REPRESENTATION THEORY OF FINITE GROUPS – FOR
MD131
JOHN BOURKE
0. Introduction
• You will be familiar with group actions on sets G → End(X). In
this course we deal with groups acting linearly on vector spaces G →
Gl(V ), which amounts to viewing group elements as matrices/linear
transformations.
• Primarily, we will look at complex vector spaces and finite dimensional
representations of finite groups – there is a very nice structure theory
for these.
• The aim is to have lots of examples – note: in the future, I would like
to add more examples and pictures.
• We give a thorough fairly category theoretic treatment of tensor products
and homs of representations, and their relationship – using this to
understand characters of groups.
• We then classify the irreducible representations of the symmetric group.
• At the end, we touch on Hopf algebras and their representations, and
observe that this more general setting captures the nice tensor-hom
structure of representations of groups.
• Books that I am using:
(1) Representations and characters of groups by James and Liebeck.
Very gentle book that we used at Warwick when I was there. Lots
of examples.
(2) Representation theory by Fulton and Harris for the approach to
the inner product of characters.
(3) The symmetric group – representations, combinatorial algorithms,
and symmetric functions by Sagan for the stuff about the symmetric
group.
(4) Quantum groups: a path to current algebra by Street for the Hopf
algebra stuff.
Remark 0.1. The students had all taken a course in category theory, so
in the end I made the course more category theoretic than I had originally
planned.
Date: May 20, 2019.
2000 Mathematics Subject Classification. Primary: 18D05, 18C10, 55U35.
1
2 JOHN BOURKE
1. Basics of representation theory
1.1. Matrix representations. Let F be a field. By Gld(F) we mean the
group of invertible d-dimensional matrices, with group operation given by
matrix multiplication.
Definition 1.1. Let G be a group. A d-dimensional matrix representation
of G over F is a homomorphism of groups ρ : G → Gld(F).
In other words, for each g ∈ G we have an invertible matrix ρ(g), and
these satisfy ρ(g)ρ(h) = ρ(gh) and ρ(1) = I, the identity matrix.
Example 1.2. The trivial representation of a group G → Gl1(F) sends
each element to (1).
Example 1.3. Let Cn = g, gn
= 1 be the cyclic group of order n. We
will calculate all the complex 1-dimensional representations of Cn – that is,
homomorphisms ρ : Cn → C. Letting ρ(g) = c we must have ρ(gn
) = cn
= 1
so that c is an n-th complex root of unity. There are precisely n of these
– cos(2kπ/n) = isin(2kπ/n) for k = 0, n − 1 – and so there are exactly n
such representations. For instance, in the case C4 we have {1, i, −1, −i} are
the roots of unity.
Example 1.4. Let D8 be the dihedral group a, b : a4
= b2
= 1, b−1
ab =
a−1
. This group has 8 elements and describes the symmetries of the square
– which are generated by a rotation of order 4 and a reflection.
Let
A =
0 1
−1 0
B =
1 0
0 −1
be the matrices representing such a rotation, and a reflection through the
x-axis. Now defining ρ : Cn → Gl2(F) by a, b → A, B we obtain a matrix
representation of Cn upon checking that A4
= B2
= I and B−1
AB = A−1
.
Writing down the 8 matrices that you get, we see that no two are the same –
equivalently, ker(ρ) = 1 – which is to say that ρ is a faithful representation.
The following is easy.
Lemma 1.5. If ρ : G → Gln(F) is a representation and T ∈ Gn(F) then
the assignment G → GlnF : g → T−1
ρ(g)T is a representation.
Proof. T−1
ρ(gh)T = T−1
ρ(g)ρ(h)T = T−1
ρ(g)TT−1
ρ(h)T and T−1
ρ(1)T =
T−1
IT = T−1
T = I.
We can get lots of new examples like this.
Example 1.6. Consider D8 again, and its representation from Example 1.4.
Choose
T = 1/
√
2
1 1
i −i
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 3
to diagonalise A. We get
T−1
AT =
i 0
0 −i
T−1
BT =
0 1
1 0
This yields another (complex) matrix representation of D8.
Definition 1.7. Two matrix representations ρ, σ : G Gld(F) are equivalent
if there exists T ∈ Gld(F) such that T−1
ρ(g)T = σ(g) for all g ∈ G.
Remark 1.8. It is easy to see that this is an equivalence relation on the
set of d-dimensional matrix representations over F.
1.2. G-modules. Matrices correspond to linear transformations, up to a
choice of basis. For V be a vector space over F let Gl(V ) denote the group
of invertible linear transformations of V .
Definition 1.9. Let G be a group. A G-module/representation of G is a
homomorphism of groups ρ : G → Gl(V ) for some V .
Equivalently, a G-module is specified by
• a vector space V ,
• for each g ∈ G and v ∈ V an element ρ(g)(v) ∈ V , which we usually
just write as g.v or even gv, such that:
• g(v + w) = gv + gw and g(λv) = λ(gv) for λ ∈ F,
• (gh)v = g(h(v)) and 1v = v.
As we know from first year linear algebra, any choice of basis B for
a d-dimensional vector space V induces an isomorphism of groups [−]B :
Gl(V ) ∼= Gld(F). This sends a linear transformation X : V → V to the corresponding
d-dimensional matrix [X]B. Since [−]B is a group isomorphism,
sending a G-module ρ to the composite group homomorphism
[−]B ◦ ρ : G → Gl(V ) → Gld(F)
gives rise to a bijection between G-module structures on V and d-dimensional
matrix representations of G. Let us record this.
Proposition 1.10. For a d-dimensional vector space V , there is a bijection
between G-module structures on V and d-dimensional matrix representations
of G. Given a G-module ρ the corresponding matrix representation [ρ]B has
value [ρ(g)]B at g ∈ G.
Remark 1.11. Of course V = kd
has a standard basis – we typically view
a matrix representation as a representation with respect to this basis via
g.v = ρ(g)(v) where v ∈ kd
.
In particular, G-modules of finite degree and matrix representations of
G are essentially the same thing. As for classical linear algebra, the use
of vector spaces rather than just matrices allows a tidier development of
4 JOHN BOURKE
the theory, and we will usually work with G-modules as opposed to the
corresponding matrix representations.
Definition 1.12. Let G be a group and V, W be G-modules over F. A
homomorphism of G-modules is a linear transformation θ : V → W such
that θ(g.v) = g.θ(v) for all g ∈ G, v ∈ V . It is an isomorphism of G-modules
if θ : V → W is an invertible linear transformation.
As for other kinds of algebraic structures, we only really care about
G-modules up to isomorphism. The following exercise expressed the relationship
between isomorphism of G-modules and equivalence of matrix
representations.
Exercise 1.13. Let (V, ρ) and (W, σ) be G-module structures, and B and C
bases of V and W of the same cardinality d. Then V and W are isomorphic
as G-modules if and only if the matrix representations [ρ]B and [σ]C are
equivalent.
Remark 1.14. It is routine to see that if θ : V → W and ϕ : W → U are
homomorphisms of G-modules then so is the composite linear transformation
ϕ ◦ θ : V → U, and the identity 1 : V → V is certainly a homomorphism
of G-modules. It follows that G-modules and their homomorphisms form
a category ModF (G), which comes equipped with a forgetful functor U :
ModF (G) → V ectF .
Example 1.15. Let G be a group acting on a set X, and let FX be
the free F-vector space with basis X – that is, the elements of FX are
F-linear combinations λ1x1 + . . . + λnxn where the xi are elements of X.
Thus the elements of X form a natural basis for FX. We can define a
G-module structure on FX by defining the linear transformations g.− :
FX → FX by extending the functions g.− : X → X linearly; that is,
g(λ1x1 + . . . + λnxn) = λ1gx1 + . . . + λngxn. This is clearly a G-module.
G-modules arising in this way are called permutation representations.
Example 1.16. For instance, D8 acts on the four corners {w, x, y, z} of
the square
x
w z
y
so that we obtain a D8-module with basis {w, x, y, z}. Since the rotation a of
order 4 permutes these to x, y, z, w the matrix representation ρ(a) ∈ GlF (4)
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 5
has value 



0 0 0 1
1 0 0 0
0 1 0 0
0 0 1 0




Example 1.17. The regular G-module k[G] is obtained, using the above
construction, from the left action of G on itself – that is, by g.h = gh. In
particular k[G] is the vector space with basis G. For g ∈ G and λ1g1 + . . . +
λngn ∈ k[G] we define
g(λ1g1 + . . . + λngn) = λ1gg1 + . . . + λnggn.
This representation is of central importance, since for k = C, it contains all
irreducible modules as submodules.
Example 1.18. Consider C3 = g|g3
= 1 . Its regular representation, with
standard basis g, g2
, g3
= e , consists of the three matrices
ρ(g) =


