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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.

Incomplete argument deletedUse induction on $|G|$. Let $N$ be a minimal normal subgroup of $G$, so $N$ is an elementary abelian $p$-group for some prime $p$. Apply inductive hypothesis to $G/N$ to get $HN$ and $H^gN$ conjugate by an element of $\langle H,H^g \rangle$. So now we can assume that $HN=H^gN$.

Then $g \in N_G(HN)$ and so the Frattini property implies that $H$ and $H^g$ are conjugate in $HN$, and hence we can assume that $g \in HN$ and $HN=G$.

So $H$ acts irreducibly on $N$, and hence either $G=H$ and we are done, or $H$ is a complement of $N$ in $G$ and a maximal subgroup of $G$. Then either $H=H^g$ or $\langle H,H^g \rangle=G$ and we are done.

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.

Incomplete argument deleted.

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 of $G$, so $N$ is an elementary abelian $p$-group for some prime $p$. Apply inductive hypothesis to $G/N$ to get $HN$ and $H^gN$ conjugate by an element of $\langle H,H^g \rangle$. So now we can assume that $HN=H^gN$.

Then $g \in N_G(HN)$ and so the Frattini property implies that $H$ and $H^g$ are conjugate in $HN$, and hence we can assume that $g \in HN$ and $HN=G$.

So $H$ acts irreducibly on $N$, and hence either $G=H$ and we are done, or $H$ is a complement of $N$ in $G$ and a maximal subgroup of $G$. Then either $H=H^g$ or $\langle H,H^g \rangle=G$ and we are done.

deleted 385 characters in body
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Derek Holt
  • 37.4k
  • 4
  • 95
  • 149

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 followsIncomplete argument deleted.

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.

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.

Incomplete argument deleted.

Post Deleted by Derek Holt
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Derek Holt
  • 37.4k
  • 4
  • 95
  • 149

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.