Reminder: throughout this series, *G* is a finite group and *K* is a field. All *K*-vector spaces are assumed to be finite-dimensional over *K*.

## G4. Maschke’s Theorem

If is a *K*[*G*]-submodule, it turns out *V* is isomorphic to the direct sum of *W* and some other submodule *W’*.

Maschke’s Theorem. Suppose char(K) does not divide #G (e.g. char(K) = 0). If is a K[G]-submodule, then there’s a submodule such that:This gives .

As a consequence, if is a nontrivial submodule, then one can always decompose ; since *V* is finite-dimensional, this process must eventually terminate.

Definition. A K[G]-module V≠0 is said to besimpleif it has no submodule other than 0 and itself. The corresponding representation is then said to beirreducible.

Thanks to Maschke’s theorem, every *K*[*G*]-module is a direct sum of simple modules.

**Proof of Maschke’s Theorem**.

Let *X* be a *K*-subspace of *V* such that and , which is always possible by linear algebra. Take the projection map Now define the linear map:

We claim: *q* is *K*[*G*]-linear; it suffices to prove that *q*(*hv*) = *h*·*q*(*v*) for all But

Next claim: if then *q*(*w*)=*w*. Indeed since *W* is a *K*[*G*]-submodule of *V*, so for every Hence *q*(*w*) is 1/|*G*| times a sum of |*G*| copies of *w*.

To complete the proof, we’ll show that *W’* = ker(*q*) satisfies the conditions of the theorem:

- Since
*q*is*K*[*G*]-linear, its kernel*W’*is a*K*[*G*]-submodule of*V*. - If , then
*q*(*v*) = 0 since on the other hand, since we have*q*(*v*)=*v*so*v*=0. Thus, - For any , write . Since we have also. So and we get Thus ♦

**Example**

Consider acting on by permuting the coordinates (see example 3 in G1). The 1-dimensional subspace *W* spanned by (1, 1, 1) is clearly a *G*-invariant subspace. Applying the construction in the above proof gives us the complementary submodule

## G5. Schur’s Lemma

Assume char(*K*) = 0. By Maschke’s theorem, every *K*[*G*]-module is a direct sum of simple modules. The following result is handy:

Schur’s Lemma.Let V and W be simple K[G]-modules.

- If f : V → W is a K[G]-linear map, then f=0 or f is an isomorphism.
- Suppose K =
C. If f : V → V is a K[G]-linear map, then it’s a scalar multiple of the identity.

**Proof**.

Suppose *f* : *V* → *W* is non-zero. Its kernel is a submodule of *V* so it’s either 0 or *V*; since *f*≠0, we have ker(*f*) = 0. Likewise, its image is a submodule of *W* so it’s either 0 or *W*; since *f*≠0, we have im(*f*) = *W*. Thus *f* is bijective.

Let λ be an eigenvalue of *f* with eigenvector *v*, i.e. *f*(*v*) = λ*v*. Then ker(*f* – λ) is non-zero so it must be whole of *V*, i.e. *f* = λ·1_{V}. ♦

Another way of looking at Schur’s lemma: for *K *= **C** and simple **C**[*G*]-modules *V*, *W*,

From now onwards, we let *K* = **C**.

## G6. The Space of Fixed Vectors.

For any **C**[*G*]-module *V*, consider the space:

the space of all vectors fixed by *G*. The following result is critical.

Lemma. We have:the average of the traces of acting on V.

**Proof**.

Take the averaging map (which is **C**-linear):

We claim: for any , we have . Indeed, for any , we have

On the other hand, if then *gw* = *w*, which gives us *p*(*w*) = *w*. Just as in the proof of Maschke’s theorem, this shows and If we pick a basis of ker(*p*) and of *V ^{G}*, then the matrix for

*p*is:

## G7. Introducing Characters

Since the trace of *g*‘s action is important, let’s define:

Definition. Let be a representation of G. Itscharacteris the functionLooking at V as a

C[G]-module, we have

From G6, we see that Suppose *V* and *W* are **C**[*G*]-modules. Let’s consider the various constructions in G3.

