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 
 A: 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.  
A: Suppose you know every finite group of odd order is solvable. Let us be given a non-abelian simple group $G$. The commutator subgroup of $G$ is a normal subgroup of $G$, and since $G$ is simple, this is either $G$ or trivial. But it cannot be trivial since $G$ is non-abelian. This means the commutator subgroup of $G$ is $G$. Thus the derived series for $G$ is constant at $G$ and $G$ is not unsolvable, which the hypothesis then forces the order to be even.
Conversely, suppose we know every finite simple non abelian group has even order. Let $G$ be a group of odd order. If $G$ is abelian, then it is clearly solvable. If $G$ is not abelian, since $G$ has odd order, the hypothesis forces $G$ is be non-simple. So let $N$ be a proper non-trivial normal subgroup of $G$. Note $|N|$ and $|G/N|$ divide $|G|$ and are therefore odd. We can use an induction argument to ensure $N$ and $G/N$ are solvable, which makes $G$ solvable.
