4
$\begingroup$

Can't find a flaw in that proof:

Induction by the length of derived series.

Base: if $[G, G]=e$ then the group is abelian...

Assume that statement is true for n-1.

We have group $G$ with the derived series length $n$, and a subgroup $H$.

$[G, G]=G'$ has derived series of length $n-1$, so $H \cap G'$ is finitely generated.

$H/(H\cap G')$ is a subgroup of finitely generated abelian subgroup $G/G'$. So $H$ is finitely generated.

  • 0
    Well, where do you think the problem might be? If $H/(H\cap G^{\prime})$ is finitely generated then it isn't obvious that $H$ is. I would start by looking here, if I were you.2012-02-10
  • 0
    @User1729: If he'd proven that $H\cap G'$ is finitely generated, then the finite generation of both $H/(H\cap G')$ and $H\cap G'$ would imply that $H$ is finitely generated: more generally, if $N$ and $G/N$ are finitely generated, then so is $G$: pick $g_1,\ldots,g_m$ whose images generated $G/N$, and $m_1,\ldots,m_k$ that generated $N$. Given $g\in G$, white $gN=g_{i_1}^{a_1}\cdots g_{i_{\ell}}^{a_{\ell}}N$, hence $g=g_{i_1}^{a_1}\cdots g_{i_{\ell}}^{a_{\ell}}n$ for some $n$, and now express $n$ in terms of $m_1,\ldots m_k$. The problem was elsewhere.2012-02-10

3 Answers 3