[ Note: the contents of this article overlap with a previous series on character theory. ]

Let *K* be a field and *G* a finite group. The **group algebra** *K*[*G*] is defined to be a vector space over *K* with basis , where “*g*” here is an abstract symbol for each element *g* of *G*. Thus, *K*[*G*] has dimension |*G*| over *K*. Now *K*[*G*] has a ring structure obtained from group multiplication, where the algebra product σ⋅τ is precisely the group product, i.e. an element of *G*. For example, multiplication in **Q**[*S*_{3}] gives

(1∙(1,2) – 3∙ (2,3)) ∙ (2∙ (1,3) + 4*e*) = 2∙(1,3,2) + 4∙(1,2) – 6∙(1,2,3) – 12∙(2,3).

**Note**

The most important aspect of the group algebra is that its modules *M* correspond to group homomorphisms , of *G* to the group of invertible *K*-linear maps *M**→M*; such a ρ is called a **representation** of *G* over *K*. Indeed, if *M* is a *K*[*G*]-module, then considering how elements *g* ∈ *K*[*G*] acts on *M* (for *g* ∈ *G*), we obtain a homomorphism

Conversely, if we have a group homomorphism , then this gives *M* the structure of a *K*[*G*]-module by letting () take *m* ∈ *M* to

Note that a homomorphism of *K*[*G*]-modules *f* : *V* → *W* is a linear map of vector spaces such that *f*(*g⋅v*) = *g**⋅f*(*v*) for any *g* in *G* and *v* in *V*. Such a map is also called an **intertwining map** (since it commutes with every element of *G*).

**Example 1**

Every group has a **trivial representation**, corresponding to *G* → *K** which maps all *g* to 1. As a *K*[*G*]-module representation, this is *V* := *K* and takes every *v* in *V* to

**Example 2**

*V* := *K*[*G*] is a module over the ring *K*[*G*]; this corresponds to the **regular representation**. E.g. if *G* = {*e*, *g*, *g*^{2}} is the cyclic group of order 3, then the group homomorphism is given by:

**Example 3**

Let *K* = **R**, *G* = *S*_{3}, *M* = **R**^{3} and let be the homomorphism taking to the linear map

E.g. *g* = (1,3,2) takes (*x*, *y*, *z*) to (*y*, *z*, *x*) and *h* = (1, 2) takes (*x*, *y*, *z*) to (*y*, *x*, *z*). The inverse of *g* is necessary for the subscripts to ensure that for any Now *M* is a module over **R**[*S*_{3}], where, e.g. *a*(1, 2) + *b*(1, 3, 2) takes the point (*x*, *y*, *z*) to (*ay*+*by*, *ax+bz*, *az+bx*) so its corresponding matrix is assuming elements of *M* are written as column vectors.

## Main Result

Theorem. If the characteristic of K does not divide |G|, then K[G] is semisimple.

**Proof**

We shall prove: for each left ideal *I* of *K*[*G*], there is a left ideal *J* such that *I* ⊕ *J* = *K*[*G*] (i.e. *I* ∩ *J* = 0 and *I* + *J* = *K*[*G*]). Assuming this holds, since *K*[*G*] is of finite dimension, it must have a minimal left ideal *I*_{1}. By our assumption, there exists * J_{1}* such that

*I*

_{1}⊕

*J*

_{1}=

*K*[

*G*]. Again,

*J*

_{1}must have a simple submodule

*I*

_{2}etc . Eventually, we obtain

*K*[

*G*] as a direct sum of minimal left ideals.

Now suppose *I* is a left ideal of *K*[*G*]. This is a vector subspace, so we can write *K*[*G*] = *I* ⊕ *W* for a subspace *W*. Let *p* : *K*[*G*] → *K*[*G*] be the projection *I* ⊕ *W* → *I*, so that *p*^{2} = *p* and *p* has image *I* (see # later). Let us now define the *K*-linear map:

for

[ Note that this is well-defined since |*G*| is invertible in *K*. ] We have the following properties of *q*:

- The image of
*q*is in*I*: indeed image of*p*lies in*I*and*g⋅I*⊆*I*for any*g*in*G*. *q*(α) = α for all α in*I*: indeed, and since*p*(α) = α for all α in*I*, the result follows. Together with property 1, this means im(*q*) =*I*.*q*^{2}=*q*: for any α in*K*[*G*], we have*q*(α) in*I*, so by property 2,*q*(*q*(α)) =*q*(α).*q*is a*K*[*G*]-module homomorphism: clearly*q*is*K*-linear; furthermore for any*h*in*G*, we have

Since *q*^{2} = *q*, *q* is a projection and we have *K*[*G*] = ker(*q*) ⊕ im(*q*) = ker(*q*) ⊕ *I *(see # later). And since *q* is a *K*[*G*]-module homomorphism, its kernel is a submodule, and we have proven our assumption. ♦

[ # Note: along the way, we used the fact that a *K*-linear map *p* : *V* → *V* such that *p*^{2} = *p* gives us *V* = ker(*p*) ⊕ im(*p*). The proof of this is an easy exercise. ]

From our classification of semisimple rings, we obtain the

Corollary. If char(K) does not divide |G|, then K[G] is isomorphic to a finite product of matrix rings over division rings. Writing and letting gives us,

where k is the number of isomorphism classes of simple K[G]-modules. [ Note: the simple K[G]-modules are also called

irreducible representationsof G. ]