# $G$ solvable $\implies$ composition factors of $G$ are of prime order.

I'm trying to prove the equivalency of the following definitions for a finite group, $G$:

(i) $G$ is solvable, i.e. there exists a chain of subgroups $1 = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_s = G$ such that $G_{i+1}/ \ G_i$ is abelian for all $0 \leq i \leq s-1$

(ii) $G$ has a chain of subgroups $1 = H_0 \trianglelefteq H_1 \trianglelefteq \cdots \trianglelefteq H_s = G$ such that $H_{i+1} /\ H_i$ is cyclic for all $0 \leq i \leq s-1$.

(iii) All composition factors of $G$ are of prime order.

(iv) G has a chain of subgroups $1 = N_0 \trianglelefteq N_1 \trianglelefteq \cdots \trianglelefteq N_t = G$ such that each $N_i$ is a normal subgroup of $G$, and $N_{i+1}/ \ N_i$ is abelian for all $0 \leq i \leq t-1$.

It was simple to show that (iii) $\implies$ (ii) and (ii) $\implies$ (i). Now, I'm trying to prove that (i) $\implies$ (iii), and I'm getting stuck.

I know that If $G$ is an abelian simple group, then $|G|$ is prime, so all I need to show is that each $G_{i+1}/ \ G_i$ in the definition of solvable is simple.

If I knew that $1 = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_s = G$ were a composition series, then each $G_{i+1}/ \ G_i$ would be simple by definition. But the definition of solvable doesn't assume a composition series. I know every group has a unique composition series, but I'm not sure how to connect that to the subgroups in the definition of solvable.

More generally, I guess I just don't understand how being abelian connects to being of simple or of prime order.

Any pushes in the right direction would be appreciated!

• Indeed, the factors in the subnormal series in the definition of solvable do not need to be simple. But are you familiar with refinements of subnormal series? – Tobias Kildetoft Sep 14 '15 at 9:10

Hint: Use induction on the order of $G$:
If $G_{s-1}$ is a proper normal subgroup of $G$ such that $G/G_{s-1}$ is abelian, then there are two cases:
• Either $1 < G_{s-1}$; then by induction, the composition factors of $G_{s-1}$ and $G/G_{s-1}$ have prime orders.
• Or $1 = G_{s-1}$ in which case $G$ is itself abelian.