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I'm trying to prove that if $N$ is a normal subgroup of $G$, with $N$ and $G/N$ solvable, then $G$ is solvable.

Proving that $G/N$ is abelian would of course suffice, but I'm not sure if that's a necessary condition or not. I suppose it's possible that I could find some normal subgroup $K$ in $G$ such that $N\subseteq K\subseteq G$ with both $G/K$ and $K/N$ abelian, but I can't see how to go about constructing it.

I've tried a couple things with the isomorphism theorems, and looking at properties preserved under homomorphisms and such, but nothing's panned out.

Can anyone help me here, I'm a bit stuck, thanks.

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$G/N$ need not be abelian. Consider $G$ to be the symmetric group of degree 4, $N=1$. There is no normal subgroup $K$ such that $K/N$ and $G/N$ are abelian. This is no big deal though, as you can always form the long chains of normal subgroups indicated in the two answers. –  Jack Schmidt Aug 2 '12 at 13:39

2 Answers 2

up vote 6 down vote accepted

If $~1=U_0\le U_1\le \cdots \le U_n=G/N~$ and $~1=V_0\le V_1\le \cdots \le V_m=N$ are subnormal series with abelian factor groups, then consider the series induced by "superimposing" the former over latter:

$$1=V_0\le V_1\le \cdots\le V_m=N=U_0'\le U_1'\le \cdots \le U_n'=G$$

where $U_i'$ are the unique subgroups of $G$ such that $U_i'/N=U_i$ (given by the lattice theorem).

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Thanks for your help. –  cactuar Aug 2 '12 at 3:59

This is a good to know that:

Theorem: $G$ is a group and $N\unlhd G$ then, $\forall n\in \mathbb N, \big(\frac{G}{N}\big)^{(n)}=\frac{G^{(n)}N}{N}$

wherein $G^{(1)}=G'$ and $G^{(i)}=\big(G^{(i-1)}\big)', i\geq 2$. Definition tells there are non-negative integer $n,m$ such that $N^{(n)}=\{e\}$ and $\big(\frac{G}{N}\big)^{(m)}=\{N\}$ because $N$ and $\frac{G}{N}$ both are soluble. Using the above theorem, we have $$\{N\}=\bigg(\frac{G}{N}\bigg)^{(m)}=\frac{G^{(m)}N}{N}\Longrightarrow G^{(m)}\leq N $$ $$G^{(m)}\leq N \Longrightarrow \big(G^{(m)}\big)^{(n)}\leq N^{(n)}=\{1\}\Longrightarrow G^{(m+n)}\leq N^{(n)}=\{1\} $$ or $G$ is soluble gorup.

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Great! (...of course!) +1 –  amWhy Mar 12 '13 at 1:52

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