# Profinite non-rigidity of arithmetic groups

Amir Y. Weiss Behar Einstein Institute of Mathematics
Edmond J. Safra Campus (Givat-Ram)
The Hebrew University of Jerusalem
9190401
Israel<EMAIL_ADDRESS>
###### Abstract.

We show that for a typical high rank arithmetic lattice $\Gamma$, there exist
finite index subgroups $\Gamma_{1}$ and $\Gamma_{2}$ such that
$\Gamma_{1}\not\simeq\Gamma_{2}$ while
$\widehat{\Gamma_{1}}\simeq\widehat{\Gamma_{2}}$. But there are exceptions to
that rule.

##  1\. Introduction

Let $\Gamma$ be a finitely generated residually finite group. We say that
$\Gamma$ is profinitely-rigid if whenever
$\widehat{\Lambda}\cong\widehat{\Gamma}$ for some finitely generated
residually finite group $\Lambda$, then $\Lambda\cong\Gamma$. Here
$\widehat{\Gamma}$ (resp. $\widehat{\Lambda}$) denotes the profinite
completion of $\Gamma$ (resp. $\Lambda$).
Up until recently, the only profinite rigid groups were ”small” (i.e. without
non-abelian free subgroups). Recently, in a groundbreaking work, Bridson,
McReynolds, Reid and Spitler gave first examples of ”big” groups which are
profinitely rigid, among them are some fundamental groups of hyperbolic
3-manifolds [3] and some triangle groups [2].
Arithmetic subgroups of semisimple Lie groups need not be profinitely rigid
([1],[7]). A well known open problem asks:

For $n\geq 2$, is $SL_{n}(\mathbb{Z})$ profinitely rigid?

While we will not answer this question, we will show that there are finite
index subgroups of these groups (at least when $n\geq 3$) which are not
profinitely rigid. In fact, we will show a much more general result:

######  Theorem 1.1 (Main Theorem).

Let $k$ be a number field and $\mathbf{G}$ be a connected, simply connected,
absolutely almost simple $k$-linear algebraic group of high $\infty$-rank such
that $\mathbf{G}(k)$ satisfies the congruence subgroup property and
$\Gamma\subseteq\mathbf{G}(k)$ an arithmetic subgroup.

  1. (1)

Unless $\mathbf{G}$ is of type $G_{2},F_{4}$ or $E_{8}$ and $k=\mathbb{Q}$,
$\Gamma$ has infinitely many pairs of finite index subgroups $\Gamma_{1}$ and
$\Gamma_{2}$ which are not isomorphic but their profinite completions are.

  2. (2)

The exceptional cases are truly exceptional and in these cases, there are no
such pairs at all. In fact, if
$\Gamma_{1},\Gamma_{2}\subseteq\mathbf{G}(\mathbb{Q})$ are arithmetic
subgroups with isomorphic profinite completions then $\Gamma_{1}$ and
$\Gamma_{2}$ are isomorphic.

Note that we show that each such $\Gamma$ has a finite index subgroup
$\Gamma_{1}$ which is not profinitely rigid by showing that
$\widehat{\Gamma_{1}}\cong\widehat{\Gamma_{2}}$ for some $\Gamma_{2}$
commensurable to it. This complements results of [7],[1] and [9] which give
examples of non-commensurable arithmetic groups which are profinitely
isomorphic.
To illustrate our methods, let us now present them only for
$\Gamma=SL_{4}(\mathbb{Z})$ (methods A & B) and
$\Gamma=SL_{2}(\mathbb{Z}[\sqrt{2}])$ (method C).

Method A:

Using the centre of the simply connected form: Let $2\neq p,q$ be two
different primes, and let $\Lambda$ be the principle congruence subgroup
corresponding to $pq$. Let $\rho_{p}$ be the element of
$\widehat{SL_{4}(\mathbb{Z})}\cong\prod_{r}SL_{4}(\mathbb{Z}_{r})$ which is
$1$ at the places $r\neq{p}$ and $-1$ at the place $p$, similarly define
$\rho_{q}$. Define $\Gamma_{1}:=\langle\widehat{\Lambda},\rho_{p}\rangle\cap
SL_{4}(\mathbb{Z})$ and
$\Gamma_{2}:=\langle\widehat{\Lambda},\rho_{q}\rangle\cap SL_{4}(\mathbb{Z})$.
Then
$\widehat{\Gamma_{1}}\cong\widehat{\Lambda}\times\mathbb{Z}/2\mathbb{Z}\cong\widehat{\Gamma_{2}}$
but $\Gamma_{1}$ and $\Gamma_{2}$ cannot be isomorphic. For details see 3.1.

Method B:

Using a non-trivial Dynkin automorphism: Let $2,3\neq p,q$ be two different
primes. Consider the following maximal parabolic subgroups of
$SL_{4}(\mathbb{Z}/p\mathbb{Z})$.

$P_{1,p}:=\left\\{\begin{pmatrix}\ast&\ast&\ast&\ast\\\ &\ast&\ast&\ast\\\
&\ast&\ast&\ast\\\ &\ast&\ast&\ast\\\ \end{pmatrix}\right\\},\quad
P_{2,p}:=\left\\{\begin{pmatrix}\ast&\ast&\ast&\ast\\\ \ast&\ast&\ast&\ast\\\
\ast&\ast&\ast&\ast\\\ &&&\ast\\\ \end{pmatrix}\right\\}$

Similarly define the maximal parabolic subgroups $P_{1,q}$ and $P_{2,q}$ of
$SL_{4}(\mathbb{Z}/q\mathbb{Z})$. Now, let $\Gamma_{1}$ be the congruence
subgroup corresponding to $P_{1,p}\mod p$, $P_{1,q}\mod q$ and is trivial
$\mod 3$, and $\Gamma_{2}$ to be the congruence subgroup corresponding to
$P_{1,p}\mod p$, $P_{2,q}\mod q$ and is trivial$\mod 3$. Then the profinite
completions of $\Gamma_{1}$ and $\Gamma_{2}$ are isomorphic via the
automorphism of $\widehat{SL_{4}(\mathbb{Z})}$ which is the non-trivial Dynkin
automorphism at the place $q$ and the identity elsewhere, but they themselves
cannot be isomorphic. For details see 4.1.

