TITLE: Torsion - subgroup and quotient
QUESTION [9 upvotes]: Let $G$ be a finitely generated pro-$p$ group.
Let $T$ be the set of all torsion elements in $G$.


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.

REPLY [2 votes]: 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.