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$\newcommand{\GL}{\mathrm{GL}}\newcommand{\dbZ}{\mathbb{Z}}\newcommand{\dbF}{\mathbb{F}}\newcommand{\dbN}{\mathbb{N}}$ Hello.

I have a small question regarding closed subgroup of $\GL_n(\dbZ_p)$, which I tried floating on MSE but did not get a response. As this question arose from my research I figured it might be suitable to ask it here.

Fix $n\in \dbN$ and let $p$ be a fixed prime. Let $\eta:\GL_n(\dbZ_p)\to\GL_n(\dbF_p)$ denote the coordinate-wise reduction-map.

Suppose we have two closed (with respect to $p$-adic topology) and infinite subgroups $G, H\in\GL_n(\dbZ_p)$, which have the same image under $\eta$, i.e. $\eta(G)=\eta(H)$.

What can said in this case about $G$ and $H$? I know one can not expect them to be equal (e.g. $G=1+p M_n(\dbZ_p)$ and $H=1+p^k M_n(\dbZ_p)$ have the same image mod $p$).

Can it be proved for example that in such a case $G$ and $H$ are commensurable? perhaps under some additional assumptions regarding these groups? $\newcommand{\GG}{\mathsf{G}}\newcommand{\HH}{\mathsf{H}}$

The main focus of interest is for the case where $G$ and $H$ arise as the group of $\dbZ_p$ points of (smooth) algebraic groups $\GG,\HH$, defined over $\dbZ_p$.

I remark in this case there is no assumption that either group has good reduction modulo $p$ and that I am intentionally looking at coorindate-wise reduction (for which I don't know of any functorial interpretation), rather than the better behaved operation of reduction mod $p$ of the group schemes $\GG$ and $\HH$ (i.e. the $\dbF_p$-points of the image of the functor $\newcommand{\spec}{\mathrm{Spec}} \GG\mapsto \GG\times_{\spec\dbZ_p}\spec\dbF_p$, and similarly for $\HH$).

I would very much appreciate any sort of idea or refernece anyone might have to offer on the subject.

Thank you.

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    $\begingroup$ They will not be commensurable in general. Take e.g. $G$ to be the diagonal matrices in $\mathbf{SL}_2$ and $H$ to be the conjugate of $G$ by $\begin{pmatrix} 1 & p \\ 0 & 1 \end{pmatrix}$. Then $G$ and $H$ have the same image mod $p$ but $G \cap H$ is trivial. $\endgroup$ Commented Nov 23, 2016 at 10:21
  • $\begingroup$ So maybe the right question is: is some conjugate of $H$ commensurable with $G$, since the condition is blind to conjugation by elements of $1 + pM_{n}(\mathbb{Z}_{p})$. ( also, I think that if the groups remain irreducible (mod $p$), then the groups upstairs must be conjugate by an element of $1 + pM_{n}(\mathbb{Z}_{p})$ if they are conjugate at all). $\endgroup$ Commented Nov 23, 2016 at 15:44
  • $\begingroup$ I should have said "absolutely irreducible" above. $\endgroup$ Commented Nov 23, 2016 at 15:53
  • $\begingroup$ Hi @GeoffRobinson - is absolutely irreducible the same as absolutely simple? $\endgroup$
    – kneidell
    Commented Nov 23, 2016 at 16:07
  • $\begingroup$ Yes, it is indeed. $\endgroup$ Commented Nov 23, 2016 at 16:15

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