Exercise 3.4.12(Equivalent statements of the Feit-Thompson Theorem)

Proof. (i) $\Rightarrow$(ii): Assume that every group of odd order is solvable.

Let $G$ be a simple group of odd order. Then $G$ is a solvable simple group.

Since $G$ is simple, its only composition factor is $G$.

Since $G$ is solvable, by Exercise 8, all composition factors of $G$ are of prime order. Hence the order of $G$ is odd.

(ii) $\Rightarrow$(i): Conversely, assume that the only simple groups of odd order are those of prime order.

Let $G$ be a group of odd order $n$. Let

be a composition series of $G$.

Then

is odd, therefore all indices $|G_{i+1}:G_i|$ ( $0\le i<s$) are odd, i.e., all composition factors $G_{i+1}∕G_i$ are simple of odd order. By hypothesis, the only simple groups of odd order are those of prime order. Hence the order of $G_{i+1}∕G_i$ is prime, therefore $G_{i+1}∕G_i$ is cyclic of prime order, for all $i\in [[0,s[[$. By Exercise 8, $G$ is solvable.

The following are equivalent:

(i)

Every group of odd order is solvable

(ii)

the only simple groups of odd order are those of prime order.