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Possible Duplicate:
A Nontrivial Subgroup of a Solvable Group
How to find a normal abelian subgroup in a solvable group?

Could someone help me with this proof? Let $G$ be a solvable group and $H$ a nontrivial normal subgroup of $G$. Then $H$ contains a nontrivial normal abelian subgroup of $G$. The proof that I found in my book is: Since $G$ is solvable $G^{(n)}=1$ for some $n>0$, where $G^{(n)}$ is the derived subgroup. Let $m$ be the minimum positive integer such that $H\cap G^{(m)}=1$. Then $H\cap G^{(m-1)}\neq1$ and are true the following relations: $H\cap G^{(m-1)}\cong [(H\cap G^{(m-1)})G^{(m)}/G^{(m)}]\leq G^{(m-1)}/G^{(m)}$. Why are these relations true? And from these finally $H\cap G^{(m-1)}$ is a nontrivial normal abelian subgroup of $G$ contained in $H$.I will be grateful for any help. Thanks!!

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