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Let $G$ be a finitely generated pro-$p$ group. Let $T$ be the set of all torsion elements in $G$.

  1. Is it possible for $T$ to be a non-closed subgroup?

Anyway,

  1. Can $G/\overline{\langle T\rangle}$ have torsion?

For any finitely generated abelian (more generally, powerful) pro-$p$ group $G$, I know that 1. and 2. have negative answers: $T$ is finite and $G/T$ is torsion-free.

Thanks.

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  • $\begingroup$ I think $G=\ker(GL_2({\mathbb Z}_2)\to GL_2({\mathbb F}_2))$ is a pro-2 group for which (1) holds. The conjugates of $\scriptstyle\begin{pmatrix}1&0\cr0&-1\end{pmatrix}$, i.e. reflections, seem to generate topologically a finite index subgroup with plenty of non-torsion elements (maybe even the whole of $G$?) Don't know about (2) though. $\endgroup$
    – Tim Dokchitser
    Commented Jun 11, 2011 at 11:51
  • $\begingroup$ Tim, but do the torsion form a subgroup? I don't think so. Michael is not asking whether the torsion as a set is not closed but he also requires it to be a subgroup. $\endgroup$
    – Yiftach Barnea
    Commented Jun 12, 2011 at 17:54
  • $\begingroup$ @Yiftach: You are right, thanks! I missed the word "subgroup" $\endgroup$
    – Tim Dokchitser
    Commented Jun 13, 2011 at 8:05

3 Answers 3

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It seems to me that the answer on both questions is Yes.

Let $F$ be a finitely generated free non-abelian pro-$p$ group, $N=\overline{F^{\prime\prime}}$ and $K=\overline {[N,F]}$. Then $N/K$ is torsion free and of infinite rank. Let $a_1,a_2,\ldots$ be a free $\mathbb Z_p$-generating set of $N/K$. Put $G_1=F/\langle K, a_1^{p},a_2^{p^2},\ldots\rangle$. Let $a=\prod a_i$ and let $\bar a$ be its image in $G_1$. Then $\bar a$ is not a torsion element but lies in the clousure of the subgroup of torsion elements. This answers the first question.

Let $G_2$ be the quotient of $G_1\times \langle z\rangle $ by the subgroup generated by $z^p\bar a$. Then the image of $z$ in $G_2$ is not a torsion element.

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  • $\begingroup$ @Andrei: Many thanks! Though, I must say I need some time to fully understand your interesting example. Do you think it is possible to do something similar when the group is nilpotent? (the case I'm in fact interested in). Thanks again. $\endgroup$
    – Michael
    Commented Jun 15, 2011 at 4:06
  • $\begingroup$ @Michael: If $G$ is finitely generated and nilpotent, then every closed subgroup is also finitely generated and a subgroup generated by a finite number of torsion elements is finite. Thus if $G$ is nilpotent and finitely generated, then $\langle T\rangle=T $ is finite. $\endgroup$
    – Andrei Jaikin
    Commented Jun 15, 2011 at 9:02
  • $\begingroup$ @Andrei: To prove your first claim, should I check that the factors of the (topological) lower central series are (topologically) finitely generated from basic commutators, or is there a simpler argument? About the second claim, how did you choose (topological) generators of $\overline{T}$ in $T$? Sorry about these questions, I'm just beginning to learn profinite groups. $\endgroup$
    – Michael
    Commented Jun 16, 2011 at 2:12
  • $\begingroup$ @Michael: Yes, you prove the first claim by induction on the nilpotency class using that the factors of the (topological) lower central series are (topologically) finitely generated. The subgroup $\overline{\langle T\rangle }$ is topologically generated by a finite subset of $T$. Since these elements are torsion, then $\overline{\langle T\rangle }$ is finite. $\endgroup$
    – Andrei Jaikin
    Commented Jun 16, 2011 at 9:38
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Here's just a small remark and a belief. (I'd leave this as a comment, but I don't have enough rep.).

Assume $T$ is a subgroup. If $T$ is closed then the quotient in (2) is equal to $G/T$ which is torsion-free.

I believe that this is the situation if your $G$ is nilpotent. The experts here may provide an argument for this.

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  • $\begingroup$ @Alves: Sure! And that was my first belief too! Unfortunately I'm having a hard time to prove this. $\endgroup$
    – Michael
    Commented Jun 14, 2011 at 1:41
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Here is a partial answer to (1). If $T$ is a subsgroup, then it is normal. if we assume that it is finitely generated, then I believe that Corolalry 1.8 from the Nikolov and Segal's preprint http://arxiv.org/PS_cache/arxiv/pdf/1102/1102.3037v4.pdf implies that $T$ is closed. (This is true for more general case than $G$ is pro-$p$.)

So the question is reduced to the case when $T$ is not finitely generated. So for instance one can ask is it possible for $c_p \times c_{p^2} \times c_{p^3} \times \cdots$ to be embedded in a finitely generated pro-$p$ group $G$ so that the torsion of $G$ lies in $c_p \times c_{p^2} \times c_{p^3} \times \cdots$? I am not sure what the answer is.

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  • $\begingroup$ @Yiftach: Thanks for the idea and for the reference! I'm thinking about them too. $\endgroup$
    – Michael
    Commented Jun 14, 2011 at 1:43

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