Irreducible representations; definition & basic remarks

Posted: February 8, 2011 in Representations of Finite Groups
Tags: , ,

Throughout G is a finite group. For the definition of the group algebra k[G], where k is a field, see this post.

Remark 1. If \rho : G \longrightarrow \text{GL}(V) is a representation of G, then V is a \mathbb{C}[G]-module. Conversely, if V is a \mathbb{C}[G]-module, then \rho : G \longrightarrow \text{GL}(G) defined by \rho(g)(v)=gv is a representation of G.

Proof. Let g \in G and v \in V. We define

gv = \rho(g)(v). \ \ \ \ \ \ \ \ \ \ (*)

The above definition is extended to every element  x = \sum_{g \in G} x_g g \in \mathbb{C}[G] and v \in V by

xv = \sum_{g \in G} x_g \rho(g)(v).

Proving that (*) defines a \mathbb{C}[G]-module structure (also called a G-module) over V is easy: we only need to show that for any g, g_1, g_2 \in G and v_1,v_2 \in V we have

g(v_1+v_2) = gv_1+gv_2 and (g_1g_2)v = g_1(g_2v).

Both of the above are trivially true: the first one holds because \rho(g) \in \text{GL}(V) and thus \rho(g) is linear. The second one holds because \rho is a group homomorphism and thus

(g_1g_2)v = \rho(g_1g_2)v=\rho(g_1) \rho(g_2)v = \rho(g_1)(g_2v)=g_1(g_2v).

For the converse, we need to show that \rho(g) \in \text{GL}(V) for all g \in G and that \rho is a group homomorphism. It is clear that \rho(g) is linear. Also, if \rho(g)(v)=0, then gv=0 and hence v=g^{-1}(gv)=0. So \rho(g) \in \text{GL}(V). Finally, \rho is a group homomorphism because if g_1,g_2 \in G and v \in V, then

\rho(g_1g_2)(v)=(g_1g_2)v = g_1(g_2v)=g_1 \rho_2(g)(v)=\rho(g_1) \rho(g_2)(v). \ \Box

Definition. A representation \rho : G \longrightarrow \text{GL}(V) is called irreducible if V, with the structure defined in Remark 1, is a simple \mathbb{C}[G]-module. If \rho is not irreducible, it is called reducible.

Remark 2. Every degree one representation of G is irreducible.

Proof. Let \rho : G \longrightarrow \text{GL}(V), where V \cong \mathbb{C}, be a degree one representation of G. Then, since a \mathbb{C}[G]-submodule of V is obviously a \mathbb{C}-submodule of V and \dim_{\mathbb{C}} V = 1, V has no non-trivial \mathbb{C}-submodule. So V is a simple \mathbb{C}[G]-module and hence \rho is irreducible. \Box

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