2
$\begingroup$

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 $G$.

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.