The Group Algebra (I)

[ 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 $\{g : g\in G\}$ , 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[S₃] gives

(1∙(1,2) – 3∙ (2,3)) ∙ (2∙ (1,3) + 4e) = 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 $\rho : G \to GL_K(M)$ , 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 $G \to GL_K(M).$

Conversely, if we have a group homomorphism $\rho : G\to GL_K(M)$ , then this gives M the structure of a K[G]-module by letting $\sum_{g\in G} c_g\cdot g$ ( $c_g \in K$ ) take m ∈ M to $\sum_{g\in G} c_g (\rho(g))(m).$

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 $\sum_{g\in G} c_g\cdot g$ takes every v in V to $\sum_{g\in G} c_g v.$

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, g²} is the cyclic group of order 3, then the group homomorphism $\rho : G \to GL_K(K^3)$ is given by:

$\rho(e) = \begin{pmatrix} 1 &0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}, \quad \rho(g) = \begin{pmatrix} 0 & 0 & 1\\ 1&0&0 \\ 0&1&0\end{pmatrix}, \quad \rho(g^2) =\begin{pmatrix} 0&1&0 \\ 0&0&1 \\ 1&0&0\end{pmatrix}.$

Example 3

Let K = R, G = S₃, M = R³ and let $\rho : G \to GL(M)$ be the homomorphism taking $g \in S_3$ to the linear map

$\rho_g : M\to M, \qquad (x_1, x_2, x_3) \mapsto (x_{g^{-1}(1)}, x_{g^{-1}(2)}, x_{g^{-1}(3)}).$

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 $\rho_g \rho_h = \rho_{gh}$ for any $g, h\in S_3.$ Now M is a module over R[S₃], 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 $\begin{pmatrix} 0 & a+b & 0 \\ a & 0 & b \\ b & 0 & a\end{pmatrix},$ 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 I₁. By our assumption, there exists J₁ such that I₁ ⊕ J₁ = K[G]. Again, J₁ must have a simple submodule I₂ 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 p² = p and p has image I (see # later). Let us now define the K-linear map:

$q : K[G] \to K[G], \quad q(\alpha) := \frac 1 {|G|} \sum_{g\in G} g\cdot p(g^{-1} \alpha)$ for $\alpha \in K[G].$

[ 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, $g^{-1}\alpha \in I$ and since p(α) = α for all α in I, the result follows. Together with property 1, this means im(q) = I.
q² = 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

$h\cdot q(h^{-1}\alpha) = \frac 1 {|G|} \sum_{g\in G} hg\cdot p(g^{-1}h^{-1}\alpha)\stackrel{x=gh}{=} \frac 1{|G|} \sum_{x\in G} x\cdot p(x^{-1}\alpha) = q(\alpha) \implies q(h\alpha) = h\cdot q(\alpha)$

Since q² = 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 p² = 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 $K[G] \cong \prod_{i=1}^k M_{n_i}(D_i)$ and letting $m_i = [D_i : K]$ gives us

$|G| = \dim_K K[G] = \sum_{i=1}^k n_i^2 m_i$ ,

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

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