0 0 1
1 0 0
0 1 0

 ρ(g2
) =


0 1 0
0 0 1
1 0 0

 ρ(e) =


1 0 0
0 1 0
0 0 1


1.3. Other perspectives – algebras, modules and categories. The
regular G-module k[G] has, in fact, more structure than a k-vector space –
it is a k-algebra. A k-algebra is a ring (A, ., 1.+, 0) together with an element
λa for each λ ∈ k such that
• the abelian group (A, +, 0) is a k-vector space;
• the multiplication a.b is bilinear: that is, it is linear on the right
a.(b+c) = a.b+a, c and a.λb = λa.b and similarly on the left (a+b).c =
ab + ac and λa.b = λ(a.b).
A homomorphism of k-algebras ρ : A → B is one that preserves all of the
structure.
In the case of kG we have defined the k-vector space structure already.
The multiplication is given by extending the group multiplication bilinearly,
that is:
(Σg∈Gλgg).(Σh∈Gθhh) = Σg,h∈Gλgλh(gh)
and it has unit e.
If V is a k-vector space then the set End(V ) of (not-necessarily invertible)
linear transformations from V to itself is a k-algebra, where the
multiplication is given by composition of linear transformations.
A representation of a k-algebra A on V as a homomorphism of k-algebras
ρ : A → End(V ).
Proposition 1.19. Representations of the k-algebra k[G] are in bijection
with representations of the group G.
6 JOHN BOURKE
Proof. Let ρ : k[G] → End(V ) be a representation. Then for each g ∈ G
we have ρ(g) ∈ End(V ). Note that since ρ preserves multiplication ρ(g) is
invertible; thus ρ(g) ∈ Gl(V ) and we have a group representation ρ|G : G →
Gl(V ) by restriction. Since the linear map ρ is determined by its value on the
basis G, we observe that the map sending algebra representations to group
representations is injective. To see that it is surjective, let ρ : G → Gl(V ) be
a group representation. Then each ρ(g) ∈ End(V ). Since k[G] is the vector
space with basis G the function ρ : G → End(V ) extends uniquely to a
linear transformation ρ : k[G] → End(V ), defined by ρ(λ1g1 +. . .+λngn) =
λ1ρg1 + . . . + λnρgn. This preserves the k[G]-algebra structure since ρ
preserves the group structure – and restricting it to a group representation
obviously gives back ρ : G → Gl(V ), as required.
In this way, we see that the study of representation of algebras generalises
that of representations of groups. Some of the results developed herein for
group representations can be developed for more general algebras, but we
will refrain from this generalization, for the sake of simplicity.
Proposition 1.20. Representations of a k-algebra A are precisely representations
of its underlying ring UA – that is, left UA-module structures.
Proof. Consider an A-module – i.e. algebra homomorphism ρ : A →
End(V ). This is, in particular, a ring homomorphism ρ : A → End(V ) →
Ab(V, V ) – thus V is a UA-module. On the other hand suppose V is a
UA-module. That is, we have a ring homomorphism θ : A → Ab(V, V ).
Now we have a ring homomorphism .1 : k → UA : λ → λ.1 and the
composite ring homomorphism k → UA → Ab(V, V ) shows that V is a
k-vector space – with action λ.v = θ(λ.1).v. In fact, considering V with
this vector space structure, the function θ : A → End(V ) is an algebra
homomorphism since it satisfies θ(λ.a) = θ((λ.1).a) = θ(λ.1)θ(a) = λ.θ(a).
The two constructions are inverse.
1.4. A categorical point of view. Given a group G let BG be the
category with one object • and BG(•, •) = G. Then a functor ρ : BG →
Set is precisely a group action – that is, we have a set ρ(•) = X and
functions ρ(g) : X → X for each g ∈ G satisfying ρ(g)ρ(h) = ρ(gh) and
ρ(e) = idX. Indeed the functor category [BG, Set] is exactly the category of
G-sets and equivariant maps – the natural transformations capture precisely
the equivariance. The regular action of G on itself is the representable
BG(•, −) : BG → Set, which sends g ∈ G to the function g.− : G → G.
Similarly, a functor ρ : BG → Vect from BG to the category of vector
spaces is precisely a G-representation. Indeed the functor category
[BG, Vect] is exactly the category of G-modules – that is, the natural
transformations between functors capture the morphisms of G-modules.
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 7
The forgetful functor U : Vect → Set has a left adjoint k[−] : Set →
Vect, which sends the set X to the vector space k[X] with basis X. In
fact, given a group action ρ : BG → Set the corresponding permutation
representation of G is simply the composite functor k[−] ◦ ρ : BG →
Set → Vect where k[−] : Set → Vect. In fact we get an adjunction
k[−]∗ U∗ : [BG, Set] [BG, Vect] by composition in this way.
We can also understand representations of k-algebras from the categorical
point of view. The tensor product ⊗ of k-vector spaces equips the category
Vect of k-vector spaces admits a symmetric monoidal closed structure,
whose unit is k. A k-algebra A is exactly a monoid in the monoidal category
(Vect, ⊗, I). The left adjoint F : Set → Vect is strong monoidal – in
particular, F(X × Y ) ∼= FX ⊗ FY . It follows that it takes monoids to
monoids – thus a group G is sent to a k-algebra k[G]. In fact, k[G] inherits
Hopf algebra structure from the group structure on G – we will come back
to this at the end of the course.
2. Irreducible G-modules
2.1. Reducibility and irreducibility.
Definition 2.1. Let V be a G-module. A vector subspace W ⊆ V is a
submodule if v ∈ W =⇒ gv ∈ W for all v ∈ V, g ∈ G.
Example 2.2. Each G-module V has two standard submodules – V itself
and the zero module {0}. These two submodules are called trivial.
Example 2.3. Consider the regular G-module k[G]. If G is finite, so that
G = {g1, . . . , gn} then these elements form a basis of k[G]. Then we can
consider the 1 − d subspace k[g1 + . . . + gn] of k[G] spanned by g1 + . . . gn.
then
g(λ(g1 + . . . + gn)) = λ(gg1 + . . . + ggn) = λ(g1 + . . . + gn)
since left multiplication by g permutes the elements of G. Thus k[g1+. . .+gn]
is a submodule of k[G].
Example 2.4. More generally, if G acts on a finite set X = {x1, . . . , xn},
then the corresponding permutation representation k[X] has submodule
k[x1 + . . . + xn]. For instance, the symmetric group Sn acts naturally on
{1, . . . , n} so that k[1 + . . . + n] ⊆ k[1, . . . , n] is an Sn-submodule.
Definition 2.5. A G-module is reducible if it contains a non-trivial submodule.
A G-module which is non-zero and not reducible is said to be
irreducible.
Remark 2.6. A matrix representation ρ : G → Gld(k) is said to be
reducible if the corresponding G-module g.v = ρ(g)v is reducible. What
does this mean in matrix terms?
8 JOHN BOURKE
It is reducible iff kd
has a G-submodule U < kd
. The subspace U =
u1, . . . um extends to a basis B = u1, . . . , um, v1, . . . , vl of kd
. Let T be
the base change matrix from the standard basis to B. In this basis the
corresponding matrix representation [g]B = T−1
ρ(g)T has the form
[g]B =
A(g) B(g)
0 C(g)
where each A(g) is of dimension m < d.
In this way, we see that a matrix representation is reducible if and only
if it is equivalent to matrix representation admitting a block decomposition
of the above form.
Example 2.7. Consider again the matrix representation of D8 given by
the rotation and reflection.
A =
0 1
−1 0
B =
1 0
0 −1
Since this is a 2-dimensional representation, a non-trivial submodule must
consist of a 1-d subspace v such that gv = λv ⊆ v for each g ∈ G – thus
v must be an eigenvector for both A and B.
The rotation matrix has no real eigenvectors – it doesn’t preserve any
straight lines, so the representation is irreducible over R. The reflection B
through the x-axis has (1, 0) and (0, 1) as eigenvectors.
The rotation A does have complex eigenvalues +i, −i, which we can find
using the characteristic polynomial, and associated eigenvectors (1, i) and
(1, −i). But A and B have no common eigenvectors – thus the representation
is also irreducible over C.
Alternatively, consider this as a representation of C2
given by a(x, y) =
(y, −x) and b(x, y) = (x, −y). Then if we have an eigenvector (α, β),
b(x, y) = (x, −y) forces α or β to be 0. But then a(x, y) = (−y, x) forces
the other one to be zero – so no 1-d subspace.
2.2. Kernels, images and direct sums. The following is routine.
Proposition 2.8. Let θ : V → W be a homomorphism of G-modules. Then
ker(θ) ≤ V and im(θ) ≤ W are G-submodules.
• Let U, V ≤ W be subspaces of a vector spaces. The sum U + V ≤ W
is the subspace {u + v : u ∈ U, v ∈ V }. We say that it is a (internal)
direct sum and write
W = U ⊕ V
if each element of W is of the form u + v for unique u ∈ U, v ∈ V . This
is equivalent to saying that U + V = W and U ∩ V = {0}.
• If U, V ≤ W are G-submodules then U + V is a G-submodule since
g(u + v) = gu + gv. In this case, if W = U ⊕ V , we call it a direct sum
of G-submodules.
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 9
Definition 2.9. A projection/idempotent is a homomorphism of vector
spaces p : V → V such that p2
= p.
Lemma 2.10. If p : V → V is a projection then V = ker(p)⊕im(p). Each
direct sum of subspaces arises from a unique projection in this way.
Proof. Write v = pv + (pv − v). This is unique since if v ∈ ker(p) ∩ im(p)
then pv = 0, but on the other hand since v = px we have pv = v, so that
v = 0.
Given a direct sum V = U ⊕ W the associated projection p : V → V is
defined by p(u + v) = u.
Corollary 2.11. If the G-module morphism p : V → V is a projection,
then V = ker(p) ⊕ im(p) is a direct sum of G-submodules.
Proof. Since ker(p) and im(p) are themselves G-submodules this follows
from the above.
Theorem 2.12 (Maschke’s theorem). Let G be a finite group and k a field
such that char(K) does not divide |G| – e.g. if k is infinite. If V is a
G-module and U ≤ V a proper G-submodule, then there exists W ≤ V with
V = U ⊕ W.
Proof. Firstly, since U is a subspace of V we can choose linearly independent
vectors to extend to a basis where V = U ⊕ W0 – this corresponds to giving
the projection p : V → V (meaning p2
= p) such that im(p) = U and
ker(p) = W0. (Of course, p need not itself be a homomorphism of Gmodules,
unless W0 is a G-submodule.)
What we will do is modify p to a homomorphism q of G-modules with
image U and satisfying q2
= q – then ker(q) ≤ V will a G-submodule and
V = im(q) ⊕ ker(q) = U ⊕ ker(q) as required.
For each g ∈ G we consider the linear map pg : v → g−1
p(gv) obtained
by conjugating p with the action of g and its inverse. As a composite of
three linear maps each pg is linear. We then define
q =
1
|G| g∈G
pg
as the “average of these maps”. Let us check that q is a homomorphism.
(2.1) q(hv) =
1
|G| g∈G
g−1
p(g.hv)
Since each element of G can be written uniquely as gh−1
for g ∈ G, this is
equally the sum
1
|G| g∈G
(gh−1
)−1
p((gh−1
).hv) = h.
1
|G| g∈G
g−1
p(gv) = hq(v)
10 JOHN BOURKE
where we use that V is a G-module in the first equation.
Thus ker(q) ≤ V is a G-submodule of V .
For surjectivity of q consider v ∈ U. Then
q(u) =
1
|G| g∈G
g−1
p(g.u) =
1
|G| g∈G
g−1
(g.u) =
1
|G| g∈G
u =
|G|
|G|
u = u
where in the second equality we use that gu ∈ U and hence is invariant
under p.
This gives one direction of im(q) = U – for the inclusion im(q) ⊆ U we
use that, since p takes its image in U, each g−1
p(gv) belongs to U.
Since qv ∈ U and qu = u for u ∈ U we get that q2
= q.
Then by the lemma, we have that V = ker(q) ⊕ im(q) = ker(q) ⊕ U.
Remark 2.13. The matrix version of Maschke’s theorem says that if a
matrix representation ρ : G → Gld(k) is equivalent to one with matrices of
the form
[g]B =
A(g) B(g)
0 C(g)
where each A(g) is of fixed dimension m < d, then it is equivalent to one
with matrices of the form
[g]B =
A(g) 0
0 B(g)
for the same dimension m < d.
Exercise 2.14. If F = Zp the cyclic field of characteristic p, show that
there is a counterexample to the above. Add detail.
Example 2.15. Consider the permutation module of the symmetric group
S3 acting on V = C[v1, v2, v3]. We have the S3-submodule U = v1 +v2 +v3 .
Let us use the formula of Maschke’s theorem to find a S3-submodule W
with V = U ⊕ W.
Firstly let V = U ⊕ W0 for W0 = v1, v2 . Then the corresponding
projection p : V → V onto U sends v1 and v2 to 0 and v3 to v1 + v2 + v3.
Observe that (12)p(12)v1 = (12)p(v2) = 0. Also (132)p(123)v1) = 0,
whilst (23) and e also result in 0. On the other hand (132) and (13) result in
v1 +v2 +v3. Thus v1 → 2/6(v1 +v2 +v3) = 1/3(v1 +v2 +v3), whilst it is easy
to see that the same is true for v2 and v3 – thus θ : vi → 2/6(v1 + v2 + v3)
is the resulting homomorphism. Then V = U ⊕ ker(θ). Then ker(θ) =
v1 − v2, v3 − v2 and this is the complementary S3-submodule.
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 11
3. Decomposing G-modules
Theorem 3.1. Let G be a finite group and k a field such that char(k) does
not divide |G| – e.g. if k is infinite. Then each finite dimensional G-module
admits a decomposition
V = V1 ⊕ . . . ⊕ Vn
as a direct sum of irreducible G-submodules.
Proof. It is by induction on the dimension of the vector space V . If
dim(V ) = 1 it is trivial since each 1-module is irreducible. Otherwise
suppose it is true for dimensions lower than that of V . If V is irreducible
are done – otherwise it has a non-trivial submodule U, so that by Maschke’s
theorem V = U ⊕ W. Then since dim(U), dim(W) < dim(V ) we get
V = (U1 ⊕ . . . Uk) ⊕ (V1 ⊕ . . . ⊕ Vl) as required.
Remark 3.2. It is possible to prove, at this level of generality, that the
above decomposition is unique in the following sense: namely,
• if V = W1 ⊕ . . . ⊕ Wm is a direct sum of irreducible submodules, then
n = m and there is a bijection ρ : n → n such that Vi
∼= Wρi for each
i ∈ {1, . . . , n}.
This is an instance of the Jordan-Holder theorem. In the case that k = C
we will prove a stronger result about the form of this decomposition, given
in Theorem 4.5 below, which implies the above uniqueness.
Lemma 3.3 (Schur’s lemma). Let θ : V → W be a homomorphism of
irreducible G-modules.
(1) Then θ = 0 or θ is an isomorphism.
(2) If k is algebraically closed and V = W of finite dimension then there
exists λ ∈ C such that θ(v) = λv for all v ∈ V .
Proof. If θ = 0 then ker(θ) = 0 and im(θ) = 0. Then since V is irreducible
ker(θ) = V and since W is irreducible im(θ) = W. Therefore θ is both
injective and surjective: an isomorphism.
For the second part – if θ = 0 it is trivial. Since k is algebraically closed θ
must have an eigenvalue – i.e. the polynomial det(θ −λ.Id) = 0 over k must
have a solution. Hence there exists λ = 0 and v non-zero with θ(v) = λ.v.
Therefore the G-module map θ − λ.Id : V → V has a non-zero kernel, and
so its kernel must be the entirety of V . Therefore θ = λ.Id.
Proposition 3.4. The regular module k[G] is the free G-module on the set
with one element 1.
Proof. Let 1 = {•} be the singleton set – we have the map e : 1 → k[G] :
• → e. Let V be a G-module and v : 1 → V : • → v. We must show that
12 JOHN BOURKE
there exists a unique G-module map θ : k[G] → V such that the triangle
k[G]
θ
!!
1
e
OO
v
// V
commutes. This triangle says that θ(e) = v. If θ is to be a homomorphism
we must have θ(g.e) = g.θ(e) = v – thus we4 are forced to define θ(g) = e –
since the elements of G form a basis for the vector space k[G] there is then
a unique linear transformation θ : k[G] → V satisfying θ(g) = g.v – it sends
λ1g1 + . . . + . . . λngn to λ1(g1.v) + . . . + λn(gn.v). It is clearly a G-module
map.
Corollary 3.5. Let G be a finite group and k a field such that char(k)
does not divide |G| – e.g. if k is infinite. Each short exact sequence of
G-modules splits.
Proof. Consider
0 // U
j
// V
k // W // 0
short exact. By Maschke’s theorem V = im(U) ⊕ W∗
for some G-module
W∗
. Then we have the projection
V = im(U) ⊕ W → im(U) : v + w → v
and its composite with the isomorphism j−1
: im(U) ∼= U gives a splitting
for j. Therefore the short exact sequence splits.
Corollary 3.6. Let G be a finite group, such that char(k) does not divide
|G|. Let k[G] = U1 ⊕ . . . Un have each Ui irreducible. Then each irreducible
G-module U is isomorphic to one of the Ui.
Proof. Let v = 0 ∈ V . By the universal property of k[G] there exists a
unique G-module map θ : k[G] → V with θ(e) = v. Since V is irreducible,
we must have that im(θ) = V . Now we then have ker(θ) ≤ k[G]. Since
0 → ker(θ) → k[G] → im(θ) → 0 is a short exact sequence, and each short
exact sequence of G-modules splits we have k[G] ∼= ker(θ)⊕V . In particular
V is then a direct summand of k[G] and so isomorphic to a submodule of
k[G].
It thus suffices to prove the claim in the case that U is a submodule of k[G].
If Ui is trivial it is clear. Otherwise, let pi : k[G] → Ui : u1 + . . . + un → ui
denote the projection on to each factor. Since Ui is non-zero one of the
composites U → k[G] → Ui must be non zero, and so – by the first part of
Schur’s lemma – an isomorphism.
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 13
Corollary 3.7. Let G be a finite group, such that char(k) does not divide
|G|. Then there are only finitely many irreducible G-modules over k, up to
isomorphism.
4. Finer results on decompositions when k = C
Terminology 4.1. We call V1, . . . , Vm a complete set of irreducible Gmodules
no two are isomorphic, and each irreducible G-module is isomorphic
to one of them.
Our goal now is to take a closer look at the decomposition of a G-module
into its irreducible constituents. The following is an immediate consequence
of Schur’s lemma.
Definition 4.2. Let V and W be G-modules. We write Homk[G](V, W) ⊂
V ect(V, W) for the vector subspace of G-module homomorphisms from V
to W. Warning – this is not a sub-G-module in general, though if G is
commutative, it is.
Corollary 4.3. Let V and W be irreducible G-modules. Then
dim(HomC[G](V, W)) = 1
iff V and W are isomorphic and 0 otherwise.
Proof. To say that dim(HomC[G](V, W)) = 0 is to say that the only homomorphism
from V to W is the zero homomorphism. By Schur’s lemma is
equivalent saying that V and W are non-isomorphic.
If dim(HomC[G](V, W)) = 1 then there exists a non-zero homomorphism
from V to W which is then an isomorphism by Schur’s lemma. On the
other hand if ϕ : V → W is an isomorphism of G-modules, then postcomposition
by ϕ gives an isomorphism of vector spaces HomC[G](V, V ) ∼=
HomC[G](V, W)). Now dim(HomC[G](V, V )) = 1 by Schur’s lemma Part
2 – that is, each non-zero linear endomorphism is a scalar multiple of the
identity – hence dim(HomC[G](V, W)) = 1 too.
Proposition 4.4. (1) For all G-modules we have
Homk[G](V1 ⊕ . . . ⊕ Vn, W) ∼= Homk[G](V1, W) ⊕ . . . ⊕ Homk[G](Vn, W)
and
Homk[G](V, W1 ⊕ . . . ⊕ Wn) ∼= Homk[G](V, W1) ⊕ . . . ⊕ Homk[G](V, Wn)
.
(2) We have an isomorphism of vector spaces Homk[G](k[G], V ) ∼= V .
Proof. The first part is standard and simply expresses the fact that direct
sums of modules are biproducts. The universal property of k[G] as the
free G-module on 1 asserts that restriction along e : 1 → k[G] induces a
bijection Homk[G](k[G], V ) → Set(1, V ) ∼= V . This composite bijection
14 JOHN BOURKE
sends θ to the value θ(e) and is clearly a linear transformation. Therefore
it is an isomorphism of vector spaces.
Theorem 4.5. Let V be a finite dimensional G-module over C.
(1) Then V admits a decomposition V = V1 ⊕ . . . Vn as a direct sum of
irreducible submodules.
(2) Each irreducible G-module W appears in this decomposition, up to
isomorphism, exactly dim(HomCG(V, W)) times.
(3) In particular, let U1, . . . , Um be a complete set of irreducible G-modules.
Then V ∼= Ud1
1 ⊕ . . . ⊕ Udm
m where di = dim(HomCG(V, Ui)).
Proof. (1) This is just Theorem 3.1.
(2) We have
HomC[G](V, W) = HomC[G](V1 ⊕ . . . Vn, W) ∼=
HomC[G](V1, W) ⊕ . . . ⊕ HomC[G](Vn, W)
By Schur’s lemma dim(HomC[G](Vi, W)) = 1 iff Vi is isomorphic to
W and 0 otherwise. Hence, taking dimensions on either side of the
above isomorphism gives dim(HomCG(V, W) = Σi:Vi
∼=W 1, which is the
number of i for which Vi
∼= W, as claimed.
(3) Using the previous part, the direct sum V = V1 ⊕ . . . Vn is a direct sum
(V11 ⊕ . . . ⊕ V1m1 ) ⊕ (V21 ⊕ . . . ⊕ V2m2 ) ⊕ . . . ⊕ (Vn1 ⊕ . . . ⊕ Vnmn )
of irreducibles where we gather the isomorphic G-modules in the decomposition
inside the same bracket and where mi = dim(HomCG(V, Vi1)).
Since all of those within a single bracket are isomorphic to a common
Ui, and none outside are, this gives V ∼= Ud1
1 ⊕ . . . ⊕ Udm
m .
Specialising the above to deal with the case of the regular G-module, we
obtain a still finer result.
Theorem 4.6. Let C[G] = V1 ⊕ . . . Vn be a decomposition with each Vi
irreducible.
(1) Then each irreducible G-module W appears in this decomposition, up
to isomorphism, exactly dim(W) times.
(2) In particular, let U1, . . . , Um be a complete set of irreducible G-modules.
Then C[G] ∼= U
dim(U1)
1 ⊕ . . . ⊕ U
dim(Um)
m .
Proof. This follows immediately from the above given that we have an
isomorphism of vector spaces HomCG(CG, W) ∼= W for all W.
The following result is very useful for calculating the irreducible representations
of a given group.
Theorem 4.7. Let G be a finite group and U1, . . . Um a complete set of
irreducible G-modules over C. Then |G| = Σi=1,...mdim(Vi)2
.
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 15
Proof. This result follows from the second part of the above, on comparing
the dimension of the left and right hand side.
A simple consequence of the above results is the following.
Proposition 4.8. A finite group is abelian if and only if all of its complex
irreducible representations are 1-dimensional.
Proof. Let G be abelian and consider an irreducible G-module V . Let
g ∈ G. Then g.− : V → V is a homomorphism of G-modules since
g.(h.v) = (g.h).v = (h.g).v = h.(g.v). Therefore, by Schur’s lemma, there
exists λ such that g.v = λ.v for all v ∈ V . But then each 1-dimensional
subspace v ≤ V is a G-submodule – hence, by irreducibility we have
v = V , as required.
Let V be a 1-d representation. Since endomorphisms of V are just
multiplication by a scalar, their composition is commutative – hence given
g, h ∈ G we have (gh).v = (hg).v. Now let C[G] = V1 ⊕ . . . ⊕ Vn be
a decomposition as a sum of 1-dimensional representations. (We know
that C[G] is a direct sum of irreducibles). Then for gh.(v1 ⊕ . . . ⊕ vn) =
(gh.v1 ⊕ . . . ⊕ gh.vn) = (hg.v1 ⊕ . . . ⊕ hg.vn) = hg.(v1 ⊕ . . . ⊕ vn). Since
gh.− = hg.− : C[G] → C[G] and C[G] is a faithful G-module we conclude
that gh = hg.
Let us calculate the 1-dimensional representations of a finite abelian
group. Firstly, recall:
Proposition 4.9. For a finite abelian group G = Cn1 × . . . × Cnm it
has n1 × . . . × nm non-isomorphic irreducible complex representations –
all one dimensional. They are given by the assignments (gi1
1 , . . . , gim
m ) →
(λ1
i1
, . . . , λm
im
) where the λi are ni’th roots of unity.
Proof. The first claim follows from the fact that all of its irreducible reps
are 1-dimensional, and Theorem 4.7.
The above functions G → C are clearly homomorphisms and indeed
the only such – such each element of order k must be sent to an k’th
root of unity. They are also distinct – or, equally, non isomorphic. For
isomorphism of 1-dimensional representations on the same space means
equality λ.(g.1v) = g.2(λ.v) = λ(g.2v) so g.1v = g.2v. Therefore they form
a complete set of irreducible representations of G.
Example 4.10. Representations of D8. By the above theorem, the possible
dimensions of the irreducible representations of any group of order
8 are 1, 1, 1, 1, 1, 1, 1, 1 or 1, 1, 1, 1, 2. In fact, we saw that D8 has an irreducible
2 − d representation earlier. Hence it must have four irreducible
1-dimensional representations. Thus D8 has 1 irreducible 2-d representation
and 4 irreducible non-isomorphic 1-d representations.
16 JOHN BOURKE
What are the four 1-dimensional representations? To give a 1-dimensional
representation is to give a homomorphism D8 → C – since C is abelian we
obtain a unique factorisation
D8
p

