Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $n-1$

share|improve this question
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

1 Answer 1

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

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

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.