for
all a, or else H would not be normal. So, all elements of H (namely e and g) commute with
all elements of G, so that H is a subset of Z (G).
67. Prove that A5 cannot have a normal subgroup of order 2.
SOLUTION:
Suppose that A5 has a normal subgroup of order 2, call it H. Then H is a subset of Z (A5 )
by 9.66. Recall from problem 5.46 that the center of Sn is trivial for all n 3. An argument
similar to the one presented there shows that A5 , a subgroup of S5 , also has a trivial center.
Since Z (A5 ) is trivial, it can have no subset of order 2, contradicting the assumption that A5
has a normal subgroup of order 2.
68. Let G be finite and let |H| be odd with |G : H| = 2. Show that the product of all the
elements of G (taken in any order) cannot belong
to H.
69. Let G be a group. If H = g 2 | g G is a subgroup of G, prove that it is a normal
subgroup of G.
SOLUTION:
2
Observe that xg 2 x1 = xgx1 .
70. Suppose H G. If |H| = 4 and gH has order 3 in G/H, find a subgroup of order 12
in G.
SOLUTION:
We begin with the following general result.
Proposition. Let G be a group and H G. For any g G the set
gi H
K=
iZ
is a subgroup of G.
Proof 1: We use the one-step subgroup test. We begin by noting that K = since H K
The conclusion of the problem now follows easily. Since gH has order 3, the cosets H, gH
and g 2 H are distinct, and any other coset of the form g i H is one of these. Therefore
g i H = H gH g 2 H
iZ
and the latter set contains exactly 12 elements since |H| = 4. The proposition tells us this set
is a subgroup of G, so were finished.
71. Prove that A4 Z3 has no subgroup of order 18.
(p.210)
9. Prove that the mapping from G H G given by (g, h) g is a homomorphism.
What is the kernel? This mapping is called the projection of G H onto G.
10. Let G be a subgroup of some dihedral group. For each x G, define
+1 if x is a rotation,
(x) =
1 if x is reflection.
Note that f is onto, since if a + kZ Z/kZ, then there is an equivalence class a + nZ Z/nZ
for which f (a + nZ) = a + kZ.
f is a homomorphism, since f ((a + nZ) + (b + nZ)) = f ((a + b) + nZ) = (a + b) + kZ =
(a + kZ) + (b + kZ) = f (a + nZ) + f (b + nZ) .
ker f = {a + nZ | f (a + nZ) = 0 + kZ}
= {a + nZ | a 0 (mod k)}
= k
Hence, by the First Isomorphism Theorem, Zn / k
= Zk .
13. Prove that (A B) / (A {e})
B.
=
SOLUTION:
The mapping : A B B which is given by (a, b) = b is a homomorphism form A B
onto B with kernel
ker =
=
=
=
{(a, b) A B | (a, b) = eB }
{(a, b) A B | b = e}
{(a, e) | a A}
A {e}
23 + ker
23 + {0, 10, 20}
{23 + 0, 23 + 10, 23 + 20}
{3, 13, 23}
We know that |a| divides 20 which is the order of 1 Z20 . We know that Im () Z8 , and
so by Lagranges theorem, |a| divides the order of |Im ()|, so |a| | 8. Hence |a| divides both 8
and 20, and so there are 3 possibilities for |a|, namely 1, 2, 4. This gives the four possibilities
a = 0, 2, 4, 6 Z8 , which already means that there are no homomorphisms from Z20 onto Z8 ,
since for any of these as a + a + + a will never be odd. On the other hand, each of the 4
possibilities gives rise to a homomorphism, by inspection of the following explicit formulas:
21. If is a homomorphism from Z30 onto a group of order 5, determine ker .
SOLUTION:
The order of ker is |ker | = 5.
22. Suppose : G G is an epimorphism and that G = 8. Prove that has an element
of order 8. Generalize.
SOLUTION:
It is given that G has an element of order 8. Let us call this element g. So we get |g| = 8.