ρ
""
(D8)ab
ˆρ
// C
through the abelianization of D8. This gives a bijection, obtained by
restricting along the quotient map D8 → (D8)ab, between the irreducible
representations of the abelianization and the irreducible 1-dimensional
representations of D8. In the abelianization we force ab = ba. Since
b−1
ab = a3
this forces a = a3
and so a2
= 1. In this way, we see that
(D8)ab
∼= C2 × C2 with quotient map p : D8 → (D8)ab
∼= C2 × C2 : a →
(1, 0), b → (0, 1) so that the irreducible representations of D8 are obtained
from the four irreducible representations of C2 × C2 obtained by restricting
along p – namely ρ(a) =+
− 1 and ρ(b) =+
− 1.
5. Tensor products, internal homs and duals
5.1. Tensor products. Let V and W be vector spaces. The tensor product
V ⊗ W can be described in various ways. I will describe a few and discuss
a few standard facts about the tensor product, and their relationship with
homs, and duals.
(1) Let Bil(V, W; A) be the set of bilinear maps F : V × W → A – recall
that F is bilinear if each F(v, −) : V → A and F(−, w) : V → A are
linear maps. Then V ⊗ W classifies bilinear maps out of V × W: more
precisely, there exists a bilinear map θ : V × W → V ⊗ W with the
universal property that: given any bilinear map F : V × W → A there
exists a unique linear map ˜F : V ⊗ W → A such that
V × W
θ

F
##
V ⊗ W ˜F
// A
This characterises the tensor product up to a unique isomorphism of
vector spaces.
(2) Explicitly, V ⊗ W, can be described as the quotient F(V × W)/ of
the free vector space on the underlying set of V × W quotiented by
the relations:
• λv ⊗ w = λ(v ⊗ w) = v ⊗ λw;
• v ⊗ (w1 + w2) = v ⊗ w1 + v ⊗ w2;
• (v1 + v2) ⊗ w = v1 ⊗ w + v2 ⊗ w
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 17
wherein we write v ⊗ w for a basis element of F(V × W). In these
terms we have θ(v, w) = v ⊗ w and the above equations simply assert
that θ is, indeed, bilinear.
(3) If V has basis vi and W basis wj then V ⊗ W has a basis consisting of
the symbols vi ⊗ wj – in particular, if V has dimension n and W has
dimension m then V ⊗ W has dimension n × m.
(4) Given linear maps T : V → V and S : W → W represented by matrices
A and B the corresponding matrix representing T ⊗S is the Kronecker
product
A ⊗ B =


a11B . . . a1nB
...
...
...
an1B . . . annB


In particular, it clearly follows that Tr(A ⊗ B) = Tr(A)Tr(B).
(5) The tensor is associative and symmetric up to natural isomorphism:
we have (A ⊗ B) ⊗ C ∼= A ⊗ (B ⊗ C) and A ⊗ B ∼= B ⊗ A. The
second isomorphism follows easily from the universal property. For the
first isomorphism, one should show that both sides have the universal
property of representing trilinear maps. A similar argument show that
k ⊗ A ∼= A ∼= A ⊗ k so that we have a kind of (up to isomorphism)
commutative monoidal structure on Vect – this is called a symmetric
monoidal category or symmetric tensor category.
We now explain how to lift the tensor product of G-modules to G-modules.
Let BilG(V, W; A) be the set of bilinear maps F : V × W → A satisfying
g.F(v, w) = F(gv, gw) for all v ∈ V and w ∈ W.
Proposition 5.1. (1) The tensor product V ⊗ W becomes a G-module
when we define g(v ⊗ w) = gv ⊗ gw.
(2) It has the universal property that G-Mod(V ⊗ W, A) ∼= BilG(V, W; A).
Proof. Consider the diagram
V × W
θ

gV ×gW
// V × W
θ

V ⊗ W g.−
// V ⊗ W
and observe that the composite θ(gV × gW ) : (v, w) → gv ⊗ gw is bilinear.
Therefore we obtain a unique linear map g.− = gV ⊗gW : V ⊗W → V ⊗W
which is given by g(v ⊗ w) = gv ⊗ gw. That this action is compatible with
the group structure of G follows from the universal property of V ⊗ W.
For the second part, we observe that a G-module map θ : V ⊗ W → A
corresponds to a bilinear map ϕ : V × W → A, with the extra property
18 JOHN BOURKE
that the diagram
V × W
ϕ

gV ×gW
// V × W
ϕ

A g.−
// A
commutes – this says exactly that θ is a G-bilinear map.
Remark 5.2. Using this universal property (as well as its ternary version)
we can see that the isomorphisms (A ⊗ B) ⊗ C ∼= A ⊗ (B ⊗ C) and
A ⊗ B ∼= B ⊗ A of vector spaces lift to isomorphisms of G-modules. The
isomorphisms k ⊗ A ∼= A ∼= A ⊗ k also lift to isomorphisms of G-modules
when k is equipped with the trivial G-module structure.
5.2. Homs and their relationship with tensor products. Given vector
spaces V and W we have the hom vector space [V, W] = Vect(V, W)
– its structure is given by pointwise addition and scalar multiplication of
linear maps. It is closely related to the tensor product of vector spaces by
the fact that we have a bijection
Vect(V ⊗ W, A) ∼= Vect(V, [W, A])
natural in each variable. This follows from the fact that we have an
isomorphism Bil(V, W; A) ∼= Vect(V, [W, A]) sending the bilinear
F : V × W → A
to the linear map
˜F : V → Vect(W, A) : v → (w → F(v, w)).
Using this, we can see that the composite isomorphism
Vect(V ⊗ W, A) ∼= Bil(V, W; A) ∼= Vect(V, Vect(W, A))
sends F : V ⊗ W → A to the map
V → Vect(W, A) : v → (w → F(v ⊗ w)).
Remark 5.3. In fact, the tensor product and hom determine one another
up to unique isomorphism in this way – that is, the bijection says that we
have adjoint functors − ⊗ W [W, −].
We have seen that if V and W are G-modules then V ⊗ W is naturally a
G-module. We would now like to describe a G-module structure on [V, W]
so that the natural bijection
Vect(V ⊗ W, A) ∼= Vect(V, [W, A])
lifts to a natural bijection
G-Mod(V ⊗ W, A) ∼= G-Mod(V, [W, A]).
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 19
This means that we would like to equip [W, A] with a G-module structure
such the square of vector spaces on the left commutes
V ⊗ W
gV ⊗gW

