## Semiprimitivity of k[G] for finite groups G

Posted: January 9, 2011 in Group Algebras, Noncommutative Ring Theory Notes
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Theorem (Maschke, 1899) Let $k$ be a field and let $G$ be a finite group of order $n.$ Let $R:=k[G].$ Then $R$ is semiprimitive if and only if $char(k) \nmid n.$

Proof. Let $G = \{g_1, \cdots , g_n \}$ where $g_1=1.$ Suppose first that $char(k) \nmid n$ and consider the algebra homomorphism $\rho : R \longrightarrow End_k(R)$ defined by $\rho(r)(x)=rx$ for all $r,x \in R.$ Define $\alpha : R \longrightarrow k$ by $\alpha(r) = Tr(\rho(r))$ for all $r \in R.$ Note that here $Tr(\rho(r))$ is the trace of the matrix corresponding to the linear transformation $\rho(r)$ with respect to the ordered basis $\{g_1, \cdots , g_n \}.$ We remark a few points about $\alpha$:

1) $\alpha(1)=n$ because $\rho(1)$ is the identity map of $R.$

2) If $1 \neq g \in G,$ then $\alpha(g)=0.$ The reason is that $\rho(g)(g_i)=gg_i \neq g_i$ for all $i$ and thus the diagonal entries of the matrix of $\rho(g)$ are all zero and so $\alpha(g)=Tr(\rho(g))=0.$

3) If $r$ is a nilpotent element of $R,$ then $\alpha(r)=0$ because then $r^m=0$ for some $m$ and thus $(\rho(r))^m = \rho(r^m)=0.$ So $\rho(r)$ is nilpotent and we know that the trace of a nilpotent matrix is zero.

Now let $c \in J(R).$ Since $R$ is finite dimensional over $k,$ it is Artinian and hence $J(R)$ is nilpotent. Thus $c$ is nilpotent and therefore, by 3), $\alpha(c)=0.$ Let $c = \sum_{i=1}^n c_i g_i,$ where $c_i \in k.$ Then

$0=\alpha(c)=\sum_{i=1}^n c_i \alpha(g_i)=c_1n,$

by 1) and 2). It follows that $c_1=0$ because $char(k) \nmid n.$ So the coefficient of $g_1=1$ of every element in $J(R)$ is zero. But for every $i,$ the coefficient of $g_1=1$ of the element $g_i^{-1}c \in J(R)$ is $c_i$ and so we must have $c_i=0.$ Hence $c = 0$ and so $J(R)=\{0\}.$

Conversely, suppose that $char(k) \mid n$ and put $x = \sum_{i=1}^n g_i.$ Clearly $xg_j = x$ for all $j$ and hence

$x^2=x\sum_{j=1}^n g_j = \sum_{j=1}^n xg_j=\sum_{j=1}^n x = nx = 0.$

Thus $kx$ is a nilpotent ideal of $R$ and so $kx \subseteq J(R).$ Therefore $J(R) \neq \{0\}$ because $kx \neq \{0\}. \ \Box$