1
$\begingroup$

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 $n-1$

  • 1
    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.2012-10-07
  • 0
    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?2012-10-07
  • 0
    Well, what do you get when you take the image of $[G^{(n-2)},G^{(n-2)}]$ in $G/G^{(n-1)}$?2012-10-07
  • 0
    I don't know your opinion. homomorphism: $G \to G/G^{(i-1)}$2012-10-07
  • 0
    $[G^{(n-1)},G^{(n-1)}]=G^{(n-2)}$, so $[G^{(n-1)},G^{(n-1)}]$ goes to the identity in $G/G^{(n-2)}$. Now can you prove that $G^{(k)}/G^{(n-1)}=(G/G^{(n-1)})^{(k)}$?2012-10-07
  • 0
    I think $[G^{(n-2)},G^{(n-2)}]=G^{(n-1)}$2012-10-07
  • 0
    Oh yes, sorry, I misspoke. I meant: $[G^{(n−2)},G^{(n−2)}]=G^{(n−1)}$, so $[G^{(n−2)},G^{(n−2)}]$ goes to the identity in $G/G^{(n−1)}$. Now can you prove that $G^{(k)}/G^{(n−1)}=(G/G^{(n−1)})^{(k)}$?2012-10-07
  • 0
    Yes, by natural homomorphism $\pi: G \to G/G^{(n-1)}$ and property $(\pi(G))^{(k)}=\pi(G^{(k)})$. And what is next step?2012-10-07
  • 0
    oh, $(G/G^{(n-1)})^{(n-1)}=G^{(n-1)}/G^{(n-1)}=1$2012-10-07
  • 0
    Yeah, there you go. You've got it.2012-10-07
  • 0
    very nice, thank you about everything2012-10-07

1 Answers 1