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Let G be an infinite group. Suppose that the commutator subgroup [G,G] of G and the abelianization of G are finitely generated. Does this imply that G is finitely generated?

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2 Answers 2

In fact if $G$ is a group and $N$ is a normal subgroup. Then $G/N$ and $N$ finitely generated implies $G$ is finitely generated.

Let $G$ be an infinite group and $N$ a normal subgroup of $G$. Suppose $N$ is generated by $h_1,...,h_n$ and $G/N$ is generated by $k_1 +N,...k_m +N$. We aim to show that $h_1,...,h_n,k_1,...,k_m$ generate $G$. Given any $g\in G$, $g +N\in G/N$ can be written as a product of $k_1+N,....,K_m+N$. So there exists an element $a\in G$ which is a product of $k_1,...,k_m$ such that $ga^{-1}\in N$. Now since $N$ is finitely generated $ga^{-1}$ can be written as a product of $h_1,...,h_n$. Therefore $g$ is a product of $k_1,,,.k_m,h_1,,,h_n$.

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Yes. A sketch of the proof: Suppose that $H$ is the Abelianization of $G$. Pick generators $h_i$ for $H$. Pick generators $g_j$ for $[G,G]$, the kernel of the map $\pi : G \to H$. Pick preimages $h_i' \in \pi^{-1}(h_i)$. Show that the $g_j$ and the $h_i'$ generate $G$.

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This also works when $[G, G]$ is replaced by any normal subgroup $N$ such that $N$ and $G/N$ are both finitely generated. –  Chris Eagle Sep 7 '11 at 9:10

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