$G$ 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!

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.


Can you figure out how the claim follows in each of the two cases?
