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.



Irreducible mod-p representation of a semidirect product with trivial p-core
The following is a direct proof that any extension of $G$ by $V$ splits. It is taken from [a joint paper of mine with Tim Dokchitser](http://arxiv.org/abs/1103.2047), where the proof starts in the last paragraph of page 12.


First, note that for any $n\in\mathbb{N}$, $H^n(H,V)=0$, since it is killed by $|H|$ and $|V|$, which are coprime. This implies that the inflation map
$$
H^2(G/H,V^H)\longrightarrow H^2(G,V)
$$
is an isomorphism. But also, either $V$ is trivial, in which case what we want to prove is obvious, or $V^H=0$. Indeed, since $H$ is normal, $V^H$ is a subrepresentation. But $C$ is a $p$-group, so it has a fixed vector in $V^H$, which is then fixed under all of $G$. This contradicts irreducibility of $V$, unless $V$ is the trivial representation.
So $H^2(G,V)=0$, as required.