Method C:

Using the number field: Let $2\neq p,q$ be two different primes such that $2$
is a square in $\mathbb{Q}_{p}$ and $\mathbb{Q}_{q}$, hence $p$ and $q$ splits
completely in $\mathbb{Q}[\sqrt{2}]$. Set $\mathcal{O}=\mathbb{Z}[\sqrt{2}]$
and $\mathfrak{p}_{1},\mathfrak{p}_{2}|p$,
$\mathfrak{q}_{1},\mathfrak{q}_{2}|q$ to be the primes lying over $p$ and $q$
respectively. Let $\Gamma_{1}$ be the principle congruence subgroup
corresponding to $\mathfrak{p}_{1}\mathfrak{q}_{1}$, and $\Gamma_{2}$ be the
principle congruence subgroup corresponding to
$\mathfrak{p}_{2}\mathfrak{q}_{1}$. As $SL_{2}(\mathcal{O})$ has trivial
congruence kernel, $\widehat{SL_{2}(\mathcal{O})}\cong\left(\prod_{l\neq
p,q}SL_{2}(\mathcal{O}_{l})\right)\times\prod_{i=1}^{2}\left(SL_{2}(\mathcal{O}_{\mathfrak{p}_{i}})\times
SL_{2}(\mathcal{O}_{\mathfrak{q}_{i}})\right)$, where for a prime $l\neq p,q$,
$\mathcal{O}_{l}$ is the completion of $\mathcal{O}$ with respect to the
primes lying over $l$. Then the profinite completions, $\widehat{\Gamma_{1}}$
and $\widehat{\Gamma_{2}}$ are isomorphic via the automorphism of
$\widehat{SL_{2}(\mathbb{Z}[\sqrt{2}])}$ which is the transposition of the
places $\mathfrak{p}_{1}$ and $\mathfrak{p}_{2}$. But $\Gamma_{1}$ and
$\Gamma_{2}$ cannot be isomorphic. For details see 3.2.

The main theorem is proved by generalizing the above methods to more general
arithmetic lattices (In fact, only methods A and C are really needed).
The paper is organized as follows: after some preliminaries in §2, we will
generalize methods A and C in §3 deducing the first part of the main theorem.
In §4, we will elaborate on method B and finally in §5 we will prove that the
exceptional cases are true exceptions, concluding the main theorem. In §6, we
will give a stronger and more general version of the main theorem, stating
that it holds for $S$-arithmetic groups and not merely for arithmetic groups.
Moreover, one can get any (finite) number of non-isomorphic subgroup with
isomorphic profinite completions (not just pairs).

Acknowledgments. This work is a part of the author’s PhD thesis at the Hebrew
University. For suggesting the above topic and for providing helpful guidance,
suggestions and ideas I am deeply grateful to my advisors Alexander Lubotzky
and Shahar Mozes. During the period of work on this paper I was supported by
the European Research Council (ERC) under the European Union’s Horizon 2020
research and innovation programme (grant agreement No 882751) and by the ISF-
Moked grant 2019/19.

##  2\. Preliminaries

Throughout we assume that $k$ is a number field. The set of places of $k$ is
denoted by $V(k)$, it is the union of the set of archimedean places
$V_{\infty}(k)$ and the set of finite places $V_{f}(k)$. The completion of $k$
at $v\in V(k)$ is denoted by $k_{v}$. Let $\mathcal{O}_{k}$ denote the ring of
integers of $k$ and for a finite place $v\in V_{f}(k)$, denote by
$\mathcal{O}_{k,v}$ the ring of integers of $k_{v}$. The ring of finite adeles
$\mathbb{A}_{k}^{f}=\prod^{*}_{v\in
V_{f}(k)}k_{v}:=\\{(x_{v})_{v}\in\prod_{v\in
V_{f}(k)}k_{v}:\,x_{v}\in\mathcal{O}_{k,v}\text{ for all but finitely many
places }\\}$ is the restricted product over all the finite completions of $k$.
If $k$ is clear from the context, we will omit the letter $k$ from all the
above.
Let $\tilde{\mathbf{G}}$ be a connected, simply connected, absolutely almost
simple $k$-linear algebraic group, with a fixed faithful $k$-representation
$\rho:\tilde{\mathbf{G}}\to GL(n_{\rho})$. A subgroup
$\Gamma\subseteq\tilde{\mathbf{G}}(k)$ is called arithmetic if it is
commensurable with $\tilde{\mathbf{G}}(\mathcal{O})$ (see [12] and [14] for
more details about arithmetic groups). We will usually write $\mathbf{G}$ for
the adjoint form of $\tilde{\mathbf{G}}$ (which is the universal form), and by
$\pi:\tilde{\mathbf{G}}\to\mathbf{G}$ the universal covering map, it is a
central isogeny, and $\mathbf{C}:=\ker\pi=\mathcal{Z}(\tilde{\mathbf{G}})$ is
a finite group. The $V_{\infty}(k)$-rank of $\tilde{\mathbf{G}}$ is
$\text{rank}_{V_{\infty}(k)}\tilde{\mathbf{G}}:=\sum_{v\in
V_{\infty}(k)}\text{rank}_{k_{v}}\tilde{\mathbf{G}}$, where
$\text{rank}_{k_{v}}\tilde{\mathbf{G}}$ is the dimension of a maximal
$k_{v}$-split torus, $\tilde{\mathbf{G}}$ is said to have high $\infty$-rank
if its $V_{\infty}(k)$-rank is $\geq 2$.

We will use Margulis’ superrigidity in a rather delicate manner. The
particular version we use is the following:

######  Theorem 2.1 (Margulis’ superrigidity).

Assume $\tilde{\mathbf{G}}$ is of high $\infty$-rank, and let
$\Gamma_{1},\Gamma_{2}\subseteq\tilde{\mathbf{G}}(k)$ be arithmetic subgroups.
Assume further that $\Gamma_{i}\cap\mathbf{C}(k)={1}$. If
$\varphi:\Gamma_{1}\to\Gamma_{2}$ is an isomorphism, then there exists a
unique $k$-automorphism $\Phi$ of $\tilde{\mathbf{G}}$ and a unique
automorphism $\sigma$ of $k$ such that
$\varphi(\gamma)=\Phi(\sigma^{0}(\gamma))$ for every $\gamma\in\Gamma_{1}$,
where $\sigma^{0}$ is the automorphism of $\tilde{\mathbf{G}}(k)$ induced by
$\sigma$.