Since is an onto homomorphism, there is an element g G so that (g) = g. By a theorem
we know that | (g)| = |g| divides |g|. This means 8 divides |g| because |g| = 8. So we obtain
that |g| = 8t for some positive integer t. Then if we consider the element g t in G, we get
t 8
g
= g t g t g t (8 copies of t)
= g 8t
= e.
SOLUTION:
For each k with 0 k n1, the mapping : 1 (1) = k determines a homomorphism.
28. Suppose : S4 Z2 is a homomorphism. Determine ker . Determine all homomorphisms from S4 to Z2 .
29. Suppose G is finite and : G Z10 is a homomorphism. Prove that G has normal
subgroups of indexes 2 and 5.
SOLUTION:
By the First Isomorphism Theorem, we have G/ ker
= Im Z10 . So the possibilities for
the order of G/ ker are |G/ ker | = 1, 2, 5, or 10.
30. Suppose : G Z6 Z2 is an epimorphism and |ker | = 5.
31. Suppose : U (30) U (30) is a homomorphism with ker = {1, 11}. If (7) = 7,
find 1 (7).
SOLUTION:
We have
1 (7) = 7 ker () = 7 {1, 11} = {7, 77 mod 30} = {7, 17}.
32. Find a homomorphism : U (30) U (30) with ker = {1, 11} and (7) = 7.
SOLUTION:
First U (30) = {1, 7, 11, 13, 17, 19, 23, 29}. Then 1 (1) = ker = {1, 11} and 1 (7) =
{7, 17} . Next, what is 1 (11)? We know that 1 (11) {1, 7, 11, 17} = . We know that
19has order 2 in U (30). The orders of order untouched elements are |13| = 4 , |17| = 4, |19| =
2, |23| = 4, |29| = 2. Since the order of 19 is not divisible by 4, there are two choices for (19).
Either (19) = 19, or (19) = 29. In the first case,1 (19) = 19 ker = 19 {1, 11} = {19, 29}
and so in the second case (19) = 29, we have 1 (29) = 29 ker = 29 {1, 11} = {29, 19}
33. Suppose : U (40) U (40) is a homomorphism with ker = {1, 9, 17, 33}. If
(11) = 11, find 1 (11).
SOLUTION:
We have
1 (11) = 11 ker
= 11 {1, 9, 17, 33}
= {11, 19, 27, 3}
34. Find a homomorphism : U (40) U (40) with ker = {1, 9, 17, 33} and (11) = 11.
SOLUTION:
It is possible that nothing maps to 11. However, suppose that (a) = 11. Then any element
of a ker = {a, 19a, 17a, 33a} also maps to 11 (where the multiplication is in U (40).
35. Prove that the mapping
: Z Z Z
(a, b) (a, b) = a b
is a homomorphism. Determine ker . Describe 1 (3).
SOLUTION:
Since
((a, b) + (c, d)) =
=
=
=
it follows that is a homomorphism.
((a + c, b + d))
(a + c) (b + d)
ab+cd
(a, b) + (c, d) , (a, b) , (c, d) in Z Z
{(a, b) | (a, b) = 3}
{(a, b) | a b = 3}
{(a, b) | a = b + 3}
{(a + 3, a) | a Z} .
: C C
x (x) = x6
is a homomorphism. What is ker ?
SOLUTION:
Since (xy) = (xy)6 = x6 y 6 = (x) (y) , x, y C , is a homomorphism. Next
ker = {x C | (x) = 1}
= x C | x6 = 1
= ei/6 = cos + i sin
6
6
We will use the one-step subgroup test to prove that H is indeed a subgroup of G. First of
all, H = ; since (e) = e = (e) implies that e H. Now, if a, b H then
ab1 = (a) b1 = (a) (b)1 = (a) (b)1 = (a) b1 = ab1
implying that ab1 H. Therefore H is a subgroup of G.