F // A
gA

V ⊗ W F
// A
V
gV

˜F // [W, A]
g[W,A]

V ˜F
// [W, A]
if and only if the square on the right above commutes. To see how we
should do this, consider a diagram as on the left below.
V ⊗ W
R⊗S

F // A
T

V ⊗ W
F
// A
V
R

˜F // [W, A]
[W,T]

[W, A ]
V ˜F
// [W , A ]
[S,A ]
OO
V
R

˜F // [W, A]
[S−1,T]

V ˜F
// [W , A ]
This commutes if and only the diagram in the middle commutes – here
[W, T] means postcomposition by T, and [S, A ] means precomposition by S.
In the case that S is invertible, this can be rephrased as the commutativity
of the square on the right. Finally this tells that we should define the hom
G-module as follows.
Proposition 5.4. Let V and W be G-modules. Then [V, W] becomes
a G-module, where an element g acts as [g−1
V , gW ] : [V, W] → [V, W].
Equationally, this says (gf)(v) = g.f(g−1
(v)).
In terms of commutative diagrams it says
V
f
// W
g.−

V
g−1.−
OO
g.f
// W
Proof. Certainly each g.f ∈ Vect(V, W) as it is a composite of three linear
maps. It is easy to see that g.− is itself a linear map, and ((hg).f)(v) =
hg.f.(g−1
h−1
)(v) = h(gfg−1
)h−1
(v) = (h.(g.f))(v).
It then follows from the above discussion that we have a bijection
G-Mod(V ⊗ W, A) ∼= G-Mod(V, [W, A])
natural in each variable, as required.
20 JOHN BOURKE
Remark 5.5. Observe that the above construction of a hom certainly
doesn’t work if we are considering representations of a general algebra or
monoid – in order to lift the hom from vector spaces, we very much used
that the elements of a group are invertible. More generally, it can be done
for an algebra which is Hopf – see the final section for an outline of the
construction in that setting.
5.3. Homs via tensors and duals.
Remark 5.6. The dual representation V ∗
= [V, k] is the special case of
the above, on taking W = k with the trivial G-module structure. In
particular, the action on the dual space is defined by (gf)(v) = f(g−1
v) or
just g.− = Vect(g−1
, k) : Vect(V, k) → Vect(V, k).
Recall that if V is finite dimensional with basis E = e1, . . . , en then V ∗
has basis E∗
= (e1)∗
, . . . , (en)∗
where (ei)∗
: V → k is the map sending ei
to 1 and all other basis vectors to 0.
For all V, W the evaluation map
V ⊗ [V, W] → W : a ⊗ f → fa
is clearly a G-module map. In particular, this gives us a map ev : V ⊗V ∗
→ k
sending f ⊗ v → f(v).
We would like to prove that for V, W finite dimensional G-modules we
have V ⊗ W ∼= [V, W].
The following proof is based on a nice argument of M´aria ˘Simkov´a in the
exercises, and simpler than the one I originally gave in class. We have the
composite G-module map
V ⊗ W ⊗ V
∼= // V ⊗ V ⊗ W
ev⊗1
// k ⊗ W
∼= // W
which sends f ⊗ w ⊗ v to f(v)w. Using the natural bijection G-Mod(V ⊗
W ⊗ V, W) ∼= G-Mod(V ⊗ W, [V, W]) this corresponds to a G-module map
ϕ : V ⊗ W → [V, W].
Proposition 5.7. For V, W finite dimensional G-modules the map ϕ :
V ⊗ W → [V, W] is invertible.
Proof. Let (vi)i∈I and (wj)j∈J form bases of V and W respectively. Then
the elements v∗
i ⊗ wj form a basis of V ⊗ W. On the other hand, the
elements χi
j of [V, W] defined by χi
j(vi) = wj, and 0 otherwise, form a basis
of [V, W]. It is easy to see that ϕ(v∗
i ⊗ wj) = χi
j; thus ϕ is a bijection on
basis elements, and so an isomorphism of vector spaces.
Remark 5.8. In future, it would be nice to consider the relationship with
duals in a symmetric monoidal category.
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 21
6. Characters of groups
In the present section, we will assume that G is a finite group, that k = C
and that all G-modules are finite dimensional. Let A be a n × n-matrix.
The trace tr[A] = Σi=1,...naii is the sum of the diagonal entries.
Exercise 6.1. Use Proposition 5.7 and the evaluation map V ∗
⊗ V → C
to give a matrix-free formulation of the trace.
Proposition 6.2. Some standard properties of the trace are as follows.
(1) tr(A + B) = tr(B + A),
(2) tr(AB) = tr(BA),
(3) tr(B−1
AB) = tr(A),
(4) tr(A ⊕ B) = tr(A) + tr(B),
(5) tr(A ⊗ B) = tr(A) × tr(B).
Proof. The proof of the first two is elementary – whilst the third follows
from the second. The fourth and fifth hold since the direct sum and tensor
product of matrices are of the form
A ⊕ B =
A 0
0 B
A ⊗ B =


a11B . . . a1nB
...
...
...
an1B . . . annB


respectively.
Definition 6.3. Let V be a G-module with basis B of dimension n. Then
each g ∈ G gives rise to a matrix [g]B ∈ Gln(C). The character of V is the
function χ : G → C : g → tr[g]B.
The character of V is independent of the choice of basis, since if C is a
second basis, then [g]C = T−1
[g]BT for the base change matrix, and then
by (3) above, the claim follows.
Terminology 6.4. Let V be a G-module with character χ.
(1) The degree of χ is the dimension of V – this is equally the value χ(1)
of the identity element.
(2) The character χ is said to be irreducible if V is irreducible.
Proposition 6.5.
Isomorphic G-modules have the same character.
Proof. If V and W are isomorphic, with bases B and C, then the matrix
representations [g]B and [g]C are equivalent – that is, T−1
[g]BT = [g]C for
an invertible T. Hence tr[g]B = tr[g]C.
We will prove a converse to the above: that is, if the characters of V and
W are equal then V and W are isomorphic. The strategy is as follows.
22 JOHN BOURKE
(1) We introduce an inner product −, − on the complex vector space
[G, C] of functions from G to C. This will have the property that
dim(HomCG(V, W)) = χV , χW . A consequence of Schur’s lemma is
hen that distinct irreducible characters χ and ϕ are orthonormal.
(2) Now by Theorem 4.5 the G-module V is isomorphic to direct sum
V ∼= ⊕i=1,...,nV di
i for Vi a complete set of irreducibles, and di =
dim(HomCG(V, Vi)). But since dim(HomCG(V, Vi)) = χ, χi where
the χi are the irreducible characters corresponding to the Vi this tells
us that V can be recovered from its character χ.
The approach we take heavily uses the results of the preceding section
on tensor products, homs and duals of representations. Before that, lets
derive a few simple properties of characters.
Proposition 6.6. Let G be a finite group and V a finite dimensional Gmodule
over C. If g is order n then there is a basis B of V such that [g]B
is a diagonal matrix, all of whose values are n-th roots of unity.
Proof. Consider the cyclic group g ≤ G. Then viewing V as a g -module,
and since this group is abelian, it splits as a direct sum V = V1 ⊕ . . . ⊕ Vn
of 1-d submodules. Taking a basis v1, . . . , vn we have gvi = λivi for λi a
n’th root of unity. This proves the claim.
Proposition 6.7. Basic facts about characters of finite groups.
(1) If x and y are conjugate elements of G then χ(x) = χ(y).
(2) If x has order n then χ(x) is a sum of n’th roots of unity.
(3) We have χ(x−1
) = χ(x).
Proof. (2) For V a G-module with basis B and y = g−1
xg then χ(y) =
tr[y]B = tr([g−1
xg]B) = tr([g]−1
B[x]B[g]B) = tr[x]B = χ[x], as required.
(3) By the corollary above, we know that there is a basis B of V with [x]B
a diagonal matrix whose values wi are n’th roots of unity. Taking tr[x]B
gives the result.
(4) χ(x) is a sum of n-th roots of unity w = w1 + . . . + wm. Since [x−1
]B =
[x]−1
B we have χ(x−1
) = w−1
1 +. . .+w−1
n = w1 +. . .+wn = w1 + . . . wn = w,
as required.
Proposition 6.8. Characters of sums, tensor products, duals and homs.
(1) χV ⊕W (g) = χV (g) + χW (g),
(2) χV ⊗W (g) = χV (g)χW (g)
(3) χV ∗ (g) = χV (g)
(4) χ[V,W](g) = χV (g).χW (g)
Proof. The first and second follow from the corresponding facts for the
trace function.
For (3) let E = e1, . . . , en be a basis of V . In the dual basis E∗
= e∗
1, . . . , e∗
n
to that of V , a map Vect(f, k) : Vect(V, k) → Vect(V, k) is represented by
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 23
the transpose of the matrix [f]E. Since the trace of the transpose is equal
to that of the original matrix, we conclude that Vect(g−1
, k) has trace
χV (g−1
). By the basic properties of characters, this equals the complex
conjugate χV (g).
For (4) we use the isomorphism [V, W] ∼= V ∗
⊗W of Proposition 5.7. Since
the character is isomorphism invariant, we have χ[V,W] = χV ∗⊗W . Then using
the above equations we get χV ∗⊗W (g) = χV ∗ (g)χW (g) = χV (g)χW (g).
Our next goal is to describe dim(Homk[G](V, W)) in terms of the characters
of [V, W]. To start, observe that given any G-module V we can form
the vector subspace V G
≤ V consisting of those v for which gv = v for all
g ∈ G. This has a trivial G-module structure.
Proposition 6.9. We have an equality Vect(V, W)G
= Homk[G](V, W) of
vector spaces.
Proof. To say that f : V → W is fixed under the action of all g ∈ G is to
say that f(v) = gf(g−1
v), or equally, that gf(v) = f(g(v)) all v.
The following result is closely connected to Maschke’s theorem.
Lemma 6.10 (Projection formula). Let V be a G-module.
(1) Then the linear map
p : v →
1
|G| g∈G
g(v)
is a projection with image V G
.
(2) In particular, we have
dim(V G
) =
1
|G| g∈G
χV (g)
.
Proof. To see that h.p(v) = p(v) we use that g∈G g = g∈G hg – namely,
h.p(v) = h.(
1
|G| g∈G
g(v)) =
1
|G| g∈G
hg(v) =
1
|G| g∈G
g(v) = p(v)
since h( g∈G g) = ( g∈G g). Thus im(p) ≤ V G
. On the other hand if
h.v = v all v we have pv = 1
|G| g∈G(v) = |G|
|G|
v = v. Thus im(p) = V .
Finally p2
v = ppv = pv since pv ∈ V G
and p is invariant on elements of V G
.
(2) Since we have a projection p, we can write V = V G
⊕ W and p =
idV G ⊕ 0W . Choosing any bases of V G
and W we get tr(p) = dim(V G
) so
that
dim(V G
) = tr(
1
|G| g∈G
g) =
1
|G| g∈G
χV (g)
24 JOHN BOURKE
Exercise 6.11. Show how the procedure of averaging a projection used
in Maschke’s theorem can be obtained from the projection formula by
considering the internal hom G-module [V, V ].
Combining the two preceding results we obtain
Proposition 6.12. We have dim(HomC[G](V, W)) = 1
|G| g∈G χ[V,W](g).
Finally, putting it all together we have
Theorem 6.13. dim(HomC[G](V, W)) = 1
|G| g∈G χV (g).χW (g).
6.1. The standard inner product.
Definition 6.14. Let f1, f2 : G → C. We define
f1, f2 =
1
|G| g∈G
f1(g)f2(g).
It is a standard fact that this gives an inner product on [G, C] – in other
words, that it is linear in the first variable, conjugate symmetric (meaning
f, g = g, f ) and positive definite.
Theorem 6.15. We have χV , χW = dim(HomC[G](V, W)). (In particular,
the value is always a natural number.)
Proof. Actually, what Theorem 6.13 gives is the first equation of
dim(HomC[G](V, W)) = χW , χV = χV , χW = χV , χW .
We then use conjugate symmetry followed by the fact that the value is a
natural number (being the dimension of a vector space).
Question 6.16. In class Yuriy Dupyn asked me to what extent characters
are a decategorification of representations – tensor product corresponds
to multiplication, direct sum to addition etc. I’m especially curious about
how internal homs induce inner products – can one show something like:
the set of isomorphism classes of objects in a symmetric monoidal closed
category with duals enriched in finite dimensional vector spaces admits an
inner product – or something in that direction?
Remark 6.17. Note that since χV , χW is always a natural number, the
inner product on [G, C] is in fact symmetric when restricted to characters –
it follows that it is linear in both variables when restricted to characters.
The following important result is a routine consequence of what we have
done.
Let U1, . . . , Um be a complete set of irreducible G-modules. Recall that
V ∼= Ud1
1 ⊕ . . . ⊕ Udm
m where di = dim(HomCG(V, Ui)) = χV , χUi
.
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 25
Theorem 6.18. Two G-modules V and W are isomorphic if and only if
they have the same character.
Proof. We have proven one direction already. For the other, we have
V ∼= Ud1
1 ⊕. . .⊕Udm
m where di = χV , χUi
= χW , χUi
. Hence we also have
W ∼= Ud1
1 ⊕ . . . ⊕ Udm
m , as required.
Theorem 6.19. Irreducible characters are orthonormal – that is, if V
and W are irreducible G-modules then χV , χW = 1 iff χV = χW and 0
otherwise.
Proof. We have χV , χW = dim(HomCG(V, W). By Schur’s lemma, the
rhs is 1 just when V ∼= W (equally χV = χW by the preceding) and 0
otherwise.
In fact, we can strengthen the above result to obtain a very useful tool
for checking whether a given representation is irreducible!
Theorem 6.20. A G-module V is irreducible if and only if χV , χV = 1.
Proof. One direction holds by the preceding result – we must show that if
χV , χV = 1 then χV is irreducible.
Write V ∼= Ud1
1 ⊕ . . . ⊕ Udm
m . Then χV = d1χU1 + . . . dmχUm since the
character of a direct sum is the sum of the characters. Using orthogonality of
irreducible characters and linearity in each variable, we then get χV , χV =
(d1)2
+ . . . + (dn)2
. Being a sum of natural numbers, this can equal 1 if and
only if one of the di equals 1, and all of the others are 0, as required.
Example 6.21. Consider D8 = a, b : a4
= b2
= 1, b−1
ab = a−1
. and its
2-dimensional representation by rotation and reflection matrices.
A =
0 1
−1 0
B =
1 0
0 −1
Then
A2
=
−1 0
0 −1
, A3
=
0 −1
1 0
, AB =
0 −1
−1 0
, A2
B =
−1 0
0 1
, A3
B =
0 1
1 0
so its character is given by
g 1 a a2
a3
b ab a2
b a3
b
χ 2 0 −2 0 0 0 0 0
In particular, we see that χ, χ = (22
+ (−2)2
)/8 = 8/8 = 1. This gives
another proof that the 2-dimensional representation is irreducible.
26 JOHN BOURKE
6.2. The number of irreducible characters.
Definition 6.22. A class function f : G → C is a function which is
invariant on conjugacy classes.
Each character is a class function. Let C(G) ≤ [G, C] be the vector
subspace consisting of the class functions – we will show that the irreducible
characters form a basis for C(G).
Lemma 6.23. The dimension of C(G) is equal to the number of conjugacy
classes of G.
Proof. Consider the surjective function p : G → Conj(G) : g → [g] sending
an element of g to its conjugacy class. Since a function G → C is a class
function if and only if it factors through p, restriction along p induces a
bijection p∗
: [Conj(G), C] ∼= C(G). Clearly this respects the pointwise
vector space structure and so is an isomorphism of vector spaces. The left
hand side has dimension Conj(G) – the set of conjugacy classes, and so the
right does too. (Indeed transporting the canonical basis on [Conj(G), C]
along p∗
we see that the functions G → C sending each member of a given
conjugacy class to 1 and all other elements to 0 form a basis for C(G).)
We wish to prove that the number of irreducible characters – equally
the number of isomorphism classes of irreducible G-modules – equals the
number of conjugacy classes. Firstly, we need to consider the centre of the
group algebra.
Recall that C[G] is an C-algebra with multiplication (Σjλigi)(Σjθjgj) =
Σi,jλiθjgigj. We observed earlier that a G-module V is equivalently a
C[G]-module – in particular each z = λ1g1 + . . . + λngn ∈ C[G] induces a
linear transformation z.− = λ1g1. − + . . . + λngn.− : V → V by pointwise
operations.
The centre
Z(C[G]) = {z ∈ C[G] : zw = wz ∀w ∈ C[G]}
is a vector subspace of C[G], though not necessarily a G-module. Clearly z
lies in the centre just when zg = gz for each g ∈ G – in particular, each
element of the centre Z(G) of the group G belongs to Z(C[G]) but others
do too.
Note that if g ∈ Z(G) then g.− : V → V is a G-module homomorphism
since g.(h.v) = (gh).v = (hg).v = h(gv). Arguing in the same way, we
see more generally that if z ∈ Z(C[G]) then z.− : V → V is a G-module
homomorphism.
Lemma 6.24. The centre of the group algebra Z(CG) has dimension equal
to the number of conjugacy classes of G.
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 27
Proof. Let C1, . . . , Cn be the conjugacy classes of G. We define the class
sum Ci = Σg∈Ci
g and will show that the class sums form a basis for Z(CG),
thus proving the claim.
Firstly, we show that Ci ∈ Z(CG). In order to show this, it suffices to
prove that h.Ci = Ci.h for each h ∈ G since these are the basis elements of
CG. Equivalently, hCih−1
= Ci but since h−1
. − .h : Ci → Ci permutes the
elements of the conjugacy class, the class sum is indeed invariant.
Next, we show that the Ci are linearly independent. For suppose that
ΣiλiCi = 0. This is a sum of the basis elements g ∈ G, and since the
conjugacy classes are disjoint no two basis elements are repeated – it follows
that each λi = 0.
Finally, we must show they are spanning. Let r = Σgλgg lie in the centre.
Then for all group elements h we have r = hrh−
1, which says that
Σgλgg = Σgλghgh−1
– in particular we must have λg = λhgh−1 . In other words, the function
λ : G → C : g → λg is invariant on conjugacy classes (a class function).
Therefore r = Σgλgg = Σiλgi
Ci where gi is any member of the conjugacy
class Ci.
Theorem 6.25. Let V1, . . . , Vn be a complete set of irreducible G-modules,
and χ1, . . . , χn the associated characters form a basis of C(G) – in particular
the number of irreducible characters equals the number of conjugacy classes
of G.
Proof. Since the χi are orthogonal, they are linearly independent in C(G).
Hence n ≤ l the number of conjugacy classes, since l is the dimension of
C(G). We must show that n = l. By the preceding lemma l is equally the
dimension of the centre of the group algebra Z(CG) – it suffices to prove
that dim(Z(CG)) ≤ n.
Let CG = W1 ⊕ . . . ⊕ Wn where Wi is isomorphic to a direct sum of
copies of Vi. Consider the identity group element e ∈ C[G] – then we can
write e = f1 + . . . fn where fi ∈ Wi. Let z ∈ Z(CG) and consider
z.− : V → V : v → z.v
for any G-module V . This is a G-module morphism since z.gv = (z.g)v =
(g.z)v = g(zv). In particular, by Schur’s lemma we have z.− = λi.− : Vi →
Vi for some λi ∈ C. Furthermore we then have z.− = λi : − : ⊕Vi → ⊕Vi
for any such direct sum, and since we have Wi
∼= ⊕Vi as G-modules, we
also have that z.− = λi.− : Wi → Wi.
Then z = z.e = z.(f1+. . .+fn) = λ1f1+. . .+λnfn so that dim(Z(CG)) ≤
n, as required.
6.3. Character tables.
28 JOHN BOURKE
Definition 6.26. Let G be a group. The character table of G is a square
array with columns indexed by the conjugacy classes of G and rows by the
irreducible characters, as below.