###### Proof.

Identifying $\Gamma_{1}$ and $\Gamma_{2}$ via the universal covering map
$\pi:\tilde{\mathbf{G}}\to\mathbf{G}$ as arithmetic subgroups of the adjoint
group $\mathbf{G}(k)$, Margulis’ superrigidity [11, Theorem VIII.3.6.(ii)]
implies that the isomorphism $\varphi$ can be extended to an automorphism of
$Res_{k/\mathbb{Q}}\mathbf{G}$. By the properties of the restriction of
scalars functor, such an automorphism must be of the form
$\Phi\circ\sigma^{0}$ for a $k$-automorphism $\Phi$ of $\mathbf{G}$ and an
automorphism $\sigma$ of the field $k$ [6, Proposition A.5.14]. Moreover, the
$k$-automorphism $\Phi$ of the adjoint form $\mathbf{G}$ can be interpreted as
a $k$-automorphism of the universal form $\tilde{\mathbf{G}}$, hence the
assertion of the theorem. ∎

If $\sigma$ is an automorphism of $k$, it induces a permutation of the
(finite) places of $k$, and thus an automorphism of the adelic group
$\tilde{\mathbf{G}}(\mathbb{A}_{k}^{f})$ by permuting its factors according to
$\sigma$, call this automorphism $\sigma_{\mathbb{A}}^{0}$. If $\Phi$ is a
$k$-automorphism of $\tilde{\mathbf{G}}$, it induces a unique
$k_{v}$-automorphism
$\Phi_{v}:\tilde{\mathbf{G}}(k_{v})\to\tilde{\mathbf{G}}(k_{v})$ for every
finite place $v$ of $k$ and the product
$(\Phi_{v})_{v}:\prod_{v}\tilde{\mathbf{G}}(k_{v})\to\prod_{v}\tilde{\mathbf{G}}(k_{v})$
restricts to an automorphism
$\Phi_{\mathbb{A}}:\tilde{\mathbf{G}}(\mathbb{A}_{k}^{f})\to\tilde{\mathbf{G}}(\mathbb{A}_{k}^{f})$
[14, §5]. Clearly
$\Phi(\sigma^{0}(\gamma))=\Phi_{\mathbb{A}}(\sigma_{\mathbb{A}}^{0}(\gamma))$
for every $\gamma\in\tilde{\mathbf{G}}(k)$ (We identify the group of rational
points $\tilde{\mathbf{G}}(k)$ with its diagonal embedding in the group of
adelic points $\tilde{\mathbf{G}}(\mathbb{A}_{k}^{f})$) and is unique with
this property. We thus get the following corollary:

######  Corollary 2.2.

Under the assumptions of the previous theorem. If
$\varphi:\Gamma_{1}\to\Gamma_{2}$ is an isomorphism. Then there exist unique
automorphisms $\Phi_{\mathbb{A}}$ and $\sigma_{\mathbb{A}}^{0}$ of the adelic
group $\tilde{\mathbf{G}}(\mathbb{A}_{k}^{f})$, such that
$\sigma_{\mathbb{A}}^{0}$ is induced from an automorphism of $k$, and
$\Phi_{\mathbb{A}}$ is induced from a $k$-automorphism of $\tilde{\mathbf{G}}$
with $\varphi(\gamma)=\Phi_{\mathbb{A}}(\sigma_{\mathbb{A}}^{0}(\gamma))$ for
every $\gamma\in\Gamma_{1}$.

We will also need an adelic version of Margulis’ superrigidity stated and
proven by Kammeyer and Kionke [8, Theorem 3.2]:

######  Theorem 2.3.

Let $\mathbf{G}$ be a connected, absolutely almost simple $\mathbb{Q}$-linear
algebraic group of high $\infty$-rank and
$\Gamma\subseteq\mathbf{G}(\mathbb{Q})$ an arithmetic subgroup. If
$\varphi:\Gamma\to\mathbf{G}(\mathbb{A}_{\mathbb{Q}}^{f})$ is a homomorphism
such that $\overline{\varphi(\Gamma)}$ has non-empty interior, then there
exist a homomorphism of adelic groups
$\eta:\mathbf{G}(\mathbb{A}_{\mathbb{Q}}^{f})\to\mathbf{G}(\mathbb{A}_{\mathbb{Q}}^{f})$,
and a group homomorphism
$\nu:\Gamma\to\mathcal{Z}(\mathbf{G})(\mathbb{A}_{\mathbb{Q}}^{f})$ with
finite image such that $\varphi(\gamma)=\nu(\gamma)\eta(\gamma)$ for all
$\gamma\in\Gamma$. Moreover, $\eta$ and $\nu$ are uniquely determined by this
condition.

###  2.1. Profinite groups and the congruence subgroup property

A family $(\\{G_{i}\\}_{i\in I},\\{\phi_{i,j}\\}_{i\geq j\in I})$ is an
inverse system of finite groups over the directed set $I$ if the $G_{i}$’s are
finite groups, $\phi_{i,j}:G_{i}\to G_{j}$ are homomorphisms of groups
whenever $i\geq j$ such that $\phi_{i,k}=\phi_{j,k}\phi_{i,j}$ for every
$i\geq j\geq k$ and $\phi_{ii}=id_{G_{i}}$ $\forall i$. A group $G$ is called
profinite if it is the inverse limit of an inverse system of finite groups
over some directed set. A profinite group is a compact, Hausdorff, totally
disconnected topological group, a map of profinite groups is a continuous
homomorphism of groups.

######  Example 2.4 (Profinite completion).

Let $\Gamma$ be a finitely generated group, let $\mathcal{N}$ be the set of
finite index normal subgroups of $\Gamma$, for $M,N\in\mathcal{N}$, declare
that $M\leq N$ whenever $N\subseteq M$, it is a directed set. Consider the
natural quotient homomorphisms $\phi_{N,M}:\Gamma/N\to\Gamma/M$, then the
profinite group $\widehat{\Gamma}=\varprojlim_{N\in\mathcal{N}}\Gamma/N$, is
called the profinite completion of $\Gamma$.