**Direct sum**: .**Tensor product**: .**Dual**: .

The first two equalities are easy. For the third, let’s pick a basis of *V* and consider the dual basis If *M* is the matrix for *g* : *V* → *V* with respect to *B*, the corresponding matrix for *g* : *V** → *V** with respect to *B** is given by Hence:

Since each *g* : *V* → *V* has finite order, it is diagonalisable (via expressing *g* in its Jordan canonical form) and every eigenvalue is a root of unity. With respect to a basis of eigenvectors, *g* is diagonal with entries Thus *g*^{-1} is diagonal with entries and as desired.

This gives:

**Hom**:

Now consider Hom_{C}(*V*, *W*)* ^{G}*. A

**C**-linear map

*f*:

*V*→

*W*is fixed by

*G*if and only if:

if and only if *f* is **C**[*G*]-linear. Thus and:

Together with Schur’s lemma, we have:

Corollary (Orthogonality of Characters).If V and W are simple

C[G]-modules, then:

## G8. Class Functions

Since tr(*AB*) = tr(*BA*) for square matrices of the same size (in fact, it’s even true for rectangular matrices, as long as *AB* and *BA* are both square), we have

Definition. A function is called aclass functionif for any

Thus, the character of a representation is a class function. Define an inner product between class functions as:

From G7, the set of characters of simple **C**[*G*]-modules (or irreducible representations) forms an orthonormal set. In fact, we can say more:

Theorem. The set of irreducible characters forms an orthonormal basis for the space of all class functions.

**Proof.**

It suffices to show that the irreducible characters span the space of class functions. Otherwise, there’s a class function *f* : *G* → **C** which is orthogonal to all irreducible characters. Take the element For any simple **C**[*G*]-module *V*, σ acts on *V* as a **C**-linear map. Now,

so it is in fact **C**[*G*]-linear and must be a scalar multiple of the identity. To compute this identity, we take the trace:

But *f* is orthogonal to each so *c*=0. Thus σ=0 on an irreducible representation *V* and hence on *any* representation as well. In particular, σ=0 on **C**[*G*] itself so 0 = σ·1 = σ. ♦

Corollary. The number of irreducible representations of G is the number of its conjugancy classes m.The m × m table comprising of values is called a

character table.

**Note**.

On an intuitive level, characters thus provide an excellent way to analyse non-abelian groups. Indeed, if *G* is “highly non-abelian” (e.g. simple), it has fewer conjugancy classes and thus a smaller character table. On the other hand, abelian groups have the largest possible character table since every conjugancy class has size 1.

## G9. Degrees of Irreducible Characters

If *V* is a **C**[*G*]-module, its dimension as a **C**-space is called its **degree**.

Let’s take the obvious **C**[*G*]-module, i.e. **C**[*G*] itself (on the representation side, this gives rise to the regular representation we saw in example 3, G1). Writing **C**[*G*] as a direct sum of irreducible representations , the corresponding character gives:

, where each .

To compute each *d _{i}*, we take the inner product . But is easy to compute: it takes

*e*to |

*G*| and all other to zero (since if

*g*

*≠e*, the action of

*g*on

*G*via left-multiplication has no fixed point). Thus,

We have thus proven:

Theorem. If V is an irreducible representation of G, the number of times it occurs in the regular representation is dim(V).

Corollary. . [ Follows from ]

Corollary. If G is abelian, then each . [ This follows from: size of character table is |G| × |G|. ]

**Example of a Character Table**

Here’s the character table of the symmetric group *S*_{4}. It’s 5 × 5 since the number of conjugancy classes of *S*_{4} is *p*(4) = 5. The size of each conjugancy class is written in square brackets [*s*].

In the next installation, we’ll see how this table is obtained.