. . . K . . .
. . .
...
χ . . . χK
...





The values are in the first column give the dimensions of the representations.
Remark 6.27. The values of the first column give the dimensions of the
irreducible characters.
Example 6.28 (Character table of D8). Consider D8 = a, b : a4
= b2
=
1, b−1
ab = a−1
. Its conjugacy classes are {1}, {a, a3
}, {a2
}, {b, a2
b}, {ab, a2
b}.
Its character table is








g 1 a a2
b ab
χ1 1 1 1 1 1
χ2 1 1 1 −1 −1
χ3 1 −1 1 1 −1
χ4 1 −1 1 −1 1
χ5 2 0 −2 0 0








There is one 2-dimensional representation and four 1-d representations. Note
that the character of a 1-d representation is just the original representation
– in particular it is multiplicative. We calculated these in Example 4.10.
What can we say about the character table? It is certainly a square
matrix of dimension Conj(G) the set of conjugacy classes of G. Recall that
the irreducible characters form a basis of C(G), the space of class functions
of G. Using the isomorphism C(G) ∼= Vect(Conj(G), C) we see that the
restricted irreducible characters form a basis of Vect(Conj(G), C). Hence
the matrix with these as its rows is invertible.
6.4. Row and column orthogonality relations. For irreducible characters
χi, χj we have, by their orthonormality, the equation
δi,j =
1
|G| g∈G
χi(g)χj(g).
Let Cl(g) be the conjugacy class of g ∈ G, and g1, . . . , gn a set of representatives
of the conjugacy classes. Then since characters are invariant on
conjugacy classes, the above equation becomes
δi,j =
1
|G| gk∈G
|Cl(gk)|χi(gk)χj(gk).
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 29
These are called the row orthogonality relations, since they express the
relation between different rows of the character table. Very useful are also
the column orthogonality relations, which are stated as the second part of
the following summary result.
Theorem 6.29. Consider representatives g1, . . . , gk of the conjugacy classes
of a finite group G. Then
• [Row orthogonality relations]
δi,j =
1
|G| gk∈G
|Cl(gk)|χi(gk)χj(gk).
• [Column orthogonality relations]
|G|
|Cl(gi)|
δi,j =
k=1,...,n
χk(gi)χk(gj).
Proof. We only need to prove the second case. Let
Bi,j =
|Cl(gj)|
|G|
χi(gj).
Then
(B.B
t
)ij =
k=1,...,n
|Cl(gj)|
|G|
χi(gk)χj(gk) = δij
by row orthogonality, so B.B
t
= I. By the multiplicativity of the determinant,
this implies that B is invertible, so its inverse must be B
t
. Hence
B
t
.B = I too. This says
δij = (B
t
.B)ij =
k=1,...,n
|Cl(gi)|
|G|
|Cl(gj)|
|G|
χk(gi)χk(gj)
which is equally
|G|
|Cl(gi)|
δij =
k=1,...,n
χk(gi)χk(gj) =
k=1,...,n
χk(gi)χk(gj)
using that both sides are real numbers in the last case.
These are very useful for completing character tables given partial information.
For instance consider again the character table for D8 but
with the final row and columns missing. Recall the conjugacy classes are
30 JOHN BOURKE
{1}, {a, a3
}, {a2
}, {b, a2
b}, {ab, a2
b}.








g 1 a a2
b ab
χ1 1 1 1 1 ?
χ2 1 1 1 −1 ?
χ3 1 −1 1 1 ?
χ4 1 −1 1 −1 ?
χ5 ? ? ? ? ?
















g 1 a a2
b ab
χ1 1 1 1 1 1
χ2 1 1 1 −1 ?
χ3 1 −1 1 1 ?
χ4 1 −1 1 −1 ?
χ5 ? ? ? ? ?








Firstly, the top row is the trivial character so we immediately fill in the top
row above. Now using row orthogonality, we have
0 = χ1, χ2 = 1/8(1.1 + 2(1.1) + 1(1.1) + 2(1. − 1) + 2(1.χ2(ab))
so we get 0 = 1/8(2 + 2(χ2(ab)) so χ2(ab) = −1.
Continuing in this way, we fill in as far as on the left below








g 1 a a2
b ab
χ1 1 1 1 1 1
χ2 1 1 1 −1 −1
χ3 1 −1 1 1 −1
χ4 1 −1 1 −1 1
χ5 ? ? ? ? ?
















g 1 a a2
b ab
χ1 1 1 1 1 1
χ2 1 1 1 −1 −1
χ3 1 −1 1 1 −1
χ4 1 −1 1 −1 1
χ5 2 ? ? ? ?








The we use the fact that 8 = 12
+ 12
+ 12
+ 12
+ χ5(1) to deduce that
χ5(1) = 2 – that is, the final irreducible representation has dimension 2.
Then we can calculate the last row using column orthogonality. For instance,
we have
0 = 1.1 + 1.1 + 1. − 1 + 1. − 1 + 2χ5(a) = 0 + 2χ5(a)
so χ5(a) = 0. And now we can fill in the rest.
Another useful trick for filling in the character table is the following one.
Proposition 6.30. If χ is an irreducible character and θ a 1-dimensional
character, then χ.θ is also an irreducible character.
Proof. By Theorem 6.20, we must show that χ.θ, χ.θ = 1. Now this equals
1
|G| g∈G
χ(g)θ(g)χ(g)θ(g) =
1
|G| g∈G
χ(g)χ(g)θ(g)θ(g) =
g∈G
χ(g)χ(g) = 1
where the second equation uses since θ is 1-dimensional that θ(g) is a root
of unity, and that if x is a root of unity then x = x−1
.
Example 6.31. The regular character χreg is the character of the regular
G-module C[G]. What are its values? Consider g.− : C[G] → C[G] and its
matrix representation [g] with respect to the standard basis {g1, . . . , gn}.
Then [g] is a permutation matrix – we have [g]ij = 1 if g.gj = gi and 0
otherwise. In particular, [g]ii = 0 unless [g] = e, in which case [g]ii = 1.
Therefore, we see that χreg(e) = |G| whilst χreg(g) = 0 otherwise.
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 31
7. Kan extensions, restricted and induced representations
Let H ≤ G be a subgroup. It is clear that if V is a G-module then
it becomes a H-module by restricting the action to elements of H – one
writes V ↓ H for the restricted module. Furthermore if f : V → W is a
G-module map then it can be viewed as a H-module map f : V ↓ H →
W ↓ H between the restricted H-modules – indeed we have a functor
Res : G-Mod → H-Mod.
All of this is obvious. What about the opposite direction? That is, from
a H-module can we construct a G-module and what is the best way to do
so? Category theory suggests the answer – look for a left adjoint to this
functor. Its value at a H-module V is called the induced representation
V ↑ G and is characterised by a bijection
G-Mod(V ↑ G, W) ∼= H-Mod(V, W ↓ H)
natural in W. The above formula, or actually its formulation for characters,
is often called Frobenius reciprocity.
For a group G let G• be the corresponding category with one object •
and G•(•, •) = G. The group multiplication defines composition in G•. As
noted during the first week, a G-module is a functor G• → Vect. And
indeed we can identify G-Mod with the category of functors and natural
transformations G-Mod = [G•, Vect]. The subgroup inclusion H ≤ G
defines a functor J : H• → G• so that we obtain a restriction functor
Res = [J, 1] : [G•, Vect] → [H•, Vect]
which we identify with Res : G-Mod → H-Mod.
Left adjoints to such restriction functors are called left Kan extensions.
More generally, given a functor J : A → B and a category C we obtain
a functor ResJ = [J, 1] : [B, C] → [A, C] and can ask when it has a left
adjoint. In other words, for F : A → C we should find LanJ (F) : B → C
and a natural transformation η : F → LanJ (F) ◦ J such that the induced
function
[B, C](LJ (F), G) → [A, C](F, GJ)
is invertible.
Diagramatically, this says To add.
The functor LanJ (F) : B → C is called the left Kan extension of F along
J. In fact there is a simple categorical criterion for when such left Kan
extensions exist. Firstly, recall that given b ∈ B and J : A → B we can
form the comma category J/x whose objects are pairs (a ∈ A, f : Ja → x).
A morphism h : (a, f) → (b, g) is a morphism h : a → b ∈ A such that the
32 JOHN BOURKE
triangle
Ja
Jh

