## Irreducible representations; definition & basic remarks

Posted: February 8, 2011 in Representations of Finite Groups
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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. $\Box$