TITLE: Irreducible mod-p representation of a semidirect product with trivial p-core
QUESTION [6 upvotes]: Consider the group $G=H\rtimes{}C$, where $H$ has order prime with $p$ and $C$ is cyclic of order $p^k$, and $C\rightarrow{}\mathrm{Aut}(H)$ is faithful (or equivalently $G$ has trivial $p$-core). Assume $V$ to be an irreducible (not just indecomposable) faithful representation of $G$ over $\mathbb{F}_p$.
I need a reference for the following (supposedly well known?) fact:
Proposition:
as $\mathbb{F}_p[C]$-module $V$ is free (that is isomorphic to $\mathbb{F}_p[C]^n$ for some n).
This can be proved going to the algebraic closure $\overline{\mathbb{F}_p}$, and decomposing $V$ into irreducible $H$-representations. Then it's possible to see (I omit the details) that $C$ should act with orbits of cardinality $p$ on the irreducible $H$-factors of $V$, and hence that $V$ is induced from an irreducible $H$-representation.
In particular I would like to have an easy way to say that every extension of $G$ by $V$ is split, this is an easy consequence of the proposition above and Schur-Zassenhaus theorem. And I would be surprised if this was not well-known as well.

REPLY [5 votes]: Here is an easy non-cohomological proof of the splitting question. The set-up is this. We have a group $\Gamma$ having a minimal normal subgroup $V$, where $V$ is a $p$-group. Also, $\Gamma/V = G$, and $G$ has a normal $p'$-subgroup $H$. In the original question, $H$ was complemented by a cyclic $p$-group in $G$, but we do not need to assume that. Also, in the original question, $V$ was a faithful $G$-module, but we need a much weaker assumption: that $H$ acts nontrivially on $V$. We want to show that $\Gamma$ splits over $V$.
Write $H = K/V$ and by Schur-Zassenhaus, let $X$ be a complement for $V$ in $K$. Let $N = N_\Gamma(X)$. We argue that $N$ is the desired complement for $V$ in $\Gamma$. First, $KN = \Gamma$ by a Frattini argument, using the fact that all complements of $V$ in $K$ are conjugate to $X$ in $K$. It follows that
$\Gamma = VN$. Now $V \cap N$ is normal in $G$ since $V$ is abelian. Since $V$ is minimal normal in $\Gamma$, we have either $V \cap N = 1$, as wanted, or
$V \subseteq N$. The latter would imply that $X$ centralizes $V$, and this is not the case since $H$ acts nontrivially.