The profinite completion $\widehat{\Gamma}$ and the set $\mathcal{C}(\Gamma)$
of isomorphism classes of the finite quotients of $\Gamma$ hold the same
information in following manner:

######  Theorem 2.5.

[18, Theorems 3.2.2 & 3.2.7] If $\Gamma$ and $\Lambda$ are two finitely
generated residually finite groups then
$\mathcal{C}(\Gamma)=\mathcal{C}(\Lambda)$ if and only if
$\widehat{\Gamma}\cong\widehat{\Lambda}$

There is a natural map $\iota:\Gamma\to\widehat{\Gamma}$ given by
$\gamma\mapsto(\gamma N)_{N}$, this map is injective if and only if $\Gamma$
is residually finite, in this case we identify $\Gamma$ with its image
$\iota(\Gamma)$. The pair $(\widehat{\Gamma},\iota)$ satisfies a universal
property: $\iota(\Gamma)$ is dense in $\widehat{\Gamma}$, and for every
profinite group $P$, and every homomorphism $\varphi:\Gamma\to P$, there
exists a unique homomorphism of profinite groups
$\hat{\varphi}:\widehat{\Gamma}\to P$ such that
$\widehat{\phi}\circ\iota=\varphi$.

There is a strong connection between the finite index subgroups of
$\widehat{\Gamma}$ and those of $\Gamma$:

######  Proposition 2.6.

[18, Proposition 3.2.2] Let $\Gamma$ be a finitely generated residually finite
group, then there is a one-to-one correspondence between the set $\mathcal{X}$
of all finite index subgroups of $\Gamma$ and the set $\mathcal{Y}$ of all
finite index subgroup of $\widehat{\Gamma}$, given by

$\displaystyle X\mapsto\overline{X},\quad X\in\mathcal{X}$ $\displaystyle
Y\mapsto Y\cap\Gamma,\quad Y\in\mathcal{Y}$

where $\overline{X}$ denote the closure of $X$ in $\widehat{\Gamma}$.
Moreover, this bijection preserves normality, index and quotients.

######  Example 2.7 (Congruence completion).

Let $\Gamma$ be an arithmetic subgroup of $\tilde{\mathbf{G}}(k)$. Consider
the set $\mathcal{C}$ of all congruence subgroups, i.e. subgroups that contain
$\Gamma[\mathcal{I}]:=\Gamma\cap\ker(\phi_{\mathcal{I}}:\tilde{\mathbf{G}}(\mathcal{O}_{k})\to\tilde{\mathbf{G}}(\mathcal{O}_{k}/\mathcal{I}))$
for some ideal $\mathcal{I}\triangleleft\mathcal{O}_{k}$, where
$\phi_{\mathcal{I}}$ is the reduction map $\mod\mathcal{I}$. As in the
profinite completion, $\mathcal{C}$ is a directed set by the inverse of
inclusion, and one can form the Congruence completion $\overline{\Gamma}$ of
$\Gamma$ with respect to this inverse system.

Thus, there is a surjective map $\widehat{\Gamma}\to\overline{\Gamma}$ between
the profinite completion and the congruence completion. Call $C(\Gamma)$, the
kernel of this map, the congruence kernel. The group $\Gamma$ is said to have
the congruence subgroup property if the congruence kernel $C(\Gamma)$ is a
finite group. It is not difficult to see that the congruence subgroup property
is actually a property of the ambient group $\tilde{\mathbf{G}}$ and the field
$k$.
It was conjectured by Serre [19] that if
$\text{rank}_{V_{\infty}(k)}(\tilde{\mathbf{G}})\geq 2$ and
$\Gamma\subseteq\tilde{\mathbf{G}}(k)$ is an arithmetic subgroup then
$C(\Gamma)$ is trivial or isomorphic to a subgroup of the roots of unity of
$k$. The conjecture has been proven in many instances, including for example,
all the isotropic cases [17] and all anisotropic groups of type
$B_{n}$,$C_{n}$,$D_{n}$ (except for some triality forms of $D_{4}$),
$E_{7}$,$E_{8}$,$F_{4}$ and $G_{2}$ ([14, Ch.9], [16]).

###  2.2. A number theoretic lemma

######  Lemma 2.8.

Let $\tilde{\mathbf{G}}$ be a connected, simply connected, absolutely almost
simple $k$-linear algebraic group. There exist infinitely many finite places
$v\in V_{f}(k)$ such that $\tilde{\mathbf{G}}$ splits over $k_{v}$. Moreover,
one can assume that for these places,
$\mathbf{C}(\mathcal{O}_{v})=\mathbf{C}(\mathbb{C})$.

###### Proof.

There exists a finite Galois field extension $k^{\prime}/k$ such that
$\tilde{\mathbf{G}}$ splits over $k^{\prime}$ and
$\mathbf{C}(k^{\prime})=\mathbf{C}(\mathbb{C})$. By Chebotarev’s density
theorem [13, Corollary 13.6], there exist infinitely many primes
$\mathfrak{p}\subseteq\mathcal{O}_{k}$ that splits completely in $k^{\prime}$.
In particular, if such a prime lies under a prime
$\mathfrak{p^{\prime}}\subseteq\mathcal{O}_{k^{\prime}}$ then $k_{v}\cong
k^{\prime}_{v^{\prime}}$ where $v$ and $v^{\prime}$ are the places
corresponding to the primes $\mathfrak{p}$ and $\mathfrak{p^{\prime}}$
respectively. Thus, for such a place $v$, $\tilde{\mathbf{G}}$ splits over
$k_{v}$.
Moreover, as the centre $\mathbf{C}(k^{\prime})$ is finite, for all but
finitely many places,
$\mathbf{C}(k^{\prime})=\mathbf{C}({\mathcal{O}_{k^{\prime},v^{\prime}}})$. So
there exist infinitely many places $v\in V_{f}(k)$ with
$\mathbf{C}(\mathcal{O}_{k,v})=\mathbf{C}(\mathbb{C})$ and
$\tilde{\mathbf{G}}$ splits over $k_{v}$. ∎

##  3\.  First part of the main theorem - Existence

In this section we will prove the first part of theorem 1.1, the existence
part, it will follow from the two theorems below.

######  Theorem 3.1.

Let $\Gamma\subseteq\tilde{\mathbf{G}}(k)$ be an arithmetic subgroup. Assume
that $\tilde{\mathbf{G}}$ has the congruence subgroup property and type
different then $E_{8},F_{4}$ or $G_{2}$. Then there exist two non-isomorphic
finite index subgroups $\Gamma_{1},\Gamma_{2}\subseteq\Gamma$ with isomorphic
profinite completions.

