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Let $G$ and $H$ be two finitely generated groups, and let $W = G \wr H$ be the wreath product of $G$ and $H$. Show that $W$ is finitely generated. In class today, we were showed this and told that it was obvious. However, I do not see how it is obvious. How is this obvious?

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Take the generators of $G$ together with the generators of $H$. Much harder question is when this is finitely presented... – user641 Apr 11 '12 at 0:17
@Alex: is your comment directed at the point that you can either take the direct product or the direct sum in the definition of the wreath product? Of course it must be the latter if you want to have any chance of being finitely generated and then the result is true. – t.b. Apr 11 '12 at 3:04
Well, it is not exactly obvious but it is not hard. I suggest you work through the simplest example: the Lamplighter Group $L = \mathbb{Z}/2\mathbb{Z} \wr \mathbb{Z}$. If you manage to prove that the generators of $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}$ generate $L$ then the general case should be rather easy to do. – t.b. Apr 13 '12 at 1:28
@Bernard: You should specify if you are talking about the "restricted wreath product" (the base group is the restricted direct product of $|H|$ copies of $G$, i.e., the subgroup of $\prod_{h\in H}G$ with almost all entries trivial), or the "unrestricted wreath product" (the base group is the direct product of $|H|$ copies of $G$). – Arturo Magidin Apr 13 '12 at 2:56
@Arturo: Well, in the comments Bernard says that $H$ is finite, in which case it doesn't matter. But I don't know why he says that, when it isn't in the question and, provided he means the restricted wreath product, it isn't necessary. – Tara B Apr 13 '12 at 8:02

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