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I want to prove that every finite group of odd order is solvable if and only if every finite simple non abelian group has even order.

I know that every group of odd order is solvable if and only if the only simple groups of odd order are those of prime order.

I don't see connection between the two equivalences or how to prove the first one

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    $\begingroup$ Do you know whether a simple group can be solvable or not? It's hard to get a sense of how much group theory you know. $\endgroup$ – pjs36 Dec 17 '16 at 1:55
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If every group of odd order is solvable then non-abelian simple groups have to be even.Conversely, if non-abelian groups are of even order then that means either odd order groups are abelian simple (hence solvable) or have a normal subgroup.Now suppose there exists a non-solvable odd order group $G$ with a normal subgroup $N$ of smallest order say $m$. then $N$ and $G/N$ are odd order subgroups of order less than $m$, therefore are solvable and hence $G$ is solvable.

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  • $\begingroup$ I don't understand why if every group of odd order is solvable then non-abelian simple groups have to be even $\endgroup$ – allizdog Dec 17 '16 at 1:53
  • $\begingroup$ Solvable groups can either be prime cyclic or non-simple because by definition, solvable groups are extension of prime cyclic groups.So if odd order groups are simple they have to be prime cyclic which contradicts non-abelian part of hypothesis. $\endgroup$ – Ashar Tafhim Dec 17 '16 at 1:56

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