###### Proof.

(Following Method A) Moving to a finite index subgroup, one can assume that
there exists a finite set of primes $S$ and a compact open subgroup
$\Lambda\subseteq\prod_{v\in S}\tilde{\mathbf{G}}(k_{v})$ commensurable with
$\prod_{v\in S}\tilde{\mathbf{G}}(\mathcal{O}_{v})$ such that

(1)
$\displaystyle\widehat{\Gamma}=\overline{\Gamma}\cong\Lambda\times\prod_{v\notin
S}\tilde{\mathbf{G}}(\mathcal{O}_{v}).$

Indeed, the congruence kernel $C(\Gamma)\subseteq\widehat{\Gamma}$ is finite,
so one can find a finite index subgroup $\Gamma^{\prime
P}\subseteq\widehat{\Gamma}$ of the above form, by proposition 2.6, there
exists a finite index subgroup $\Gamma^{\prime}\subseteq\Gamma$ such that
$\widehat{\Gamma^{\prime}}=\Gamma^{\prime P}$. Let
$\pi:\tilde{\mathbf{G}}\to\mathbf{G}$ be the canonical central isogeny to the
adjoint form $\mathbf{G}$, by moving again to a finite index subgroup one can
assume that $\Gamma\cap\mathbf{C}(k)=\\{1\\}$, and $\widehat{\Gamma}$ is still
of the same form as $(1)$.
Let $S_{Q}$ be the set of all rational primes lying under some valuation in
$S$, and $S_{\text{full}}=\\{v:\,v|p\text{ for some }p\in S_{Q}\\}$. By lemma
2.8 one can find two valuations $\mathfrak{p},\mathfrak{q}\notin
S_{\text{full}}$ lying over different rational primes $p$ and $q$ respectively
and such that
$\mathbf{C}(\mathcal{O}_{\mathfrak{p}})=\mathbf{C}(\mathcal{O}_{\mathfrak{q}})=\mathbf{C}(\mathbb{C})$,
and thus also elements
$1\neq\rho_{\mathfrak{p}}\in\mathbf{C}(\mathcal{O}_{\mathfrak{p}})$ and
$1\neq\rho_{\mathfrak{q}}\in\mathbf{C}(\mathcal{O}_{\mathfrak{q}})$ of the
same order. For every $v|p,q$, let $\Delta_{v}$ be a finite index subgroup of
$\tilde{\mathbf{G}}(\mathcal{O}_{v})$ with
$\mathbf{C}(\mathcal{O}_{v})\cap\Delta_{v}=\\{1\\}$. Define
$\epsilon_{\mathfrak{p}}=(\epsilon_{\mathfrak{p},w})_{w},\epsilon_{q}=(\epsilon_{\mathfrak{q},w})_{w}\in\widehat{\Gamma}$
where

$\epsilon_{\mathfrak{p},w}:=\begin{cases}1&\quad\text{if }w\neq\mathfrak{p}\\\
\rho_{\mathfrak{p}}&\quad\text{if
}w=\mathfrak{p}\end{cases},\quad\epsilon_{\mathfrak{q},w}:=\begin{cases}1&\quad\text{if
}w\neq\mathfrak{q}\\\ \rho_{\mathfrak{q}}&\quad\text{if
}w=\mathfrak{q}\end{cases}$

Now we define the following subgroups of $\widehat{\Gamma}$:

$\displaystyle\Delta=\Lambda\times\prod_{v\notin S,\,v\nmid p,\,v\nmid
q}\tilde{\mathbf{G}}(\mathcal{O}_{v})\times\prod_{v|p\text{ or
}v|q}\Delta_{v}$
$\displaystyle\Delta_{1}=\langle\Delta,\epsilon_{\mathfrak{p}}\rangle\cong\Delta\times\langle\epsilon_{\mathfrak{p}}\rangle$
$\displaystyle\Delta_{2}=\langle\Delta,\epsilon_{\mathfrak{q}}\rangle\cong\Delta\times\langle\epsilon_{\mathfrak{q}}\rangle$

Clearly $\Delta_{1}$ and $\Delta_{2}$ are isomorphic finite index subgroups of
$\widehat{\Gamma}$. By proposition 2.6, there exist finite index subgroups
$\Gamma_{i}\subseteq\Gamma$ with $\widehat{\Gamma_{i}}=\Delta_{i}$ for
$i=1,2$. We will finish the proof by showing that $\Gamma_{1}$ and
$\Gamma_{2}$ cannot be isomorphic.
Assume to the contrary that there exists an isomorphism
$\varphi:\Gamma_{1}\to\Gamma_{2}$. By corollary 2.2 there exist unique adelic
automorphisms $\Phi_{\mathbb{A}}$ and $\sigma_{\mathbb{A}}^{0}$ of
$\tilde{\mathbf{G}}(\mathbb{A}_{k}^{f})$ such that $\sigma_{\mathbb{A}}^{0}$
is induced from an automorphism of $k$ and $\Phi_{\mathbb{A}}$ is induced from
a $k$-automorphism of $\tilde{\mathbf{G}}$ such that
$\varphi(\gamma)=\Phi_{\mathbb{A}}(\sigma_{\mathbb{A}}^{0}(\gamma))$ for every
$\gamma\in\Gamma_{1}$, taking closures one deduces that
$(\tilde{\varphi}_{\mathbb{A}}\circ\sigma_{\mathbb{A}})(\widehat{\Gamma_{1}})=\widehat{\Gamma_{2}}$.
On the other hand, the induced map between the profinite completions
$\widehat{\varphi}:\widehat{\Gamma_{1}}\to\widehat{\Gamma_{2}}$ is unique with
$\widehat{\varphi}(\gamma)=\varphi(\gamma)$ for every $\gamma\in\Gamma_{1}$.
Thus, it must be that
$\widehat{\varphi}\equiv(\tilde{\varphi}_{\mathbb{A}}\circ\sigma_{\mathbb{A}})|_{\widehat{\Gamma_{1}}}$.
In particular, it implies that the $\mathfrak{p}$’th place of
$\widehat{\Gamma_{1}}$ is mapped isomorphically onto the
$\sigma(\mathfrak{p})$’th place of $\widehat{\Gamma_{2}}$. This is a
contradiction since $\sigma(\mathfrak{p})|p$, so the $\sigma(\mathfrak{p})$’th
place of $\widehat{\Gamma_{2}}$ is centerless, but the $\mathfrak{p}$’th place
of $\widehat{\Gamma_{1}}$ has a non-trivial centre. ∎

######  Theorem 3.2.

Let $\Gamma\subseteq\tilde{\mathbf{G}}(k)$ be an arithmetic subgroup. Assume
further that $\tilde{\mathbf{G}}(k)$ has the congruence subgroup property and
that $k$ is a number field of dimension $d\geq 2$ over $\mathbb{Q}$. Then
there exist two non-isomorphic finite index subgroups
$\Gamma_{1},\Gamma_{2}\subseteq\Gamma$ with isomorphic profinite completions.