f
// x
Jb
g
>>
There is a forgetful functor Px : J/x → A : (a, f) → a.
Theorem 7.1. Let J : A → B be a functor from a small category to a
locally small one, and consider F : A → C. If for each b ∈ B the colimit of
J/x
Px
// A
F // B (a, f) → a → Fa
exists, then lanJ (F) exists, and has value at x given by the above colimit
col(F ◦ Px). In particular, if C is cocomplete then the left Kan extension
always exists.
Proof. Add proof.
Theorem 7.2. • The induced representation V ↑ G exists.
• If G and V are finite dimensional then V ↑ G can be described as
follows. Let G = t1H ∪ . . . ∪ tnH decompose G into cosets of H. Write
tiV ∼= V for an isomorphic copy of V whose elements we write as tiv
where v ∈ V . Then
V ↑ G = t1V ⊕ . . . ⊕ tnV.
Given g ∈ G we should describe its action g(Σitiv). We have gti =
tσ(i)hi for a unique transversal tσ(i) and hi ∈ H. Then
g(Σitiv) = Σitσ(i)hiv.
Proof. The first part holds by the preceding theorem, on observing that the
category of vector spaces has all small colimits – these can be constructed
using (infinite) direct sums and cokernels.
For the second part, consider J : H• → G•. Now G• has a single object
•. What does the slice category J/• look like? Its objects are morphisms
g : • → • ∈ G• – that is elements of G. Morphisms h : g → g are
commutative triangles
•
h∈H

g
// •
•
g
??
where h ∈ H – that is, g h = g. Consider the functor
J
P•
// H•
V // Vect
We must show, firstly, that the colimit is t1V ⊕ . . . ⊕ tnV . For each g ∈ G
we must give a cocone map pg : V → t1V ⊕ . . . ⊕ tnV . To define this, we
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 33
use that g = ti.hi for unique ti and h ∈ H and then define pg(v) = tih.v –
note that this is a linear map since h.− : V → V is. Next, we check that
we have a cocone – namely that if g h = g as above, then the triangle left
below commutes.
V
h.−

