# Feit Thompson equivalence

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

• 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. – pjs36 Dec 17 '16 at 1:55

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.