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Let $G$ be finite group. Show that if $G$ is a solvable group, and derived length is $n$ then $G$ contains a abelian normal nontrivial subgroup $H$ with $G/H$ has derived length is $n-1$

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You must tell us what you have tried or are confused about, rather than just state the problem. Also please accept the answers to your other questions if you are happy with the solutions. Welcome to MSE. – Alexander Gruber Oct 7 '12 at 4:25
I think $G$ has a derived serie $G = G^0 \geq G^1 \geq \ldots G^n = 1$. So $H = G^{n-1}$ be a abelian normal nontrivial subgroup. But I can't show that $G/G^{n-1}$ has derived length $n-1$. Help me? – Muniain Oct 7 '12 at 4:50
Well, what do you get when you take the image of $[G^{(n-2)},G^{(n-2)}]$ in $G/G^{(n-1)}$? – Alexander Gruber Oct 7 '12 at 5:02
I don't know your opinion. homomorphism: $G \to G/G^{(i-1)}$ – Muniain Oct 7 '12 at 5:25
$[G^{(n-1)},G^{(n-1)}]=G^{(n-2)}$, so $[G^{(n-1)},G^{(n-1)}]$ goes to the identity in $G/G^{(n-2)}$. Now can you prove that $G^{(k)}/G^{(n-1)}=(G/G^{(n-1)})^{(k)}$? – Alexander Gruber Oct 7 '12 at 5:35

Hint: look to the last non-trivial element in the derived series, i.e.:

$$G> G'>G''>...>G^{(n)} >1\,\,\,,\,\,\text{then take }\,\,\,H=G^{(n)}$$

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