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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.

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up vote 6 down vote accepted

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, 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.

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well, thanks, my proof once $V$ is known to be $C$-free my proof was even shorter: the extension by $H$ split by Schur-Zassenhaus, let $\sigma$ be a lifting of generator of $C$, then $\sigma^{p^k}$ is in the socle of $V$, but replacing it with $\sigma{}v$ for a suitable $v$ we can assure $(\sigma{}v)^{p^k}=\sigma^{p^k}N(v)$ to be $0$ if $V$ is free. I expected these facts to be very standard, if you had to prove it yourself in your paper I will probably do the same. – Maurizio Monge Oct 31 '11 at 15:08
In any case thank you very much for illustrating how group cohomology can be used to prove this. – Maurizio Monge Oct 31 '11 at 15:18
Where are you getting the exactness of that inflation-restriction sequence from? It is not exact in general. For example, if $G$ is a Klein 4-group, $|H| = 2$ and $|V|=2$, then $|H^2(G/H,V^H)| = |H^2(H,V)|=2$ but $H^2(G,V)| = 8$. – Derek Holt Oct 31 '11 at 15:23
Yes, because the inflation restriction sequence $H^1(G/H,V^H) \to H^1(G,V) \to H^1(H,V)$ is exact - that one is always exact. – Derek Holt Oct 31 '11 at 21:53
I would still be interested in an answer to the original question - i.e. is the proposition true, and is there a reference? – Derek Holt Nov 2 '11 at 9:16

I am afraid the result stated as "well-known" in the problem is false as stated. This sort of thing is a question about Hall-Higman theory, as it is now-known. The 1956 (Proc LMS) Hall-Higman paper was written for a different reason, to do with the Burnside problem, but Theorem B of that paper highlights when failure, relatively rare though it is, can occur. Group theorists, especially those working in solvable groups, but also those working on fusion and transfer, have further developed the theory. Consider the case $G = {\rm SL}(2,3),$, so that $H$ is quaternion of order $8,$ and $C$ is cyclic of order $3$ (and $p =3$). Let $V$ be the natural (2-dimensional absolutely irreducible) module for $G$ over ${\rm GF}(3).$ Then $V$ is faithful and irreducible, but is not free as ${\rm GF}(3)C$-module. A similar example occurs whenever $p = 2^{n}+1$ is a Fermat prime, $C$ is cyclic of order $p$, and $H$ is an extra-special $2$-group of order $2^{2n+1}$ which admits an automorphism $c$ of order $p.$ Then $H$ has a faithful irreducible representation of degree $2^{n}$ over ${\rm GF}(p),$ and this is afforded by a module $V$ which extends to an irreducible module for the semi-direct product $HC,$ where $C = \langle c \rangle$. Since this module has dimension $p-1,$ it is certainly not free as ${\rm GF}(p)C$-module. There are similar examples with $p = 2,$ where $C$ is a cyclic $2$-group, and $H$ is an extra-special $q$-group of order $q(|C|-1)^2$, where $q$ is a Mersenne prime.

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Nice answer. And I should have known, as (if I remember correctly) the counterexamples pointed out occur as solvable subgroups of $GL(n,\mathbb{F}_q)$. – Maurizio Monge Dec 19 '11 at 0:16
Thanks. Yes, the examples I mention are all examples where $G$ is solvable. – Geoff Robinson Dec 19 '11 at 1:13

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.

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