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Derek Holt
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Personally I would try to persuade someone who does have access to the paper to e-mail it to me, but that might not be legal, so I shouldn't have said it.

I have not thought this out in detail, but I think the following approach will work.

Use induction on $|G|$. Let $N$ be a minimal normal subgroup, so $N$ is an elementary abelian $p$-group for some prime $p$. Apply the inductive hypothesis to $G/N$, and reduce to the case $G=HN$. So $H$ acts irreducibly on $N$. Then, since we can assume that $H \ne G$, $H$ is a complement of $N$ in $G$ and is a maximal subgroup of $G$. So either $H=H^g$ or $\langle H,H^g \rangle = G$, and the result follows.

Derek Holt
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