2 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS

1. The group ring k[G]

The main idea is that representations of a group G over a eld k are
the same as modules over the group ring k[G]. First I dened both
terms.
1.1. Representations of groups.
Denition 1.1. A representation of a group G over a eld k is dened
to be a group homomorphism
: G Aut
k
(V )
where V is a vector space over k.
Here Aut
k
(V ) is the group of k-linear automorphisms of V . This also
written as GL
k
(V ). This is the group of units of the ring End
k
(V ) =
Hom
k
(V, V ) which, as I explained before, is a ring with addition dened
pointwise and multiplication given by composition. If dim
k
(V ) = d
then Aut
k
(V )

= Aut
k
(k
d
) = GL
d
(k) which can also be described as
the group of units of the ring Mat
d
(k) or as:
GL
d
(k) = {A Mat
d
(k) | det(A) = 0}
d = dim
k
(V ) is called the dimension of the representation .
1.1.1. examples.
Example 1.2. The rst example I gave was the trivial representation.
This is usually dened to be the one dimensional representation V = k
with trivial action of the group G (which can be arbitrary). Trivial
action means that () = 1 = id
V
for all G.
In the next example, I pointed out that the group G needs to be
written multiplicatively no matter what.
Example 1.3. Let G = Z/3. Written multiplicatively, the elements
are 1, ,
2
. Let k = R and let V = R
2
with () dened to be rotation
by 120

= 2/3. I.e.,
() =
_
1/2

3/2

3/2 1/2
_
Example 1.4. Suppose that E is a eld extension of k and G =
Gal(E/k). Then G acts on E by k-linear transformations. This gives
a representation:
: G Aut
k
(E)
Note that this map is an inclusion by denition of Galois group.
MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS 3
1.1.2. axioms. In an elementary discussion of group representations I
would write a list of axioms as a denition. However, they are just
longwinded explanations of what it means for : G Aut
k
(V ) to be
a group homomorphism. The only advantage is that you dont need
to assume that () is an automorphism. Here are the axioms. (I
switched the order of (2) and (3) in the lecture.)
(1) (1) = 1 1v = v v V
(2) () = ()() , G ()v = (v) v V
(3) () is k-linear G (av + bw) = av + bw v, w V, a, b k
The rst two conditions say that is an action of G on V . Actions
are usually written by juxtaposition:
v := ()(v)
The third condition says that the action is k-linear. So, together, the
axioms say that a representation of G is a k-linear action of G on a
vector space V .
1.2. Modules over k[G]. The group ring k[G] is dened to be the
set of all nite k linear combinations of elements of G:

where

k for all G and a

= 0 for almost all .

For example, R[Z/3] is the set of all linear combinations
x + y + z
2
where x, y, z R. I.e., R[Z/3]

= R
3
. In general k[G] is a vector space
over k with G as a basis.
Multiplication in k[G] is given by
_

__

_
=
_

_
where c

G can be given in three dierent ways:

c

=
a

G
a

G
a

1b

Proposition 1.5. k[G] is a k-algebra.

This is straightforward and tedious. So, I didnt prove it. But I did
explain what it means and why it is important.
Recall that an algebra over k is a ring which contains k in its center.
The center Z(R) of a (noncommutative) ring R is dened to be the set
of elements of R which commute with all the other elements:
Z(R) := {x R| xy = yx y R}
Z(R) is a subring of R.
4 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS
The center is important for the following reason. Suppose that M
is a (left) R-module. Then each element r R acts on M by left
multiplication
r

r
: M M,
r
(x) = rx
This is a homomorphism of Z(R)-modules since:

r
(ax) = rax = arx = a
r
(x) a Z(R)
Thus the action of R on M gives a ring homomorphism:
: R End
Z(R)
(M)
Getting back to k[G], suppose that M is a k[G]-module. Then the
action of k[G] on M is k-linear since k is in the center of k[G]. So, we
get a ring homomorphism
: k[G] End
k
(M)
This restricts to a group homomorphism
|G : G Aut
k
(M)
I pointed out that, in general, any ring homomorphism : R S
will induce a group homomorphism U(R) U(S) where U(R) is the
group of units of R. And I pointed out earlier that Aut
k
(M) is the
group of units of End
k
(M). G is contained in the group of units of
k[G]. (An interesting related question is: Which nite groups occur as
groups of units of rings?)
This discussion shows that a k[G]-module M gives, by restriction, a
representation of the group G on the k-vector space M. Conversely,
suppose that
: G Aut
k
(V )
is a group representation. Then we can extend to a ring homomor-
phism
: k[G] End
k
(V )
by the simple formula