###### Proof.

(Following Method C) As before, by moving to a finite index subgroup we can
assume that $\Gamma\cap\mathbf{C}(k)=\\{1\\}$ and that
$\widehat{\Gamma}=\overline{\Gamma}=\Lambda\times\prod_{v\notin
S}\tilde{\mathbf{G}}(\mathcal{O}_{v})$, for some finite set of places $S$ and
$\Lambda\subseteq\prod_{v\in S}\tilde{\mathbf{G}}(k_{v})$ commensurable with
$\prod_{v\in S}\tilde{\mathbf{G}}(\mathcal{O}_{v})$. Let $S_{Q}$ and
$S_{\text{full}}$ be as before. By Chebotarev’s density theorem there exist
infinitely many rational primes that split completely in $k$ [13, Corollary
13.6], pick two such different primes $p,q\notin S_{Q}$. Say
$\mathfrak{p}_{1},...,\mathfrak{p}_{d}|p$ and
$\mathfrak{q}_{1},...,\mathfrak{q}_{d}|q$. For $i=1,2$, let
$\Gamma_{i}:=\Gamma(\mathfrak{p}_{i}\mathfrak{q}_{1})$ be the principle
congruence subgroups modulo $\mathfrak{p}_{i}\mathfrak{q}_{1}$, then obviously
$\widehat{\Gamma_{1}}\cong\widehat{\Gamma_{2}}$.
By corollary 2.2, if $\Gamma_{1}$ and $\Gamma_{2}$ are isomorphic, then there
exist unique adelic automorphisms $\Phi_{\mathbb{A}}$ and
$\sigma_{\mathbb{A}}^{0}$ of $\tilde{\mathbf{G}}(\mathbb{A}_{k}^{f})$ such
that $\sigma_{\mathbb{A}}^{0}$ is induced from an automorphism of $k$ and
$\Phi_{\mathbb{A}}$ is induced from a $k$-automorphism of $\tilde{\mathbf{G}}$
such that $\varphi(\gamma)=\Phi_{\mathbb{A}}(\sigma_{\mathbb{A}}^{0}(\gamma))$
for every $\gamma\in\Gamma_{1}$, taking closures one deduces that
$(\tilde{\varphi}_{\mathbb{A}}\circ\sigma_{\mathbb{A}})(\widehat{\Gamma_{1}})=\widehat{\Gamma_{2}}$.
As $\sigma_{\mathbb{A}}$ acts by permuting the places and
$\tilde{\varphi}_{\mathbb{A}}$ acts place-wise it must be that
$\sigma_{\mathbb{A}}:\mathfrak{p}_{1}\mapsto\mathfrak{p}_{1}$. We claim that
such $\sigma$ must be trivial, which will finish the proof.
Indeed, let $N$ be the Galois closure of $k$,
$G:=Gal(N/\mathbb{Q}),\,H:=Gal(N/k)$, let $\mathfrak{r}$ be a prime lying over
$\mathfrak{p}_{1}$ and $G_{\mathfrak{r}}$ be its decomposition group. The
correspondence $H\backslash
G/G_{\mathfrak{r}}\to\\{\mathfrak{p}_{1},....,\mathfrak{p}_{d}\\}$ given by
$H\sigma G_{\mathfrak{r}}\mapsto\sigma\mathfrak{p}_{1}$ is a one to one
correspondence (of $G$-sets), hence $G_{\mathfrak{r}}$ must be trivial. Thus,
$Aut(k)$ acts freely on the set of primes
$\\{\mathfrak{p}_{1},...,\mathfrak{p}_{d}\\}$, as needed. ∎

##  4\.  Another set of examples

The following theorem is not needed for the proof of the main theorem, but
following method B, it gives many more examples of non-profinitely rigid
arithmetic groups.

######  Theorem 4.1.

Let $\Gamma\subseteq\tilde{\mathbf{G}}(k)$ be an arithmetic subgroup. Assume
further that $\tilde{\mathbf{G}}$ has: 1) the congruence subgroup property; 2)
type $A_{n},D_{n}$ or $E_{6}$; and 3) there exists an archimedean place
$k_{v}$ such that $\text{rank}_{k_{v}}\tilde{\mathbf{G}}\geq 2$. Then there
exist two non-isomorphic finite index subgroups subgroups
$\Gamma_{1},\Gamma_{2}\subseteq\Gamma$ with isomorphic profinite completions.

Note that $A_{n},D_{n},E_{6}$ are exactly the types of Dynkin diagrams with a
non-trivial symmetry.

###### Proof.

(Following Method B) As before, by moving to a finite index subgroup we can
assume that $\Gamma\cap\mathbf{C}(k)=\\{1\\}$ and that
$\widehat{\Gamma}=\overline{\Gamma}=\Lambda\times\prod_{v\notin
S}\tilde{\mathbf{G}}(\mathcal{O}_{v})$, for some finite set of places $S$ and
$\Lambda\subseteq\prod_{v\in S}\tilde{\mathbf{G}}(k_{v})$ commensurable with
$\prod_{v\in S}\tilde{\mathbf{G}}(\mathcal{O}_{v})$.
By lemma 2.8, one can find two different primes
$\mathfrak{p},\mathfrak{q}\notin S_{\text{full}}$ (where
$S_{\text{full}}\supseteq S$ is as in the previous section) lying over
different rational primes, such that $\tilde{\mathbf{G}}$ splits over both
$k_{\mathfrak{p}}$ and $k_{\mathfrak{q}}$. Fix root systems for
$\tilde{\mathbf{G}}(k_{\mathfrak{p}})$ and
$\tilde{\mathbf{G}}(k_{\mathfrak{q}})$, and let $\Sigma$ be the set of simple
roots, identified for both groups. For $v=\mathfrak{p},\mathfrak{q}$, the
reduction maps
$\pi_{v}:\tilde{\mathbf{G}}(\mathcal{O}_{v})\to\tilde{\mathbf{G}}(\mathcal{O}_{v}/v)$
are onto, and $G_{v}:=\tilde{\mathbf{G}}(\mathcal{O}_{v}/v)$ is the split
universal Chevalley group of the same type as $\tilde{\mathbf{G}}$ over the
residue field. The root systems we fixed define corresponding root systems for
$G_{v}$.
For each of the types involved, there exist a non-trivial symmetry $s$ of the
Dynkin diagram, pick a subset $\theta\subseteq\Sigma$ which is non-invariant
under $s$, e.g.

