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.

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