_
=

()
When we say that a representation of a group G is the same as
a k[G]-module we are talking about this correspondence. The vector
space V is also called a G-module. So, it would be more accurate to
say that a G-module is the same as a k[G]-module.
Corollary 1.6. (1) Any group representation : G Aut
k
(V )
extends uniquely to a ring homomorphism : k[G] End
k
(V )
making V into a k[G]-module.
MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS 5
(2) For any k[G]-module M, the action of k[G] on M restricts to
give a group representation G Aut
k
(M).
(3) These two operations are inverse to each other in the sense that
is the restriction of and an action of the ring k[G] is the
unique extension of its restriction to G.
There are some conceptual dierences between the group represen-
tation and the corresponding k[G]-module. For example, the module
might not be faithful even if the group representation is:
Denition 1.7. A group representation : G Aut
k
(V ) is called
faithful if only the trivial element of G acts as the identity on V . I.e.,
if the kernel of is trivial. An R-module M is called faithful if the
annihilator of M is zero. (ann(M) = {r R| rx = 0 x M}).
These two denitions do not agree. For example, take the represen-
tation
: Z/3 GL
2
(R)
which we discussed earlier. This is faithful. But the extension to a ring
homomorphism
: R[Z/3] Mat
2
(R)
is not a monomorphism since 1 + +
2
is in its kernel.
1.3. Semisimplicity of k[G]. The main theorem about k[G] is the
following.
Theorem 1.8 (Maschke). If G is a nite group of order |G| = n and
k is a eld with char k n (or char k = 0) then k[G] is semisimple.
Instead of saying char k is either 0 or a prime not dividing n, I will
say that 1/n k. By the Wedderburn structure theorem we get the
following.
Corollary 1.9. If 1/|G| k then
k[G]

= Mat
d
1
(D
i
) Mat
d
b
(D
b
)
where D
i
are nite dimensional division algebras over k.
Example 1.10.
R[Z/3]

= R C
In general, if G is abelian, then the numbers d
i
must all be 1 and D
i
must be nite eld extensions of k.
6 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS
1.3.1. homomorphisms. In order to prove Maschkes theorem, we need
to talk about homomorphisms of G-modules. We can dene these to
be the same as homomorphisms of k[G]-modules. Then the following is
a proposition. (Or, we can take the following as the denition of a G-
module homomorphism, in which case the proposition is that G-module
homomorphisms are the same as homomorphisms of k[G]-modules.)
Proposition 1.11. Suppose that V, W are k[G]-modules. Then a k-
linear mapping : V W is a homomorphism of k[G]-modules if and
only if it commutes with the action of G. I.e., if
((v)) = (v)
for all G.
Proof. Any homomorphism of k[G]-modules will commute with the ac-
tion of k[G] and therefore with the action of G k[G]. Conversely, if
: V W commutes with the action of G then, for any

k[G],
we have

G
a

v
_
=

G
a

(v) =

G
a

(v) =
_

G
a

_
(v)
So, is a homomorphism of k[G]-modules.
We also have the following Proposition/Denition of a G-submodule.
Proposition 1.12. A subset W of a G-module V over k is a k[G]-
submodule (and we call it a G-submodule) if and only if
(1) W is a vector subspace of V and
(2) W is invariant under the action of G. I.e., W W for all
G.
Proof of Maschkes Theorem. Suppose that V is a nitely generated
G-module and W is any G-submodule of V . Then we want to show
that W is a direct summand of V . This is one of the characterizations
of semisimple modules. This will prove that all f.g. k[G]-modules are
semisimple and therefore k[G] is a semisimple ring.
Since W is a submodule of V , it is in particular a vector subspace
of V . So, there is a linear projection map : V W so that |W =
id
W
. If is a homomorphism of G-modules, then V = W ker
and W would split from V . So, we would be done. If is not a G-
homomorphism, we can make it into a G-homomorphism by averaging
over the group, i.e., by replacing it with =
1
n

.
MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS 7
First, I claim that |W = id
W
. To see this take any w W. Then
w W. So, (w) = w and
(w) =
1
n

1
(w) =
1
n

1
(w) = w
Next I claim that is a homomorphism of G-modules. To show this
take any G and v V . Then
(v) =
1
n

1
(v) =
1
n

=
(v)
=
1
n

1
(v) = (v)
So, gives a splitting of V as required.
1.3.2. R[Z/3]. I gave a longwinded explanation of Example 1.10 using
the universal property of the group ring k[G]. In these notes, I will
just summarize this property in one equation. If R is any k-algebra
and U(R) is the group of units of R, then:
Hom
k-alg
(k[G], R)

