# About the converse of Maschke's theorem

The Maschke's theorem say that\

Let $G$ be a finite group and $F$ a field whose characteristic does not divide $\mid G \mid$. Then every $FG$-module is completely reducible (I'm using the notation of Isaac's in the book Character Theory of finite groups). The converse of this theorem is true, but I can not prove it.\ I would like your help, if possible. Thank you. Below I outline the idea of proof.

If $kG$ were completely decomposable, then there would exist an ideal $I$ for which $kG=kv\oplus I$. I claim this would imply $$I=\left\{\sum_{g\in G} c_gg:\sum c_g=0\right\}.$$ First, we show that everything in $I$ is of this form. Suppose $x=\sum_{g\in G} c_gg\in I$. This implies that $\sum_{h\in G} h\cdot x\in I$ as well, so that (using standard sum rearrangment and reindexing tricks) $$\sum_{h\in G} h\cdot\sum_{g\in G} c_gg=\sum_{g\in G} c_g\sum_{h\in G}hg=\sum_{g\in G} c_g\sum_{h\in G} h=\left(\sum_{g\in G} c_g\right)v\in I$$ But $\left(\sum_{g\in G} c_g\right)v$ is in $kv$ as well. Since $kv\cap I=\{0\}$, this means that $\sum c_g=0$.

This proves $I\subseteq \left\{\sum c_gg:\sum c_g=0\right\}.$ The reverse inclusion must hold as well, since the both of these are vector subspaces with a dimension of $|G|-1$.

However, it cannot be the case that $kG=kv\oplus I$, since $kv\subset I$! Specifically, the sum of the coefficients of $v=\sum g_i$ is $\sum_{i=1}^{|G|}1=|G|=0$, since $p$ divides $|G|$. Thus, $kG$ cannot be completely decomposable.

• To clarify, $v$ mentioned in first sentence is defined by $v=\sum_{g\in G}g$. – Mike Earnest Sep 3 '20 at 23:39

Irrespective of the characteristic, it is not hard to prove that any $y \in FG$ for which $gy=y$ for all $g \in G$ is a scalar multiple of $x$. So the fixed submodule $\{ y \in FG : gy=y\,\forall g \in G\}$ of $FG$ is equal to $Fx$, and has dimension $1$.

$FG$ also has a submodule $I = \{ \sum_{g \in G} a_g g : a_g \in F, \sum_{g \in G} a_g = 0 \}$, which has codimension $1$ in $FG$, and $G$ acts trivially on $FG/I$.

If ${\rm char}\, F$ does not divide $|G|$, then $Fx$ and $I$ are disjoint and $FG = Fx \oplus I$.

But if ${\rm char}\, F$ divides $|G|$, then $Fx \le I$ and hence the module $FG$ has two distinct $1$-dimensional trivial composition factors, $Fx$ and $FG/I$. So if $FG$ were semisimple, then its fixed submodule would have dimension at least two, contradiction. So $FG$ is not semisimple as a module in this case.

Let $F$ be a field and $G$ be a finite group. Suppose that the characteristic of $F$ divides the order of $G$. Then $x = \displaystyle\sum_{g \in G} g \in FG$ satisfies $gx = x$ for all $g \in G$ and $x^2 = \mid G \mid x = 0$. Thus $FGx = Fx$ is a submodule of $FG$ which contains no idempotent elements.

Up to this point the evidence is very clear.

However, the assertion is that $Fx$ is not semissimple.Why?

I would like some help in order to complete this demonstration. Thanks.

• Your goal is to find an $FG$-module which is not completely reducible. Think about the special case that $F = \mathbb{F}_p$ and $G = C_p$. Does one come to mind? (The simplest example is $2$-dimensional.) – Qiaochu Yuan Feb 11 '14 at 0:24
• Note that every left ideal of a semisimple ring is generated by an idempotent (if $I \leq R$ take some complement $J$ s.t. $R = I \oplus J$, and consider the unique decomposition $1 = e_I + e_J$. Then $e_I$ is idempotent and generates $I$). Since $Fx$ is a left ideal which contains no non-trivial idempotent, there is no complement of $Fx$ in $FG$, and hence $FG$ cannot be semisimple. – Dune May 15 '14 at 16:21