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The symmetry $s$ of the Dynkin diagram induces an isomorphism $\varphi_{v}$ of
$G_{v}$ [21, Corollary to theorem 29]. For a subset $R\subseteq\Sigma$ of
simple roots, let $P_{v,R}$ be the parabolic subgroup of $G_{v}$ corresponding
to $R$, then $P_{v,\theta}$ and $P_{v,s\theta}$ are non-conjugate in $G_{v}$,
but isomorphic via $\varphi_{v}$.
Now, let $\pi:\Gamma\to G_{\mathfrak{p}}\times G_{\mathfrak{q}}$ be the
canonical reduction map modulo $\mathfrak{p}\mathfrak{q}$. Consider the
following two congruence subgroups

$\displaystyle\Gamma_{1}:=\pi^{-1}(P_{\mathfrak{p},\theta}\times
P_{\mathfrak{q},\theta});$
$\displaystyle\Gamma_{2}:=\pi^{-1}(P_{\mathfrak{p},s\theta}\times
P_{\mathfrak{q},\theta}).$

Then $\widehat{\Gamma_{1}}$ and $\widehat{\Gamma_{2}}$ are isomorphic via
$\Phi:=(\Phi_{v})_{v}$ where $\Phi_{v}\equiv id$ for $v\neq\mathfrak{p}$ and
$\Phi_{\mathfrak{p}}$ is the isomorphism of
$\tilde{\mathbf{G}}(k_{\mathfrak{p}})$ induced by the non-trivial symmetry $s$
of the Dynkin diagram. It remains to show that $\Gamma_{1}$ and $\Gamma_{2}$
cannot be isomorphic.
Assume to the contrary that there exists an isomorphism
$\varphi:\Gamma_{1}\to\Gamma_{2}$. By corollary 2.2, there exist unique adelic
automorphisms $\Phi_{\mathbb{A}}$ and $\sigma_{\mathbb{A}}^{0}$ of
$\tilde{\mathbf{G}}(\mathbb{A}_{k}^{f})$ such that $\sigma_{\mathbb{A}}^{0}$
is induced from an automorphism of $k$ and $\Phi_{\mathbb{A}}$ is induced from
a $k$-automorphism of $\tilde{\mathbf{G}}$ such that
$\varphi(\gamma)=\Phi_{\mathbb{A}}(\sigma_{\mathbb{A}}^{0}(\gamma))$ for every
$\gamma\in\Gamma_{1}$. Moreover $\Phi_{\mathbb{A}}$ is of the form
$C_{g}\cdot\omega_{\mathbb{A}}$ where $C_{g}$ is conjugation by some
$g\in\tilde{\mathbf{G}}(\overline{k})$, $\omega$ is an outer automorphism
which comes from a symmetry of the Dynkin diagram and $\omega_{\mathbb{A}}$
just acts as $\omega$ at each place. In particular
$\pi_{\mathfrak{p}}^{-1}(P_{\mathfrak{p},\theta})$ is mapped onto
$g\pi_{\mathfrak{p}}^{-1}(P_{\sigma(\mathfrak{p}),\omega\theta})g^{-1}$ and
$\pi_{\mathfrak{q}}^{-1}(P_{\mathfrak{q},\theta})$ is mapped onto
$g\pi_{\mathfrak{q}}^{-1}(P_{\sigma(\mathfrak{q}),\omega\theta})g^{-1}$. By
our choice of $\mathfrak{p}$ and $\mathfrak{q}$ it must be that
$\sigma(\mathfrak{p})=\mathfrak{p}$ and $\sigma(\mathfrak{q})=\mathfrak{q}$.
We have that
$g^{-1}\pi_{\mathfrak{p}}^{-1}(P_{\mathfrak{p},\theta})g=\pi_{\mathfrak{p}}^{-1}(P_{\mathfrak{p},\omega\theta})$,
multiplying $g$ from both sides by elements of $\tilde{\mathbf{G}}(k)$, we can
assume that conjugation by $g$ preserves the root system that was fixed in the
beginning of the proof. If $\theta\neq\omega\theta$, let $\alpha$ be a simple
root in $\omega\theta\backslash\theta$, then the action of $g$ on the one-
parameter unipotent subgroup $U_{\alpha}$ must be as scalar multiplication by
some $f_{\alpha}$ with $\text{val}_{\mathfrak{p}}(f_{\alpha})=1$. Thus, the
action on the opposite one-parameter unipotent subgroup $U_{-\alpha}$ is given
as scalar multiplication by $1/f_{\alpha}$, but then
$g^{-1}\pi_{\mathfrak{p}}^{-1}(P_{\mathfrak{p},\theta})g\nsubseteq\pi_{\mathfrak{p}}^{-1}(P_{\mathfrak{p},\omega\theta})$.
Hence $\theta$ must be equal to $\omega\theta$. The same argument imply that
$\omega\theta=s\theta$, and so $\theta=\omega\theta=s\theta$. But $\theta$ was
chosen to be non-invariant under $s$, a contradiction. ∎

##  5\.  Second part of the main theorem - The exceptional cases

######  Lemma 5.1.

Let $\mathbf{G}$ be a connected, simply connected and absolutely almost simple
$\mathbb{Q}$-linear algebraic group of type $E_{8},F_{4}$ or $G_{2}$. Then
$\mathbf{G}$ splits over $\mathbb{Q}_{p}$ for every $p$.

