## Every finite group of odd order is solvable if and only if the only simple groups of odd order have prime order

Prove (without using the Feit-Thompson Theorem) that the following are equivalent.

1. Every group of odd order is solvable.
2. The only simple groups of odd order are those of prime order.

$(1 \Rightarrow 2)$ Suppose $G$ is a simple group of composite order. If $G$ is abelian, then there exists an element $x$ which does not generate $G$. Then $\langle x \rangle \vartriangle G$ is nontrivial, so that $G$ is not simple, a contradiction. If $G$ is nonabelian, then because $G$ is solvable, there exists a composition series $1 = H_0 \vartriangleleft \cdots \vartriangleleft H_n = G$ of $G$ such that all intermediate quotients are abelian. Since $G$ itself is not abelian, the length of this composition series must be at least 2. Then $H_{n-1}$ is a nontrivial normal subgroup of $G$, so $G$ 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.)

$(2 \Rightarrow 1)$ Let $G$ be a group of odd order. We proceed by induction on the width of $G$– that is, the number of prime divisors including multiplicity. For the base case, if $|G|$ has width 1 then $G$ is simple and abelian, hence solvable. For the inductive step, suppose that every group of odd order whose width is at most $k$ is solvable, and suppose $G$ has width $k+1$. Now $|G|$ is not prime, so that $G$ is not simple. Let $N$ be a nontrivial normal subgroup of $G$. Now $N$ and $G/N$ are groups of width at most $n-1$. Moreover, $|G| = |N| \cdot |G/N|$ by Lagrange’s Theorem. Note that $1 \equiv |G| = |N| \cdot |G/N|$ modulo 2, so that $|N|$ and $|G/N|$ are both odd. Thus $N$ and $G/N$ are solvable by the induction hypothesis, hence $G$ is solvable. Thus every group of odd order is solvable.