Center of group algebras

Posted: February 8, 2011 in Group Algebras, Noncommutative Ring Theory Notes

Notation. Let k be a field and let G be a finite group. Let \mathcal{C}_1, \cdots , \mathcal{C}_m be the conjugacy classes of G. Let

c_i = \sum_{g \in \mathcal{C}_i} g \in k[G], \ 1 \leq i \leq m.

Theorem. The set \{c_1, \cdots , c_m \} is a k-basis for Z(k[G]), the center of k[G]. Thus \dim_k Z(k[G]) is equal to the number of conjugacy classes of G.

Proofc_1, \cdots , c_m are linearly independent over k because if i \neq j, then \mathcal{C}_i \cap \mathcal{C}_j = \emptyset. So we only need to show that \text{span} \{c_1, \cdots , c_m \} = Z(k[G]). Let x = \sum_{g \in G} x_g g \in k[G],  where x_g \in k for all g \in G.

Claim. x \in Z(k[G]) if and only if x_g = x_{h^{-1}gh} for all g,h \in G.

Proof  of the claim. Well, x \in Z(k[G]) if and only if xh = hx for all h \in G. Thus, x \in Z(k[G]) if and only if

\sum_{g \in G} x_g g = \sum_{g \in G} x_g h g h^{-1}. \ \ \ \ \ \ \ \ (1)

Clearly, fixing h \in G, we have \{hgh^{-1}: \ g \in G \} =G. So if we put hgh^{-1}=u, then g = h^{-1}uh and thus (1) becomes

\sum_{g \in G} x_g g = \sum_{g \in G} x_{h^{-1}gh} g. \ \ \ \ \ \ \ \ \ \ \ (2)

The claim now follows from (2).

Now, if 1 \leq i \leq m, then, by definition, g \in \mathcal{C}_i if and only if h^{-1}gh \in \mathcal{C}_i for all h \in G. So for every g,h \in G, the coefficients of g and h^{-1}gh in c_i are either both 1 or both 0. Thus, by the claim, c_i \in Z(k[G]) for all i. So \text{span} \{c_1, \cdots , c_m \} \subseteq Z(k[G]).

Conversely, if x = \sum_{g \in G} x_g g \in Z(k[G]), then, by the claim, x_g = x_{g'} for all g and g' which are in the same conjugacy class. Thus if for every i we choose a g_i \in \mathcal{C}_i, then

x = \sum_{i=1}^m x_{g_i} c_i.

So Z(k[G]) \subseteq \text{span} \{c_1, \cdots , c_m \}. \ \Box


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