Beforehand: I am not particularly algebraically educated and, especially, I do not have much background as far as free products of groups are concerned. So, it might well be that my question seems trivial (on the other hand, I could not find a quick answer to my question).

Let $(G_i)_{i=1,\ldots,n}$ be a family of copies of the cyclic group of order 2.

Then, consider the $n$-fold free product $$G := \ast_{i=1}^n G_i$$

What are the properties of $G$? In particular, is $G$ nilpotent? If yes, what is the nilpotency class of $G$? The only related theorem I am aware of is the Kurosh subgroup theorem, but I don't see a straightforward application to my case, esp. as I am looking for a central series, and therefore for normal subgroups of $G$.

  • $\begingroup$ Well, using Kurosh theorem, if there's a subgroup of that free product which is it itself the free product of a free group and some other things (conjugates of the factors), then the group cannot be nilpotent... $\endgroup$ – DonAntonio Sep 23 '13 at 16:43
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    $\begingroup$ For $n=2$, we get the infinite dihedral group, which is not nilpotent, so $G$ is not nilpotent whenever $n \ge 2$. $\endgroup$ – Derek Holt Sep 23 '13 at 17:22

For all $i$, $G'\cap G_i=1$ ($G'$ is the commutamt of $G$), since $G'$ consists of words of even lenght. By the Kurosh subgroup theorem $G'$ is free. So $G$ is an extension of a free group by an elementary Abelian $2$-group.

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    $\begingroup$ I don't follow the reasoning with $n = 2$, for example $S_3$ has a cyclic subgroup of index $2$, but $S_3$ is not nilpotent. $\endgroup$ – spin Sep 23 '13 at 17:07
  • $\begingroup$ This argument only works for when $n>2$, no? Because otherwise the free subgroup is cyclic so this argument doesn't give you anything. $\endgroup$ – user1729 Sep 24 '13 at 9:54
  • $\begingroup$ @user1729 For $n=2$ it sounds the same: $G$ is an extension of a cyclic group (= a free group of range $1$) by $\mathbb{Z}_2$. $\endgroup$ – Boris Novikov Sep 24 '13 at 10:25
  • $\begingroup$ @BorisNovikov Yeah, but that doesn't imply that it is non-nilpotent (because $\mathbb{Z}\times C_2$ also satisfies these conditions and is nilpotent.) $\endgroup$ – user1729 Sep 24 '13 at 10:28
  • $\begingroup$ @user1729 I say nothing about nilpotence. I answer the question: "What are the properties of $G$?" $\endgroup$ – Boris Novikov Sep 24 '13 at 10:32

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