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 


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