Exercise 8.4.7

Proof. We show the equivalence between the two following properties:

(FT) Every group of odd order is solvable.

(H) Every non Abelian finite simple group has even order.

(FT) $\Rightarrow$ (H)

Let $G$ be a non Abelian finite simple group. If $G$ was solvable, it would have a normal subgroup $H⊊G$, with $G∕H$ cyclic of prime order. $G$ being simple, $H=\{e\}$, hence $G\simeq G∕\{e\}=G∕H$ would be cyclic, a fortiori Abelian, which is in contradiction with the hypothesis made on $G$.

Therefore, every non Abelian finite simple group $G$ is not solvable.

If the order of $G$ was odd, $G$ would be solvable by (FT), hence $|G|$ is even. Every non Abelian finite simple group has even order.

(H) $\Rightarrow$ (FT)

Suppose (H). Let $G$ a group of odd order. By Exercise 6, as any finite group, it has a composition series

$$\{e\}=G_m\subset G_{m-1}\subset \cdots \subset G_1\subset G_0=G,$$

with $G_{i-1}∕G_i$ simple, $i=1,\cdots ,m$.

Then

$$|G|=(G:\{e\})=(G_0:G_1)(G_1:G_2)\cdots (G_{i-1}:G_i)\cdots (G_{m-1}:G_m).$$

As $|G|$ is odd, $(G_{i-1}:G_i)$ is odd for all $i,i=1,\cdots ,m$.

So $G_{i-1}∕G_i$ is a simple group of odd order. By hypothesis (H), it is then Abelian, and simple, therefore $G_{i-1}∕G_i$ is cyclic of prime order (Exercise 8.1.8). So $G$ is solvable, and this shows that (H) $\Rightarrow$ (FT).