= Hom
grp
(G, U(R))
The isomorphism is given by restriction and linear extension.
The isomorphism R[Z/3]

= R C is given by the mapping:

: Z/3 R C
which sends the generator to (1, ) where is a primitive third root
of unity. Since (1, 0), (1, ), (1, ) are linearly independent over R, the
linear extension of is an isomorphism of R-algebras.
1.3.3. group rings over C. We will specialize to the case k = C. In that
case, there are no nite dimensional division algebras over C (Part C,
Theorem 3.12). So, we get only matrix algebras:
Corollary 1.13. If G is any nite group then
C[G]

= Mat
d
1
(C) Mat
d
b
(C)
In particular, n = |G| =

d
2
i
.
Example 1.14. If G is a nite abelian group of order n then C[G]

=
C
n
.
Example 1.15. Take G = S
3
, the symmetric group on 3 letters. Since
this group is nonabelian, the numbers d
i
cannot all be equal to 1. But
8 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS
the only way that 6 can be written as a sum of squares, not all 1, is
6 = 1 + 1 + 4. Therefore,
C[S
3
]

= C C Mat
2
(C)
This can be viewed as a subalgebra of Mat
4
(C) given by
_
_
_
_
0 0 0
0 0 0
0 0
0 0
_
_
_
_
Each star () represents an independent complex variable. In this de-
scription, it is easy to visualize what are the simple factors Mat
d
i
(C)
given by the Wedderburn structure theorem. But what are the corre-
sponding factors of the group ring C[G]?
1.4. idempotents. Suppose that R = R
1
R
2
R
3
is a product of
three subrings. Then the unity of R decomposes as 1 = (1, 1, 1). This
can be written as a sum of unit vectors:
1 = (1, 0, 0) + (0, 1, 0) + (0, 0, 1) = e
1
+ e
2
+ e
3
This is a decomposition of unity (1) as a sum of central, orthogonal
idempotents e
i
.
Recall that idempotent means that e
2
i
= e
i
for all i. Also, 0 is not
considered to be an idempotent. Orthogonal means that e
i
e
j
= 0 if
i = j. Central means that e
i
Z(R).
Theorem 1.16. A ring R can be written as a product of b subrings
R
1
, R
2
, , R
b
i 1 R can be written as a sum of b central, orthogonal
idempotents and, in that case, R
i
= e
i
R.
A central idempotent e is called primitive if it cannot be written as
a sum of two central orthogonal idempotents.
Corollary 1.17. The number of factors R
i
= e
i
R is maximal i each
e
i
is primitive.
So, the problem is to write unity 1 C[G] as a sum of primitive,
central ( orthogonal) idempotents. We will derive a formula for this
decomposition using characters.
MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS 9
1.5. Center of C[G]. Before I move on to characters, I want to prove
one last thing about the group ring C[G].
Theorem 1.18. The number of factors b in the decomposition
C[G]

=
b

i=1
Mat
d
i
(C)
is equal to the number of conjugacy classes of elements of G.
For, example, the group S
3
has three conjugacy classes: the identity
{1}, the transpositions {(12), (23), (13)} and the 3-cycles {(123), (132)}.
In order to prove this we note that b is the dimension of the center
of the right hand side. Any central element of Mat
d
i
(C) is a scalar
multiple of the unit matrix which we are calling e
i
(the ith primitive
central idempotent). Therefore:
Lemma 1.19. The center of

b
i=1
Mat
d
i
(C) is the vector subspace
spanned by the primitive central idempotents e
1
, , e
b
. In particular
it is b-dimensional.
So, it suces to show that the dimension of the center of C[G] is
equal to the number of conjugacy classes of elements of G. (If G is
abelian, this is clearly true.)
Denition 1.20. A class function on G is a function f : G X so
that f takes the same value on conjugate elements. I.e.,
f(
1
) = f()
for all , G. Usually, X = C.
For example, any function on an abelian group is a class function.
Lemma 1.21. For any eld k, the center of k[G] is the set of all

G
a

so that a

is a class function on G. So, Z(k[G])

= k
c
where
c is the number of conjugacy classes of elements of G.
Proof. If

G
a

is central then

G
a

G
a

1
=

G
a

1
The coecient of
1
on both sides must agree. So
a

1 = a

I.e., a

is a class function. The converse is also clear.

These two lemmas clearly imply Theorem 1.18 (which can now be
stated as: b = c if k = C).