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!

  • $\begingroup$ 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? $\endgroup$ – 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.

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.