pg
))
qg
))
t1V ⊕ . . . ⊕ tnV
k // W
V
pg
55
qg
55
For this, let v ∈ V and consider pg (h.v). Now if g = ti.hi then since
g = g h we have g = gh−1
= ti(hih−1
). Therefore pg (h.v) = ti(hih−1
)h.v =
tihi.v = pgv as required.
Finally, suppose we have a second cocone q to W as above – we must show
that there exists a unique map of cocones making the diagram commute.
Observe that pti
(v) = tiv – thus since we must have k ◦ pti
= qti
, and k
must be linear, we must set k(Σitiv) = Σiqti
v. Thus the definition of k is
forced on us – we must show that k ◦ pg = qg for all g. For g = tihi we have
kpgv = k(tihi.v) = qti
(hi.v) = qgv where the last equation holds by virtue
of that q is a cocone. This proves that it is the colimit as claimed.
By construction of the left Kan extension, the action satisfies
V
pti
))
pgti
++
t1V ⊕ . . . ⊕ tnV
g.−
// t1V ⊕ . . . ⊕ tnV
which says that g(tiv) = tσ(i)hiv where gti = tσ(i)hi, as claimed.
Remark 7.3. The universal map of H-modules ηV : V → (V ↑ G) ↓ H is
the map pe : V → t1V ⊕. . .⊕tnV where e is the identity element of G. This
is never invertible unless H = G – the rhs has dimension dim(V ).|G|/|H|.
In particular, restriction along ηV induces a bijection G-Mod(V ↑ H, W) ∼=
H-Mod(V, W ↓ H) – this is the established adjointness. If we view both left
and right hand sides as vector spaces, with the pointwise structure, then
since ηV is linear, the bijection is in fact an isomorphism of vector spaces
G-Mod(V ↑ H, W) ∼= H-Mod(V, W ↓ H)
Theorem 7.4 (Frobenius reciprocity). For H ≤ G finite, V a H-module,
and W a G-module over C we have
χV ↑G, χW = χV , χW↓H .
34 JOHN BOURKE
Proof. Recall that A, B = dim(G-Mod(A, B)). Thus what the above says
is that dim(G-Mod(V ↑ G, W)) = dim(H-Mod(CH(V, W ↓ H)), which
follows immediately from the remark above.
Example 7.5. Consider the trivial H-module C, where H ≤ G. What is
C ↑ G? According to the above formula it is t1K ⊕ . . . tnK with action
given by gti = tσ(i) where gti = tσ(i)h some h. Another way of looking at it
is therefore that it is C[t1, . . . , tn] which is the permutation representation
corresponding to the left coset action of G on {t1, . . . , tn}.
Having already established Frobenius reciprocity, we can prove the following
formula for the character of the induced representation. Let V be a
H-module with character ϕ : H → C. Let us write ϕ ↑ G for the character
of V ↑ G. To express this, we firstly extend the definition of ϕ by defining
a function
˙ϕ : G → C : g ∈ H → ϕ(g), g /∈ H → 0
Proposition 7.6. The induced character satisfies the formula
ϕ ↑ G(g) =
1
|H| y∈G
˙ϕ(y−1
gy) .
Proof. It is easy to see that the function f : G → C on the right is a
class function. Since the irreducible characters χ1, . . . , χn form a basis of
the class functions it suffices to show that the two class functions above
have the same coefficients when written as linear combinations of the χi –
equivalently that
ϕ ↑ G, χ = f, χ
for each irreducible character χ. Now by Frobenius reciprocity, the left
hand side is equally ϕ, χ ↓ H , which equals
1
|H| x∈H
ϕ(x)χ(x) .
The right hand side is
1
|G| g∈G
f(g)χ(g) =
1
|G|
1
|H| g∈G y∈G
˙ϕ(y−1
gy)χ(g) .
Let x = y−1
gy. Then the above equals
1
|G|
1
|H| x∈G y∈G
˙ϕ(x)χ(yxy−1) =
1
|G|
1
|H| x∈H
|G|ϕ(x)χ(x) =
1
|H| x∈H
ϕ(x)χ(x) .
where in the first equality we have used that ˙ϕ is zero if x /∈ H and that
χ(yxy−1
) = χ(x).
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 35
8. Representation theory of the symmetric group
The number of irreducible representations of the symmetric group equals,
by Theorem 6.25, the number of conjugacy classes. As is well known, the
conjugacy classes of Sn are parametrised by the cycle type of its elements.
We can think of cycle types as being specified by partitions of n, which
are sequences λ = (λ1, . . . , λl) with λi ≥ λi+1 and adding to n. One writes
λ n to indicated that λ is a partition of n. For instance, the cycle type of
the permutation (123)(45)(678) corresponds to the partition (3, 3, 2). The
partition can also be represented by an array
whose rows are of length the numbers in the partition. This array is called
the shape of the partition. A Young tableaux of shape λ n is an array t of
the given shape whose entries are bijectively filled by the numbers 1, . . . , n.
For instance, one Young tableau is
2 3 1
4 7 8
5 6
Evidently if λ n there is a bijection between λ-tableaux and elements
of Sn – under this correspondence the above tableau corresponds to the
permutation
1, 2, 3, 4, 5, 6, 7, 8 → 2, 3, 1, 4, 7, 8, 5, 6
where we read across the rows first. In particular, there are n! λ-tableaux.
Furthermore, Sn acts on the set of λ-tableau in the obvious way – that
is, (πt)i,j = π(ti,j). In other words, π acts on the entries of the tableaux –
it doesn’t care about position. For instance, we have
(13). 2 3
1
= 2 1
3
As a Sn-set this is simply the left action of Sn on itself, up to isomorphism.
Two λ-tableaux s and t are said to be row equivalent if the entries in
each row coincide. For instance, the (2, 1)-tableaux
2 3
1
and 3 2
1
are row equivalent. Row equivalence is, clearly, an equivalence relation –
its equivalence classes {t} are called λ-tabloids.
36 JOHN BOURKE
Diagramatically, we remove the boxes from the rows to denote a λ-tabloid
– the single λ-tabloid below
2 3
1
corresponds to the two above λ-tableaux.
To see that the action of Sn on the set of λ-tableaux respects rowequivalence,
observe that it suffices to check it on the generators of Sn –
that is, the transpositions ρ = (ij). Let s t be row equivalent, and let i
and j appear in rows n and m of s respectively. Then i and j appear in
rows m and n and ρs respectively. But i and j appear in the same rows m
and n of t as for s and so likewise have their rows reversed in ρt. Otherwise
the two tableaux are unchanged – hence they are still row equivalent.
Because the action respects row equivalence it induces an action on the
set of λ-tabloids by π{t} = {πt}.
Definition 8.1. Let {t1}, . . . , {tk} be a complete set of λ-tabloids. We let
Mλ
= C({t1}, . . . , {tk}) denote the permutation Sn-module whose basis
elements are the λ-tabloids.
Examples 8.2. • If λ = (n) then there is only one λ-tabloid
1 2 3 . . . n
so M(n)
= C( 1 2 3. . . n ) with the trivial action.
• If λ = (1, . . . , 1) then no two λ-tableaux are row equivalent. In particular,
a λ-tabloid just amounts to an element of Sn – thus Mλ
= C{Sn}
is the regular representation – corresponding to the action of Sn on
{1, . . . , n}.
• In the case of λ = (n − 1, n) a λ-tabloid amounts to the choice of the
single element in the second row, we denote this λ-tabloid by i – in
this way we see that Mλ
= C{1, . . . , n} is an isomorph of the so-called
standard representation.
Note that the Mλ
above are not in general irreducible. However each
contains a canonical irreducible submodule Sλ
≤ Mλ
as we will see shortly.
Definition 8.3. A G-module V is said to be cyclic if it is generated by
a single element v ∈ V – in other words each w ∈ V is of the form
λ1g1v + . . . λngnv for gi ∈ G.
Proposition 8.4. Each module Mλ
is cyclic.
Proof. This follows from the fact that any λ-tabloid can be taken to any
other by a suitable permutation.
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 37
Remark 8.5. We can describe the above more algebraically. Firstly, observe
that if λ n there is a bijection between λ-tableaux and elements of Sn.
For instance, the tableau
2 1 4
5
3
corresponds to the permutation
1, 2, 3, 4, 5 → 2, 1, 4, 5, 3
where we read across the rows first. In particular, there are n! λ-tableaux.
Under this correspondence, the action of Sn on λ-tableaux corresponds
to the left action of Sn on itself: that is, π.− : Sn → Sn : θ → π ◦ θ.
What about row equivalence? The Young subgroup Sλ is defined as the
subgroup
S{1,...,λ1} × S{λ1+1,...,λ2} × . . . × S{n−λl+1,...,n} ≤ Sn
of the full symmetric group. This captures permutations which don’t
change rows, where the first λ1 correspond to the first row etc. For instance,
consider the row equivalent tableaux s and t
2 1 4
5
3
2 4 1
5
3
corresponding to the permutations
s : 1, 2, 3, 4, 5 → 2, 1, 4, 5, 3 and t : 1, 2, 3, 4, 5 → 2, 4, 1, 5, 3
Taking the permutation h : 1, 2, 3, 4, 5 → 1, 3, 2, 4, 5 we see that s.h = t.
In general, two tableaux s and t are row equivalent just when there exists
h ∈ Sλ such that s.h = t. In particular, the λ-tabloids {t1}, . . . , {tk} are
thus in bijective correspondence with the left cosets t1Sλ, . . . , tkSλ and the
action on the former corresponds to the coset action on the latter. Passing
to permutation representations, it follows that
Proposition 8.6. The permutation representation Mλ
= C({t1}, . . . , {tk})
is isomorphic to the coset permutation representation C(t1Sλ, . . . , tkSλ).
In particular, by Example 7.5, this means that Mλ
is the induced module
of the trivial representation of the subgroup Sλ.
8.1. Specht modules. We now turn to the irreducible modules Sλ
. To
this end, we need to reconsider λ-tableaux.
Definition 8.7. Let t be a λ-tableau.
(1) We define the row stabliliser Rt ≤ Sn to consist of those π which
permute only the elements within each row of t.
38 JOHN BOURKE
(2) We define the column stabliliser Ct ≤ Sn to consist of those π which
permute only the elements within each column of t.
In particular, if t has rows R1, . . . , Rk then Rt = SR1 × . . . × SRk
; if t has
columns C1, . . . , Ck then Ct = SC1 × . . . × SCk
.
4 1 2
3 5
then Rt = S4,3 × S4,1,2 × S3,5 and Ct = S4,3 × S1,5 × S2. We will be
concerned primarily with column stabilisers.
But firstly, let H ⊆ Sn be an arbitrary subset. We can then define an
element
H−
=
π∈H
sgn(π)π ∈ C(Sn)
in the group algebra. As the key instance of this, let
kt = C−
t =
π∈Ct
sgn(π)π ∈ C(Sn).
Because Mλ
is a Sn-module, we obtain kt.− : Mλ
→ Mλ
.
Definition 8.8. Let t be a tableaux. The associated polytabloid is the
element et = kt({t}) ∈ Mλ
.
In the above example, we have kt = e − (4, 3) − (1, 5) + (4, 3)(1, 5). Hence
et = 4 1 2
3 5
− 3 1 2
4 5
− 4 5 2
3 1
+ 4 1 2
3 5
.
Definition 8.9. The Specht module Sλ
is the submodule Sλ
= et :
t a λ-tableau ≤ Mλ
spanned by the polytabloids, one for each λ-tableau t.
Lemma 8.10. Let t be a λ-tableau and π a permutation. We have
(1) Rπt = πRtπ−1
(2) Cπt = πCtπ−1
(3) kπt = πktπ−1
(4) eπt = πet.
Proof. These are all straightforward. For (1) we have
ρ ∈ Rπt ⇐⇒ ρ{πt} = {πt}
⇐⇒ π−1
ρπ{t} = {t}
⇐⇒ π−1
ρπ{t} ∈ Rt
ρ ∈ πRtπ−1
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 39
For (2) one argues in an identical fashion but using column equivalence
rather than row equivalence. For (3) we have
kπt =
θ∈Cπt
sgn(θ)θ =
θ∈πCtπ−1
sgn(θ)θ =
θ∈Ct
sgn(πθπ−1
)πθπ−1
=
θ∈Ct
sgn(θ)πθπ−1
= π(
θ∈Ct
sgn(θ)θ)π−1
= πktπ−1
Finally we have πet = πkt{t} = πktπ−1
π{t} = kπt{πt} = eπt.
Corollary 8.11. The Specht module Sλ
is cyclic.
Proof. Since any two tableaux are related by a permutation, it follows from
part (4) above that any two of the generating polytabloids are related by a
permutation. The claim follows.
Examples 8.12. (1) Suppose λ = (n). Then for each λ-tableau the
column stabliliser group is trivial, and we see that kt = {t} the unique
λ-tabloid
1 2 3 . . . n .
In particular S(n)
= Mn
= C( 1 2 3. . . n ) with the trivial module
structure.
(2) Suppose λ = (1, 1, . . . , 1) so that there are Sn-many λ-tabloids Mλ ∼=
C(Sn). Let t be the λ-tableau with vertical elements 1, 2, 3, 4, . . . , n.
We will prove that eπt = sgn(π)et – in particular, each polytabloid is a
scalar multiple of et, so that as a vector space Sλ
= C(et). Furthermore,
since we then have πet = eπt = sgn(π)et at the single basis element et
we see that Sλ
is isomorphism to the sign representation.
It remains to prove that eπt = sgn(π)et. Now
eπt = πet = π
θ∈Sn
sgn(θ)θ{t}
=
θ∈Sn
sgn(θ)πθ{t} =
ϕ∈Sn
sgn(π−1
ϕ)ππ−1
ϕ{t}
= sgn(π−1
)
ϕ∈Sn
sgn(ϕ)ϕ{t} = sgn(π)et
(3) Consider λ = (n − 1, 1) again, recalling our identification Mλ
=
C(1, . . . , n). Now the λ-tableau
t = i . . . j
k
with {t} = k has column stabiliser {e, (ik)} so that et = k − i. Thus
Sλ
= i−j : i = j ≤ Mλ
. It is easy to see that this spans the subspace
40 JOHN BOURKE
{c11 + . . . + cnn : ci = 0} ≤ C(1, . . . , n). A basis for this subspace
is given by the vectors i − 1 for i > 1, so that the vector space has
dimension n − 1.
This is equally the complement of the subspace C(1 + . . . + n).
8.2. The dominance ordering on λ-tableaux.
Definition 8.13. Let λ = (λ1, λ2, . . . , λl) and µ = (µ1, µ2, . . . , λm) be
partitions of n. We say λ µ (or λ dominates µ) if
λ1 + λ2 + λi ≥ µ1 + µ2 + µi
for all i ≥ 1. (For i ≥ l we set λi = 0 and likewise if i ≥ m we take µi to
be zero.)
For instance
dominates but is incomparable to
Remark 8.14. By the above, it is not a total order, though it is in fact a
lattice – we will not need this fact.
Lemma 8.15 (Dominance lemma). Let λ, µ n and and consider tableaux
tλ
and sµ
. If the elements of each row of s belong to distinct columns of t,
then λ µ.
Proof. Let i in row 1 of sµ
, If i does not appear in row 1 of tµ
, then swap i
in tλ
with the element i1 above it in row 1 of tλ
. By assumption i1 cannot
belong to row 1 of sµ
– hence we produces a new tableau tλ
1 whose first row
contains all elements in the first row of sµ
, and still satisfying the hypothesis
of the proposition.
Now, if i in row 2 of sµ
does not belong to rows 1 or 2 of tλ
1 , we permute
it with the entry above it i2 in the second row of tλ
1 to obtain tλ
2 – since i2
cannot belong to the first or second row of sµ
we obtain a new λ-tableau
which has the further property that its first two rows contain the elements
of the first two rows of sµ
.
Continuing inductively, we obtain a tableau tλ
∗ , such that for all i the
first i rows of sµ
belong to the first i rows of tλ
∗ . Therefore the number of
elements in the first i rows of sµ
must be less than or equal to the number
of elements in the first i rows of tλ
. In other words, for all i we have
µ1 . . . + µi ≤ λ1 + . . . + λi, as required.
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 41
For an example illustrating the above proof, consider
sµ
= 1 2
3 4
5
6
and tλ
= 1 4 6
3 2 5
Permuting the second column, we modify it to
1 2 6
3 4 5
which now has the property that each element of the first i rows of sµ
belongs to the first i rows of the modified λ-tableau, and so implies λ µ.
8.3. Irreducibility of the Specht modules. There is a unique complex
inner product on Mλ
defined on basis vectors {t} and {s} by {t}, {s} =
δ{t},{s}, which is linear in the first variable and conjugate linear in the second.
That is, to say
{t}
λ{t}{t},
{t}
κ{s}{s} =
{t},{s}
λ{t}κ{s} {t}, {s}
Thanks to the fact that we use conjugate linearity in the second variable,
this is positive definite. It is also Sn-invariant, in the sense that
πu, πv = u, v
for any permutation π. This follows from the fact that {t} = {s} if and
only if π{t} = π{s}.
Lemma 8.16 (Sign lemma). Let H ≤ Sn be a subgroup.
(1) If π ∈ J then πH−
= H−
π = sgn(π)H−
.
(2) For u, v ∈ Mλ
we have H−
u, v = u, H−
v .
(3) Suppose that the transposition (a, b) ∈ H. Then H−
= k(e − (a, b)) for
some k ∈ C(Sn).
(4) If t is a tableau with a, b in the same row and (a, b) ∈ H then H−
{t} =
0.
Proof. (1)
πH−
= π
θ∈H
sgn(θ)θ =
θ∈H
sgn(θ)πθ =
θ
sgn(π−1
θ)ππ−1
θ
= sgn(π)
θ
sgn(θ)θ = sgn(π)H−
.
42 JOHN BOURKE
Similarly
H−
π = (
θ∈H
sgn(θ)θ)π =
θ∈H
sgn(θ)θπ =
θ
sgn(θπ−1
)θπ−1
π
= sgn(π)
θ
sgn(θ)θ = sgn(π)H−
.
(2)
H−
u, v =
θ∈H
sgn(θ)θu, v =
θ∈H
sgn(θ) θu, v =
θ∈H
sgn(θ) u, θ−1
v
= u,
θ∈H
sgn(θ)θ−1
v = u,
θ∈H
sgn(θ−1
)θ−1
v = u, H−
v
where for the third equality we use Sn-invariance of the inner product.
(3) Let L = {e, (a, v)} and let H = k1L ∪ . . . ∪ knL be a transversal. Write
K = {k1, . . . , kn}. Then
H−
=
i=1,...n
(sgn(kie)kie + sgn(ki(a, b))ki(a, b))
=
i=1,...n
(sgn(ki)ki − sgn(ki)ki(a, b)) = K−
(e − (a, b))
as claimed.
(4) The assumption gives (a, b){t} = {t}. Gence H−
{t} = K−
(e −
(a, b)){t} = K−
(e{t} − (a, b){t}) = K−
({t} − {t}) = 0, using the
preceding part.
Corollary 8.17. Let t = tλ
and s = sµ
be tableaux, with λ, µ n. If
kt{s} = 0 then λ µ. If, moreover, λ = µ then kt{s} =+
− et.
Proof. Suppose a and b belong to the same row of s; if they belong to the
same column of t, then (a, b) ∈ Ct so that kt(s) = 0 by Part 4 of the sign
lemma. Therefore, by the Dominance Lemma, we have λ µ.
We know, from the first part, that elements of each row of s belong to
different columns of t. Then, by the construction in the dominance lemma,
we can find π ∈ Ct so that each element of the first i rows of s belongs to
the first i rows of πt. Since the tableaux have the same shape, this means
that πt and s are row equivalent – that is {s} = π{t}. By the above, we
then have kt{s} = kt{πt} = sgnπkt{t} =+
− et.
Corollary 8.18. Let u ∈ Mλ
and t a λ-tableau. Then kt(u) is a scalar
multiple of et.
Proof. We can write u = i ai{si} for λ-tableaux si. Then kt(u) =
i aikt{si} = i ai
+
−et, by the preceding part.
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 43
Theorem 8.19. Let U be a submodule of Mµ
. Then Sµ
⊆ U or U ⊆ (Sµ
)⊥
.
In particular, the Sµ
are irreducible.
Proof. Let u ∈ U and consider a µ-tableau t. Consider kt ∈ C(Sn). Since U
is a submodule kt(u) ∈ U, whilst the preceding result kt(u) = aet for some
a. If a = 0 then et = a−1
kt(u) ∈ U. Since Sµ
is cyclic, with Sµ
= et we
then have Sµ
≤ U.
The other possibility is that for all µ-tableaux and u ∈ U we have
kt(u) = 0. We must show that u ∈ (Sµ
)⊥
– it suffices to show that
u, et = 0 for all µ-tableaux t. Now
u, et = u, kt{t} = ktu, {t} = 0, {t} = 0
where the second equality uses the third part of the sign lemma.
Now suppose U ⊆ Sµ
. If Sµ
⊆ U then U = Sµ
; the other possibility is
that U ⊆ (Sµ
)⊥
so that U ⊆ (Sµ
)⊥
∩ Sµ
. Since −, − is positive-definite,
this intersection is zero – hence U = 0.
Proposition 8.20. Suppose θ : Sλ
→ Mµ
is a non-zero Sn-module homomorphisms.
Then λ µ, and if λ = µ then θ is multiplication by a
scalar.
Proof. Since θ = 0 there exists a basis vector et such that θ(et) = 0. Now
since Mλ
= Sλ
⊕ (Sλ
)⊥
we can extend this to a non-zero homomorphism
θ : Mλ
→ Mµ
.
Now 0 = θ(et) = θ(kt{t} = ktθ({t}) = kt( i ci{si} = i cikt{si}, so
there exists some µ-tableau kt({si}) = 0. Therefore λ µ.
Finally, by Corollary 8.18, we have that θ(et) = ktθ({t}) = cet for some
non-zero scalar c. Then θ(eπt) = θ(πet) = πθ(et) = πcet = cπet. Hence θ is
multiplication by c.
Theorem 8.21. The Sλ
for λ n form a complete list of irreducible
Sn-modules.
Proof. We have already proven that they are irreducible. Let us prove
that they are pair-wise non-isomorphic. Suppose λ = µ but θ : Sλ ∼= Sµ
.
Then, by composition, we obtain an injective homomorphism Sλ
→ Mµ
and likewise Sµ
→ Mλ
. By the preceding result we then have λ µ and
µ λ so that λ = µ – a contradiction.
Since the number of partitions equals the conjugacy classes of Sn, this
gives a complete list of the irreducible representations of Sn.
We have not said anything so far about the bases or dimensions of these
irreducible representations. At this point, I will just give a brief survey of
this topic.
Definition 8.22. A tableau is said to be standard if both its rows and
columns form increasing sequences.
44 JOHN BOURKE
For example, 1 2 6
3 4
5
is standard but 1 2 6
4 3
5
is not.
Theorem 8.23. The set {et : t is a standard λ-tableau} forms a basis for
Sλ
.
We will not prove the above, or anything else about these representations,
though there is much more to say.
Let fλ
denote the number of standard λ-tableaux. Then since the number
of elements of a group equals the sum of the squares of the dimensions of
its irreducible representations, we have as an immediate consequence of the
above theorem
Corollary 8.24. n! = λ:λ n(fλ)2
.
This is a purely combinatorial identity which admits other interesting
proofs. An elegant proof is via the so-called Robinson-Schensted correspondence,
which draws a bijective correspondence between pairs of standard
tableaux and elements of Sn.
9. Representations of Hopf algebras
Let (C, ⊗, I) be a symmetric monoidal category – for example, the category
(Set, ×, 1) with cartesian product, the category (Ab, ⊗, Z) with tensor
product of abelian groups or (Vect, ⊗, k) with the tensor product of vector
spaces.
A monoid/algebra in C is an object A equipped with morphisms m :
A ⊗ A → A and e : l → A satisfying the associativity:
A ⊗ A ⊗ A
1⊗m

m⊗1
// A ⊗ A
m

A ⊗ A m
// A
A
##
e⊗1
// A ⊗ A
m

A
A
##
1⊗e
// A ⊗ A
m

A
It is said to be commutative if m = m ◦ s : A ⊗ A → A ⊗ A → A where
s is the symmetry.
Remark 9.1. We write as if we are in a strict monoidal category, rather
than a free one, for notational convenience.
Examples 9.2. An algebra in Set is a monoid. An algebra in (Ab, ⊗, Z)
is a ring. An algebra in (Vect, ⊗, k) is an algebra in the classical sense –
that is, an associative unital algebra.
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 45
An algebra morphism f : (A, m, e) → (B, m , e ) is a morphism f : A → B
such that
A ⊗ A
m

f⊗f
// B ⊗ B
m

A
f
// B
I
e

e

A
f
// B
Algebras and algebra morphisms form a category U : Alg(C) → C over C.
A coalgebra in C is an object A equipped with a comultiplication ∆ :
A → A ⊗ A and counit q : A → I satisfying the coassociativity and counit
equations dual to the above ones. Equivalently, it is an algebra in (Cop
, ⊗, I).
Again we have the category Coalg(C) → C.
Examples 9.3. In Set, or any cartesian monoidal category, each object X
admits a unique coalgebra structure with comultiplication ∆ : X → X ×X :
x → (x, x) and counit X → 1 the unique such map. Good exercise: check
the details of this!
A bialgebra is an object A equipped with both an algebra structure
(A, m, e) and a coalgebra structure (A, ∆, q) satisfying the further equations
A ⊗ A
m

