# $G$ is solvable if and only if it has a normal series with each factor of order a power of a prime

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.

• Are you assuming $G$ is finite? – D_S May 27 '16 at 4:37
• If you are not assuming that $G$ is finite, then you need to say what you mean by a normal series. Is the series necessarily finite? But I don't think there is any sensible definition of normal series that would make the result true for infinite groups. – Derek Holt May 27 '16 at 9:32

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.
• But each factor group is abelian so we have always $Z(L)=L$. Anyways I understand that you just keep on getting larger normal series at every step... How do you know you actually end this process at some point? I think we need to use $G$ has a composition series, but how? And if every quotient would have prime order, then the "prime power" in the statement is just a typo? Thank you. – Talexius May 27 '16 at 2:23
• If $H_{i+1}/H_i$ is abelian but not simple, then it has a nontrivial proper subgroup $K$. But $K = N/H_i$ for some subgroup $N$ of $H_{i+1}$. Now refine the inclusion $H_i \subseteq H_{i+1}$ to an inclusion $H_i \subseteq N \subseteq H_{i+1}$. Now $H_i$ is clearly normal in $N$, and also $N$ is normal in $H_{i+1}$ because $N/H_i$ is a normal subgroup of $H_{i+1}/H_i$ (since $H_{i+1}/H_i$ is abelian). – D_S May 27 '16 at 4:50
• I'm not assuming $G$ is finite. But I think every normal series can't get too much large, since there are a composition series it has maximal lenght. What do you think? I think that works. @D_S – Talexius May 27 '16 at 23:07
• $G$ doesn't have to be finite, this works as long as the factor groups are all finite. – Ege Erdil May 27 '16 at 23:29