## Center of group algebras

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

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