One way to state maschke's theorem is that for a finite group $G$ and a field $k$ the following holds. If we see $k[G]$ as a left-module over itsself then $k[G]$ is semi-simple if and only if $\text{char}(k) \nmid |G|$. Now we want to generalize it as follows. Let $G$ act transitively on a fnite set $X$ and let $F(X)$ be the free vector space of $X$ over $k$. Then we can see $F(X)$ as a left-module over $k[G]$ via the group action. It is easy to show that if $F(X)$ is semi-simple then $\text{char}(k)\nmid |X|$. However i am not sure wether the converse holds or not. Does someone know a counterexample or a proof?
$\begingroup$
$\endgroup$
Add a comment
|
$\begingroup$
$\endgroup$
3
The converse is false in general. I found a counterexample after a few tries with a computer calculation.
Let $G=A_5$ acting on transitively on $15$ points, Then the permutation module over ${\mathbb F}_4$ (a splitting field in characteristic $2$) has two indecomposable components of dimension $5$ with composition factors of dimensions $2$, $1$, and $2$.
-
$\begingroup$ Thanks for your answer. In general if $F(X)$ is not semi-simple it follows by maschke's theorem that $\text{char}(k) \mid |G|$. Now the interesting case is when exactly do we have $\text{char}(k)\nmid |X|$ aswell? (So when does it not follow trivialy from $F(X)$ being semi-simple?) $\endgroup$ – kevkev1695 Nov 29 '19 at 15:28
-
$\begingroup$ @kevkev1695,note that here $\operatorname{char}(k)=2$ but $|X|=15$. $\endgroup$ – AnalysisStudent0414 Nov 29 '19 at 15:34
-
$\begingroup$ Indeed. The question becomes whenever this exactly happens. If $\text{char}(k)\nmid |X|$ and $\text{char}(k) \mid |G|$ then clearly $\text{char}(k) \mid |G|/|X|$ which is the Order of a stabilizer $G_x$ of the group action. In particular it follows that $k[G_x]$ is not semi-simple over itsself. Can we work with that? $\endgroup$ – kevkev1695 Nov 29 '19 at 15:42
