Prove (without using the Feit-Thompson Theorem) that the following are equivalent.
- Every group of odd order is solvable.
- The only simple groups of odd order are those of prime order.
Suppose is a simple group of composite order. If is abelian, then there exists an element which does not generate . Then is nontrivial, so that is not simple, a contradiction. If is nonabelian, then because is solvable, there exists a composition series of such that all intermediate quotients are abelian. Since itself is not abelian, the length of this composition series must be at least 2. Then is a nontrivial normal subgroup of , so is not simple, a contradiction. Thus the only simple groups of odd order have prime order. (We happen to know that all such groups are simple and abelian, hence solvable.)
Let be a group of odd order. We proceed by induction on the width of – that is, the number of prime divisors including multiplicity. For the base case, if has width 1 then is simple and abelian, hence solvable. For the inductive step, suppose that every group of odd order whose width is at most is solvable, and suppose has width . Now is not prime, so that is not simple. Let be a nontrivial normal subgroup of . Now and are groups of width at most . Moreover, by Lagrange’s Theorem. Note that modulo 2, so that and are both odd. Thus and are solvable by the induction hypothesis, hence is solvable. Thus every group of odd order is solvable.