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

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