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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 $1\leq j<k$.

Cameron Buie
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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 01:12
  • 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 01:16

2 Answers2

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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.

Arturo Magidin
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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 01:34
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    @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 01:37
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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.

mez
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    There are many things that are"easy to show". However, knowing that something is "easy to show" doesn't help people who come here looking for answers, and therefore your "answer" probably helps no one. – Junglemath Jul 27 '20 at 03:50