# Subgroups of finite solvable groups. Solvable?

I am attempting to prove that, given a non-trivial normal subgroup $N$ of a finite group $G$, we have that $G$ is solvable iff both $N$, $G/N$ are solvable. I was able to show that if $N,G/N$ are solvable, then $G$ is; also, that if $G$ is solvable, then $G/N$ is. I am stuck showing that $N$ must be solvable if $G$ is.

It seems intuitive that any subgroup of a finite solvable group is necessarily solvable, as well. Is this true in general? For normal subgroups? How can I go about proving this result?

Edit: By solvable, I mean we have a finite sequence $1=G_0\unlhd ... \unlhd G_k=G$ such that $G_{j+1}/G_j$ is abelian for each $1\leq j<k$.

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Yes, it's true. What exactly is your definition of "solvable"? (I ask because there are a number of equivalent definitions, and how you prove it depends on your precise definition) –  Arturo Magidin May 14 '12 at 1:12
Thanks, Arturo. I have edited the question to include the definition I'm using. I am familiar with the derived series definition, too, but used this result to prove its equivalence with the formulation above. –  Cameron Buie May 14 '12 at 1:16

## 2 Answers

With your definition, to show that if $G$ is solvable then $N$ is solvable, let $$1 =G_0 \triangleleft G_1\triangleleft\cdots\triangleleft G_{m-1}\triangleleft G_m=G$$ be such that $G_{i+1}/G_{i}$ is abelian for each $i$.

Note: We do not need to assume that $N$ is normal; the argument below works just as well for any subgroup of $G$, not merely normal ones.

Let $N_i = G_i\cap N$. Note that since $G_i\triangleleft G_{i+1}$, then $N_i\triangleleft N_{i+1}$: indeed, if $x\in N_i$ and $y\in N_{i+1}$, then $yxy^{-1}\in N$ (since $x,y\in N$) and $yxy^{-1}\in G_i$ (since $G_i\triangleleft G_{i+1}$), hence $yxy^{-1}\in N\cap G_i = N_i$.

So we have a sequence $$1 = N_0\triangleleft N_1\triangleleft\cdots\triangleleft N_{m} = N.$$ Thus, it suffices to show that $N_{i+1}/N_i$ is abelian for each $i$.

Note that $N_{i} = N\cap G_i = (N\cap G_{i+1})\cap G_i = N_{i+1}\cap G_i$.

Now we simply use the isomorphism theorems: $$\frac{N_{i+1}}{N_i} =\frac{N_{i+1}}{N_{i+1}\cap G_i} \cong \frac{N_{i+1}G_i}{G_i} \leq \frac{G_{i+1}}{G_i}$$ since $N_{i+1},G_i$ are both subgroups of $G_{i+1}$ and $G_i$ is normal in $G_{i+1}$, so $N_{i+1}G_i$ is a subgroup of $G_{i+1}$.

But $G_{i+1}/G_i$ is abelian by assumption, hence $N_{i+1}/N_i$ is (isomorphic to) a subgroup of an abelian group, hence also abelian, as desired.

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Fantastic, and we never used the fact that $N$ is normal, or that $G$ is finite, so it generalizes fully! Thanks, again! –  Cameron Buie May 14 '12 at 1:34
@Cameron: Indeed; I was in the process of noting that $N$ need not be normal. Of course, you need it to be normal to be able to talk about $G/N$ in the other part of your argument. (-: –  Arturo Magidin May 14 '12 at 1:37

It's easy to show that a group $G$ is solvable iff one of the following holds

1. $G$ has a solvable series, (which is the definition you gave)
2. The $n$-th derived subgroup $G^{(n)}$ vanish for some $n$.

Then for any $H\le G$, we have $H^{(n)}\subset G^{(n)}$ therefore $H$ is solvable too.

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