TITLE: Does a group have a unique pro-p topology?
QUESTION [8 upvotes]: If $G$ is a residually $p$ group and $G_i$ ANY filtration (i.e. $G_i\subset G_{i-1}$ and $\cap G_i=e$) of normal 
$p$-power index subgroups, is the corresponding filtration the usual pro-$p$ filtration?
Put differently, if $H$ is any normal $p$-power index subgroup, does it contain one of the $G_i$'s?
The answer to the corresponding question for finite filtrations is of course negative, e.g. the filtration $p^i\Bbb{Z}$ of $\Bbb{Z}$ does not give the profinite topology of $\Bbb{Z}$. On the other hand for abelian groups the answer to the above question is positive.

REPLY [2 votes]: Contrary to OP's claim, this is false even for finitely generated abelian groups, including $\mathbf{Z}^2$.
Indeed, fix a dense group embedding $i:\mathbf{Z}^2\to\mathbf{Z}_p$ (namely $i(n,m)=n+mt$ for some fixed irrational $t\in\mathbf{Z}_p$). Define $G_n=i^{-1}(p^n\mathbf{Z}_p)$. Then $(G_n)$ is a decreasing sequence of subgroups with intersection reduced to $\{0\}$. Moreover $G_0/G_n$ is isomorphic to $\mathbf{Z}/p^n\mathbf{Z}$. Hence no $G_n$ is contained in $p\mathbf{Z}^2$.