# Talk:Lamplighter group

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WikiProject Mathematics (Rated Stub-class, Low-priority)
• Questions about the Lamplighter group

I got here by accident, and I'm not really an expert in the lamplighter group, and the wreath product definition didn't really help. So I got the presentation from http://arxiv.org/abs/math.GR/0312331 and worked the rest out on my own (getting it wrong on the first try).

But the definition there defines a finitely generated group, and the wreath product definition apparently defines an uncountable group, which cannot even be countably generated. If we are talking about Growth rate (group theory) (which seems to be the motivating feature for this article), that is defined with respect to a particular generating set, and it looks like that set needs to be finite. Perhaps we need to refer to the finitely supported subgroup of

${\displaystyle \bigoplus _{(-\infty ,+\infty )}\mathbb {Z} /2\mathbb {Z} ,}$

but I don't know the name for that.

${\displaystyle \bigcup _{n=0}^{\infty }\bigoplus _{(-n,+n)}\mathbb {Z} /2\mathbb {Z} }$

is only approximately correct.

A second question is a proof that the group is a solvable group. Anyone? Dan Hoey 03:07, 9 March 2007 (UTC) mod Dan Hoey 14:08, 9 March 2007 (UTC)

You are confused. When people write wreath product of infinite groups they don't mean that every pair of elements in the semidirect product is possible - they think of a finite number or non-identity elements. Thus the lamplighter group is countable. As for solvability, this is an easy exercise. Take the commutator and see for yourself. Mhym 21:03, 9 March 2007 (UTC)
Indeed I was confused: the direct product ${\displaystyle \bigoplus _{(-\infty ,+\infty )}\mathbb {Z} /2\mathbb {Z} }$ is defined to have finite support and so is countable.Dan Hoey 18:18, 11 March 2007 (UTC)
It's metabelian! Immediate as it's a semi direct product by definition! — Preceding unsigned comment added by 192.76.7.217 (talk) 09:50, 15 May 2012 (UTC)
I am certainly confused. Isn't Z/2ZZ isomorphic to the dihedral group of order 8? Maproom (talk) 17:33, 20 October 2013 (UTC)
No, as it is obviously infinite. Perhaps you are thinking about Z/2ZZ/2Z.—Emil J. 17:37, 19 May 2016 (UTC)