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I'm interested in a concrete example of an infinite metabelian quotient of the free product $C_2 * C_2 * C_2$, where $C_2$ is the cyclic group of order $2$. In particular, I would be interested in a homomorphism onto a wreath product of abelian groups. How would this look like on the standard generating set $\left\{a,b,c\right\}$ where $a,b,c$ are the generators of one copy of $C_2$ each.

What are the infinite nilpotent quotients of $C_2 * C_2 * C_2$?

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    $\begingroup$ The torsion elements in a nilpotent group form a subgroup, so there are no infinite nilpotent quotients. $\endgroup$
    – Derek Holt
    Commented Nov 27, 2013 at 12:19
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    $\begingroup$ The derived subgroup of $G=C_2^{*3}$ is free of rank 5. So the metabelianization of $G$ is an extension of $\mathbf{Z}^5$ by the abelian group of order 8 $C_2^3$. It would be nice to realize it explicitly as a cocompact proper group of the Euclidean 5-space (in a way the natural action of Sym(3) is visible). $\endgroup$
    – YCor
    Commented Nov 29, 2013 at 13:48

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Perhaps the easiest metabalian quotient of $C_2\ast C_2\ast C_2$ is $C_2\ast C_2 \cong \mathbb{Z}\rtimes C_2$ the infinite dihedral group.

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  • $\begingroup$ Thanks! How would you describe the corresponding homomorphism with respect to the generating set above in this case? $\endgroup$
    – Elisabeth Fink
    Commented Nov 27, 2013 at 12:08
  • $\begingroup$ Well the obvious one: $a\mapsto a, b\mapsto b, c\mapsto 1$. $\endgroup$
    – Johannes Hahn
    Commented Nov 27, 2013 at 13:21
  • $\begingroup$ But both generators $a$ and $b$ have order $2$. How do you make it work with $\mathbb{Z} \rtimes C_2$? $\endgroup$
    – Elisabeth Fink
    Commented Nov 27, 2013 at 13:44
  • $\begingroup$ You can take $ab$ as the generator of the infinite cyclic group. $\endgroup$
    – Derek Holt
    Commented Nov 27, 2013 at 14:10

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