## Number of non-equivalent irreducible representations (1)

Posted: February 8, 2011 in Representations of Finite Groups
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Throughout $G$ is a finite group.

Lemma. There exist a finite set of mutually non-equivalent irreducible representations $\rho_1, \cdots , \rho_r$ of $G$ such that every irreducible representation of $G$ is equivalent to $\rho_i$ for some $i.$

Proof. By the corollary in this post, $\mathbb{C}[G]$ is a semisimple ring. Thus there exist simple $\mathbb{C}[G]$-modules $V_1, \cdots , V_r$ which are mutually non-isomorphic and any simple $\mathbb{C}[G]$-module is isomorphic to $V_i$ for some $1 \leq i \leq r.$ For every $1 \leq i \leq r,$ let $\rho_i$ be the representation of $G$ corresponding to $V_i,$ i.e. $\rho_i: G \longrightarrow \text{GL}(V_i)$ is defined by $\rho_i(g)(v)=gv,$ for all $g \in G$ and $v \in V.$ Since $V_i$ is simple, $\rho_i$ is irreducible by definition. Also $\rho_i$ and $\rho_j$ are equivalent iff $i=j$ because $V_i \cong V_j$ iff $i=j.$ If $\rho : G \longrightarrow \text{GL}(V)$ is any irreducible representation of $G,$ then $V$ is a simple $\mathbb{C}[G]$-module and hence $V \cong V_i,$ for some $1 \leq i \leq r.$ That means $\rho$ is equivalent to $\rho_i. \ \Box$

Definition. The set $\{\rho_1, \cdots , \rho_r \}$ in the lemma is called a complete set of irreducible representations of $G.$

Theorem. Let $\{\rho_1, \cdots , \rho_r \}$ be a complete set of irreducible representations of $G.$ Then

1) The number $r$ is equal to the number of conjugacy classes of $G.$

2)  $|G|=\sum_{i=1}^r (\deg \rho_i)^2.$

Proof. 1) As we mentioned at the beginning of the proof of the above lemma, $\mathbb{C}[G]$ is semisimple. Thus by the Wedderburn – Artin theorem

$\mathbb{C}[G]=M_{n_1}(D_1) \times M_{n_2}(D_2) \times \cdots \times M_{n_r}(D_r), \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

for some integers $n_i \geq 1$ and some $\mathbb{C}$-division algebras $D_i.$ Since $\mathbb{C}[G]$ is finite dimensional over $\mathbb{C},$ each $D_i$ has to be finite dimensional, and hence algebraic, over $\mathbb{C}.$ Thus $D_i=\mathbb{C},$ because $\mathbb{C}$ is algebraically closed. So $(1)$ becomes

$\mathbb{C}[G] \cong M_{n_1}(\mathbb{C}) \times M_{n_2}(\mathbb{C}) \times \cdots \times M_{n_r}(\mathbb{C}). \ \ \ \ \ \ \ \ \ \ \ \ (2)$

Now, since $Z(M_{n_i}(\mathbb{C})) \cong \mathbb{C},$ taking the center of both sides in $(2)$ gives us

$Z(\mathbb{C}[G]) \cong \mathbb{C}^r.$

Thus $\dim_{\mathbb{C}} Z(\mathbb{C}[G])=r.$ We also proved in here that $\dim_{\mathbb{C}} Z(\mathbb{C}[G])$ is equal to the number of conjugacy classes of $G.$ So $r$ is equal to the number of conjugacy classes of $G.$

2) Let $\rho_i : G \longrightarrow \text{GL}(V_i).$ By $(2)$ we have $\deg \rho_i=\dim_{\mathbb{C}} V_i = n_i$ and also

$|G|=\dim_{\mathbb{C}} \mathbb{C}[G] = \sum_{i=1}^r \dim_{\mathbb{C}} M_{n_i}(\mathbb{C}) = \sum_{i=1}^r n_i^2 = \sum_{i=1}^r (\deg \rho_i)^2. \ \Box$