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(Maschke) Let $G$ be a finite group. Then the group algebra $F[G]$ is semiprime if and only if either $char(F) = 0$ or $char(F)$ is a prime number $p$ that does not divide $|G|$.

Proof : Since $F[G]$ is a finite dimensional vector space over $F $, we can define a linear functional $ \rho : F[G] → F$ by $\rho(a): = tr(L_a).$

Let $n = |G|$ and denote the elements in $G$ by $g_1, g_2, . . . ,g_n$ where $g_1 = 1$.Obviously,$\rho(g_1) = n$. If $i \geq 2$, then $L_{g_i} (g_j) = g_ig_j¸ G- \{g_j \}$ for every $j$. The matrix representation of $L_{g_i}$ with respect to the basis $\{g_1, g_2, . . . , g_n \}$ has therefore zeros on the diagonal. Consequently, $\rho(g_i) = 0$. This is a crucial observation upon which our proof is based.

Suppose now that $F[G]$ has a nonzero nilpotent ideal $I$. We want to show that then $F$ has finite characteristic $p$ which divides $n = |G|$. Pick a nonzero $a \in I$, and write $ a = \sum_{i=1}^{n} \lambda_ig_i $.Without loss of generality we may assume that $\lambda_1 =0$. Indeed, otherwise we choose $i$ such that $\lambda_i \neq 0$ and replace a by $g_i^{-1}a$, which is of course also an element from $I$. We have

$\rho(a) = \lambda_1 \rho(g_1) + \lambda_2 \rho(g_2)+ ..... +\lambda_n \rho(g_n) = n \lambda_1$.

As an element of a nilpotent ideal, $a$ is a nilpotent element. Hence $L_a$ is a nilpotent linear map, and so $\rho(a) = 0$. That is, $n \lambda_1 = 0$. Since $ \lambda_1 \neq 0$, this is possible only when $p = char(F)$ divides $n$. Conversely, assume that $p = char(F) $ divides $|G|$. Set $r = \sum_{i =1}^{n} g_i$. Since $rg_j = g_j r = r$ for every $j$, we see that the 1-dimensional space $Fr$ is an ideal of $F[G]$. As $r^2 = |G|r = 0, Fr $ is a nilpotent ideal.

Why do you have the following relationships?

a:$\rho(g_1) = n$?

b:If $i \geq 2$, then $L_{g_i} (g_j) = g_ig_j¸ G- \{g_j \}$ for every $j$?

c: The matrix representation of $L_{g_i}$ with respect to the basis $\{g_1, g_2, . . . , g_n \}$ has therefore zeros on the diagonal. Consequently, $\rho(g_i) = 0$.

2:Why can we assume that $L_a$ is a nilpotent linear map, and so $\rho(a) = 0$? (the last paragraph )

3: Why is 1-dimensional space $Fr$ an ideal of $F[G]$ and why is $Fr$ nilpotent ideal?

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  • $\begingroup$ The equality in b. is by definition. But given that you have copied over the expression in a way that makes it meaningless twice points to a need for some more thought put into this. $\endgroup$ – Tobias Kildetoft Jan 15 '18 at 15:53
  • $\begingroup$ What is $L_{a}$? $\endgroup$ – Delong Jan 16 '18 at 5:12

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