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

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

Anyway,

  
*
  
*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.
 A: 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.
A: 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.
A: 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.
