Let $G$ be a group with a composition series. Then $G$ is solvable if and only if $G$ has a normal series (which starts at $1$) and each factor group has order some power of a prime.
I think in some part of the proof we need to use that all simple abelian groups are finite and have prime order, but I don't see how yet.
Suppose $G$ is solvable. Then it has a normal series
$$1=H_0\triangleleft … \triangleleft H_n=G$$
and every factor group $H_{i+1}/H_i$ is abelian.
And we know that $G$ has a composition series, say
$$1=K_1\triangleleft … \triangleleft K_m=G$$
with $K_{i+1}/K_i$ simple.
If the first series were a composition series, by Jordan-Holder theorem every $H_{i}/H_{i+1}$ would be simple and abelian, hence of prime order (not a power of a prime, though). But surely we can't say the first series is a composition series.
I don't know what to do now. Any hint?
Thank you.
Best Answer
The idea is to see that if $ L $ is a $p$-group such that $ |L| = p^i $ with $ i > 1 $, then it has nontrivial center, and therefore $ Z(L) $ is a nontrivial normal subgroup. Let $ H_i $ and $ H_{i+1} $ be in the normal series, then $ L = H_{i+1}/H_i $ has a nontrivial normal subgroup $ Z(L) $, and by the correspondence theorem this means that there is a normal subgroup of $ H_{i+1} $ properly containing $ H_i $. In this way, you can continue to "break up" the normal series until each quotient is of prime order.
If $ Z(L) = L $, then any subgroup of $ L $ is normal, so take any element which does not generate $ L $ and take the cyclic subgroup generated by that element.