∆⊗∆
// A ⊗ A ⊗ A ⊗ A
1⊗s⊗1

A ⊗ A ⊗ A ⊗ A
m⊗m

A
∆ // A ⊗ A
I
e

e⊗e
!!
A
∆ // A ⊗ A
and
I
1

e // A
q

I
A ⊗ A
q⊗q
##
m // A
q

I
Remark 9.4. Note that since C is symmetric monoidal, the categories
Alg(C) and Coalg(C) inherit symmetric monoidal structure lifted from C.
Given algebras A and B we form A⊗B and equip it with the tensor product
structure
A ⊗ A ⊗ B ⊗ B
1⊗s⊗1
// A ⊗ B ⊗ A ⊗ B
mA⊗mB
// A ⊗ B
and
I
eA⊗eB
// A ⊗ B
46 JOHN BOURKE
Note that this does not work at all if C is not symmetric. Observe also that
it gives the usual pointwise formula for product of monoids when C = Set –
that is, (a, b) (a , b ) = (mA(a, a ), mB(b, b )).
Checking the axioms and lifting the remainder of the structure is straightforward
– in particular, the forgetful functors Alg(C), Coalg(C) → C are
strict monoidal.
The first two equations of a bialgebra say that ∆ : A → A ⊗ A is a
morphism of algebras and the second two that q : A → I is a morphism of
algebras. In other words, a bialgebra is exactly a coalgebra in the category
of algebras! Reading the diagrams vertically, we see that it is equally an
algebra in the category of coalgebras. Indeed, we have
Bialg(C) ∼= Coalg(Alg(C)) ∼= Alg(Coalg(C)) .
A Hopf algebra is a bialgebra A further equipped with an antipode map
a : A → A
such that the two composites on the top row equal that on the bottom
below
A
q
##
∆ // A ⊗ A
1⊗a
66
a⊗1
))
A ⊗ A
m // A
I
e
55
Here is another way of looking at the definition of the antipode. I will just
explain it in the case C = Vect since my goal here is to make it intuitive,
although it is easily seen to be true in any symmetric monoidal closed
category.
Firstly, if A a coalgebra and B an algebra, then the hom vector space
Vect(A, B) becomes an algebra if at f, g : A B we define
A
∆ // A ⊗ A
f⊗g
// B ⊗ B
m // B
and the unit element to be e◦q : A → k → A. This is called the convolution
algebra.
In the case that A = B, we see that the antipode is exactly an element
of the convolution algebra [A, A] which is an inverse for 1 : A → A under
the convolution product – that is, a convolution inverse of 1 : A → A.
Since inverses in an algebra are unique, observe that this shows that
antipodes are unique, if they exist.
Examples 9.5. (1) A Hopf algebra in (Set, ×, 1) is precisely a group.
The map a : A → A sends an element to an inverse – the two above
equations indeed say exactly this: that is, m(a(x), x) = e = m(x, a(x)).
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 47
(2) A Hopf algebra in Vect is, by definition, a Hopf algebra. We will
now show that the group algebra k[G] is a Hopf algebra. Here is an
abstract explanation. The functor k[−] : Set → Vect sending a set X
to the free vector space k[X] is a strong symmetric monoidal functor
– in other words, we have isomorphisms k[X] ⊗ k[Y ] ∼= k[X × Y ] and
k ∼= k[1] compatible with the associativity isomorphisms, unit and
symmetry isomorphisms. Any such functor sends algebras to algebras,
coalgebras to coalgebras, bialgebras to bialgebras and Hopf algebras
to Hopf algebras in an obvious way! Since Hopf algebras in Set are
exactly groups, it follows that if G is a group then the group algebra
k[G] is a Hopf algebra.
That is, we have algebra structure
k[G] ⊗ k[G]
∼= // k[G × G]
k[m]
// k[G] k
∼= // k[1]
k[e]
// k[G]
and coalgebra structure
k[G]
k[∆]
// k[G × G]
∼= // k[G] ⊗ k[G] k[G]
k[!]
// k[1]
∼= // k
with antipode
k[a] : k[G] → k[G].
The algebra structure is that which we have seen many times in the
course. The coalgebra structure is given by ∆k[G](Σiλigi) = Σiλi(gi ⊗gi)
and qk[G](Σiλigi) = Σiλi, whilst the antipode ak[G] : k[G] → k[G] similarly
sends Σiλigi to Σiλi(gi)−1
. Since the group G is a cocommutative
bialgebra in Set – i.e. the comultiplication ∆ is cocommutative – it
follows that k[G] is also a cocommutative bialgebra. Of course it won’t
be commutative unless G is abelian.
(3) It would be nice to add further examples here, such as the universal
enveloping algebra of a Lie algebra, the tensor algebra of a vector space
and some combinatorial examples.
9.1. Modules and comodules over a bialgebra. Since a bialgebra is
both an algebra and a coalgebra we can consider both its modules and its
comodules. Here, we will focus on modules since k[G]-modules are precisely
representations of G – see Section 1.3.
Let M be a bialgebra, and X, Y (left) M-modules, via actions x : M⊗X →
X and y : M ⊗ Y → Y . Then X ⊗ Y is a M-module with action:
M ⊗ X ⊗ Y
∆⊗1⊗1
// M ⊗ M ⊗ X ⊗ Y
1⊗s⊗1
// M ⊗ X ⊗ M ⊗ Y
a⊗b
// X ⊗ Y
Observe, that is generalises the tensor products of G-modules when
M = k[G] since it just captures the formula g.(a ⊗ b) = ga ⊗ gb.
Here are the associativity and unit equations for the module.
48 JOHN BOURKE
MMXY
1∆11

m11 // MXY
∆11

MMMXY
11sM,X 1

∆111// MMMMXY
111sM,X 1

1sM,M 111
// MMMMXY
11sMM,X 1

mm11// MMXY
1sM,X 1

MMXMY
1xy

∆111// MMMXMY
11xy

1sM,MX 11
// MMXMMY
1x1y

m1m1 // MXMY
xy

MXY
∆11 // MMXY
1sM,X 1
// MXMY
xy
// XY
MXY
∆11// MMXY
1s1

XY
e11
::
ee11
44
e1e1 //
1
**
MXMY
xy

XY
In the above, we use that ∆ is an algebra map – once in the first and
once in the second diagram. Otherwise, it is straightforward to see that the
tensor product lifts to give a monoidal structure on M − Mod preserved by
the forgetful functor to C. (If one wants to lift the symmetry to M − Mod
then one requires that the bialgebra be cocommutative.)
In fact, for C = Vect there is a bijection between
• Bialgebra structures on the k-algebra M.
• Monoidal structures on M −Mod strictly preserved by U : M −Mod →
C
In the opposite direction, if we have a lifted monoidal structure (M −
Mod, ⊗, I) how do we get the bialgebra structure. It is easy to see that
(M, m) is itself the free M-module on the unit I; hence we can form the
tensor product M-module (M, m)⊗(M, m) = (M ⊗ M, n) where we write
n the left action n : M ⊗ M ⊗ M → M ⊗ M; then the comultiplication for
the coalgebra M is given by the composite
M
1ee // MMM
n // MM.
For the counit, we use that I lifts to an M-module; its actions then gives
the counit M ∼= M ⊗ I ∼= I for the coalgebra structure.
Remark 9.6. Here is a more general approach to the above. An opmonoidal
functor (F, f, f0) : A B consists of a functor F : A → B, coherence
constraints fa,b : F(ab) → FaFb natural in a and b and a comparison f0 :
Fi → i compatible with the monoidal structure – that is, the associativity
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 49
and unit isomorphisms. Opmonoidal functors can be composed – given
F : A → B and G : B → C as above, the components
GF(ab)
Gf
// G(FaFb)
g
// GFaGFb GFi
Gf0
// Gi
g0
// i
equip GF with opmonoidal structure. A monoidal transformation between
opmonoidal functors is a natural transformation satisfying the extra condi-
tions
F(ab) G(ab)
FaFb GaGb
ηab
//
ηaηb
//
fa,b

ga,b

iB
FiA
GiA
f0
//
g0 $$
ηi

which we call the associativity, left unit and right unit conditions.
An opmonoidal monad is a monad (T, η, µ) on a monoidal category C such
that T is a opmonoidal functor and η, µ opmonoidal transformations. Now,
if A is an algebra then A ⊗ − : C → C is a monad, whose multiplication and
unit have components m⊗1 : A⊗A⊗X → A⊗X and unit e⊗1 : X → A⊗A.
The algebras for the monad are then precisely the left A-modules.
If A is furthermore, a bialgebra, then A ⊗ − is an opmonoidal monad –
henceforth I will use juxtaposition for tensor product. The structure maps
are
AXY
∆11 // AAXY
1s1 // AXAY A
q
// I
It is easy to see that this does give an opmonoidal functor, using that A is a
coalgebra. The two equations that we proved above (if we ignore the bottom
rows) describe the interaction of MXY → MXMY and XY → MXY
with muliplication and unit respectively, and the first is by far the trickiest
to check.
If T is an opmonoidal monad then the category of algebras AlgT admits
a monoidal structure such that the forgetful functor U : AlgT → C is strict
monoidal. Given T-algebras (X, x) and (Y, y) the tensor T-algebra on XY
has structure map
T(XY )
ϕ
// TXTY
xy
// XY
where ϕ is the component of the opmonoidal functor. This generalises the
tensor product of modules described above. In fact, for general monad T
on C there is a bijection between
• opmonoidal structures on the monad T.
• Monoidal structures on AlgT strictly preserved by U : AlgT → C
9.2. Antipodes and internal homs. In the case that M is a Hopf algebra
we can say a bit more – namely, we can show that the internal hom in C
also lifts to the category of modules. The main idea is this:
(1) We have seen that ∆ : M → M ⊗ M is an algebra map.
50 JOHN BOURKE
(2) Now we can also form the opposite algebra Mop
with reversed multiplication
m ◦ s : M ⊗ M → M ⊗ M → M. It turns out that if M is a
Hopf algebra then the antipode is an algebra map a : M → Mop
; note
that for groups this captures the fact that (xy)−1
= y−1
x−1
.
(3) Now, in general, if X, Y are left M-modules then the hom [X, Y ] is
an M-bimodule, or equally, an Mop
⊗ M-module. The point is that
the action is contravariant in X and covariant in Y – it is given by
(a ⊗ b.f)x = bf(ax).
(4) Now we can restrict scalars along the algebra map (a ⊗ 1) ◦ ∆ : M →
M ⊗ M → Mop
⊗ M to equip [X, Y ] with the structure of a left
H-module.
(5) Explicitly, we can give the formula as follows:
M ⊗ [X, Y ]
∆⊗1
// M ⊗ M ⊗ [X, Y ]
1⊗s
// M ⊗ [X, Y ] ⊗ M
µ1⊗a
// [X, Y ] ⊗ M
µ2
// [X, Y ]
where µ1 is the left action
M ⊗ [X, Y ]
˜y⊗1
// [Y, Y ] ⊗ [X, Y ]
◦ // [X, Y ]
sending m ⊗ f to x → mf(x)), and µ2 the contravariant action
[X, Y ] ⊗ M
1⊗˜x
// [X, Y ] ⊗ [X, X]
◦ // [X, Y ]
sending m ⊗ f to x → f(mx)).
For groups representations, this gives the usual formula for the internal
hom
(g.f)x = g(f(g−1
x)).
10. Exam
Know the main topics and results in the course and proofs of the theorems.
The main topics we covered are
(1) Basics on irreducibility: Maschkes theorem, Schur’s lemma, decomposition
of modules into irreducibles, decomposition of the regular module,
representations of abelian groups.
(2) Tensors and homs of G-modules: the relationship between homs, tensor
products and duals.
(3) Characters, their basic properties, how characters determine a representation,
orthonormality of irreducible characters, the number of
irreducible characters, character tables, the inner product and orthogonality
relations, calculating character tables, Frobenius reciprocity
formula, but not the proof using Kan extensions.
(4) Representations of the symmetric group – definitions of tableaux,
tabloids, Specht modules, dominance ordering, . . .
(5) Definitions of Hopf algebras.
REPRESENTATION THEORY OF FINITE GROUPS – FOR MD131 51
The exam will involve some theoretical work – proofs etc – as well as lots
of calculations, involving irreducible representable and character tables. It
will not be very category theoretic. I recommend doing the assignments as
practice.
Department of Mathematics and Statistics, Masaryk University, Kotl´aˇrsk´a
2, Brno 61137, Czech Republic
E-mail address: bourkej@math.muni.cz