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Hypotheses: Let $G$ be a simple finite non-abelian group. And let every proper subgroup of $G$ be abelian.

We have $M$ a maximal subgroup of $G$, with $M$ of order $2$. Can we deduce that $M$ is a Sylow 2-subgroup. If so, why?

This is part of a proof by contradiction (by assuming that $G$ is simple). But specifically, can we conclude that $M$ is Sylow from the fact that it is maximal and of order $2$?

CuriousKid7
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    Since no such groups exist, yes we can indeed conclude anything we like. – Tobias Kildetoft Nov 14 '16 at 13:39
  • @TobiasKildetoft This is part of a proof by contradiction (by assuming that $G$ is simple). But specifically, can we conclude that $M$ is Sylow from the fact that it is maximal and of order $2$? – CuriousKid7 Nov 14 '16 at 13:42
  • Such a group must be necessarily be solvable. See also http://math.stackexchange.com/questions/48197/what-can-we-say-of-a-group-all-of-whose-proper-subgroups-are-abelian. So if it is simple, then is has order a prime. – Nicky Hekster Nov 14 '16 at 13:42
  • In general, no we could not conclude that a subgroup is Sylow from being of order $2$ and maximal, since the group could potentially have order $4$. But once we add that it is non-abelian then yes. – Tobias Kildetoft Nov 14 '16 at 13:43
  • @TobiasKildetoft Could you please explain why the fact that it is non-abelian lets us do this? – CuriousKid7 Nov 14 '16 at 13:45
  • Because a group of order $4$ is abelian. And if a subgroup of order $2$ is not Sylow but maximal then the group has order $4$ by Sylow's theorems – Tobias Kildetoft Nov 14 '16 at 13:45
  • In fact the only (finite) groups with a maximal subgroup of order $2$ are the groups of order $2p$ with $p$ prime, of which there are two isomorphism classes for each $p$. – Derek Holt Nov 14 '16 at 13:45
  • @TobiasKildetoft How exactly does Sylow's theorems tell us that? – CuriousKid7 Nov 14 '16 at 13:51
  • If $M$ is not a Sylow-$2$ subgroup, it's contained in a Sylow-$2$ subgroup $S$. If $S \neq G$, then $M$ is not maximal. If $S = G$, let the order be $2^k$. If $k > 2$, then $S$ has a subgroup of order $4$ containing $M$, so $M$ is not maximal. – Daniel Fischer Nov 14 '16 at 14:01
  • @DanielFischer I'm confused as to how we know that $M$ would be contained in a Sylow-2 subgroup. – CuriousKid7 Nov 14 '16 at 15:34
  • That's one of Sylow's theorems, every $p$-subgroup is contained in some Sylow-$p$ subgroup. – Daniel Fischer Nov 14 '16 at 15:35
  • @DanielFischer Oh, right! Thanks – CuriousKid7 Nov 14 '16 at 15:41

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