###### Proof.

The $\mathbb{Q}_{p}$ forms of the group $\mathbf{G}$ are classified by the
first Galois cohomology set $H^{1}(\mathbb{Q}_{p},Aut(\mathbf{G}))$. For the
groups considered, the universal and the adjoint forms coincide, moreover
there are no symmetries for their Dynkin diagram, hence
$Aut(\mathbf{G})\cong\mathbf{G}$. As the field $\mathbb{Q}_{p}$ is local and
non-archimedean, by [14, Theorem 6.4], the Galois cohomology group
$H^{1}(\mathbb{Q}_{p},\mathbf{G})$ is trivial. Thus, there is only one (up to
an isomorphism) $\mathbb{Q}_{p}$ form for $\mathbf{G}$, in particular, this
form must be the split form. ∎

######  Remark.

If $\mathbf{G}$ is a $k$-split simple $k$-linear algebraic group, then the
automorphism group, $Aut(\mathbf{G}(k))$, of $\mathbf{G}(k)$ is completely
known. Precisely, following the notations of [21, Theorem 30], each
automorphism can be written as the product of an inner, a diagonal, a graph
and a field automorphism. We have used graph automorphisms for Method B and
field automorphisms for Method C. The group of diagonal automorphisms (modulo
the inner ones) has a connection with the centre of the universal form [21,
Exercise following theorem 30], which was used for Method A. Thus, restriction
ourselves to the exceptional cases, where $\mathbf{G}$ has type $E_{8},F_{4}$
or $G_{2}$ and $k=\mathbb{Q}$ (and also for $\mathbb{Q}_{p}$), all
automorphisms are inner.

######  Theorem 5.2.

Let $\mathbf{G}$ be a connected, simply connected and absolutely almost simple
high $\infty$-rank $\mathbb{Q}$-linear algebraic group of type $E_{8},F_{4}$
or $G_{2}$. If $\Gamma_{1},\Gamma_{2}\subseteq\mathbf{G}(\mathbb{Q})$ are two
arithmetic subgroups with isomorphic profinite completions, then $\Gamma_{1}$
and $\Gamma_{2}$ are isomorphic.

###### Proof.

Let $\Gamma_{1},\Gamma_{2}\subseteq\mathbf{G}(\mathbb{Q})$ be two arithmetic
subgroups. As noted in the preliminaries, the congruence kernel is trivial for
these groups, so one can write
$\overline{\Gamma_{i}}=\widehat{\Gamma_{i}}=\Lambda_{i}\times\prod_{p\notin
S}\mathbf{G}(\mathbb{Z}_{p})$ where $S$ is a finite set and $\Lambda_{i}$ are
commensurable with $\prod_{p\in S}\mathbf{G}(\mathbb{Z}_{p})$. Assume that
$\Phi:\overline{\Gamma_{1}}\cong\widehat{\Gamma_{1}}\xrightarrow{\sim}\widehat{\Gamma_{2}}\cong\overline{\Gamma_{2}}\subseteq\mathbf{G}(\mathbb{A})$
is an isomorphism between the profinite completions of the two. By adelic
supperrigidity 2.3, there exists a unique homomorphism of adelic groups

$\displaystyle\tilde{\Phi}:\mathbf{G}(\mathbb{A}^{f})\to\mathbf{G}(\mathbb{A}^{f})$

such that $\tilde{\Phi}|_{\Gamma_{1}}\equiv\Phi\circ\iota|_{\Gamma_{1}}$, as
$\mathbf{G}(\mathbb{A}^{f})$ is centerless by our assumption on the type of
$\mathbf{G}$. Moreover, using the uniqueness of the map, $\tilde{\Phi}$ must
be an isomorphism, and $\tilde{\Phi}|_{\overline{\Gamma_{1}}}\equiv\Phi$.
Consider the homomorphisms
$\tilde{\Phi}_{p,q}:\mathbf{G}(\mathbb{Q}_{p})\to\mathbf{G}(\mathbb{Q}_{q})$
which are the composite

$\displaystyle\mathbf{G}(\mathbb{Q}_{p})\xrightarrow{\iota_{p}}\mathbf{G}(\mathbb{A}^{f})\xrightarrow{\tilde{\Phi}}\mathbf{G}(\mathbb{A}^{f})\xrightarrow{\pi_{q}}\mathbf{G}(\mathbb{Q}_{q})$

of the inclusion in the $p$’th place, $\tilde{\Phi}$, and the projection onto
the $q$’th place. This is a continuous homomorphism between a $p$-adic group
and a $q$-adic group, so if $p\neq q$ it must be a locally constant map. So
its image is a normal countable subgroup of $\mathbf{G}(\mathbb{Q}_{q})$, in
particular it is not of finite index, and hence must be trivial [14,
Proposition 3.17]. Thus, $\tilde{\Phi}=(\tilde{\Phi}_{p,p})_{p}$ is given by
an isomorphism at each place, and $\tilde{\Phi}_{p,p}$ must be conjugation by
some $y^{\prime}_{p}\in\mathbf{G}(\mathbb{Q}_{p})$ (see the remark above).
We truncate $\tilde{\Phi}$ in the following manner,

$\text{write }y=(y_{p})_{p},\text{ where
}y_{p}=\begin{cases}y_{p}=y^{\prime}_{p}&\quad p\in S\\\ y_{p}=1&\quad p\notin
S\end{cases}.$

By our choice of $S$, conjugation by $y$ is again an isomorphism between
$\overline{\Gamma_{1}}$ and $\overline{\Gamma_{2}}$. By the Strong
Approximation theorem [14, Theorem 7.12], there exists some
$g\in\mathbf{G}(\mathbb{Q})$ with $g\in y\overline{\Gamma_{1}}$. Thus,
conjugation by $g\in\mathbf{G}(\mathbb{Q})$ is an isomorphism between
$\overline{\Gamma_{1}}$ and $\overline{\Gamma_{2}}$. We have that
$\Gamma_{i}=\overline{\Gamma_{i}}\cap\mathbf{G}(\mathbb{Q})$, which imply that
$\Gamma_{1}$ and $\Gamma_{2}$ can be conjugated by $g$, as needed. ∎

##  6\. Final Remarks

It is possible to generalize our methods even further. For example, using
Method A, we can find finite index subgroups of
$\Gamma:=SL_{2}(\mathbb{Z}[1/p])$ which are not profinitely rigid. Explicitly
(for $p\neq 2,3,5$), the following finite index subgroups of
$SL_{2}(\mathbb{Z}[1/p])$ are non-isomorphic, but their profinite completions
are:

$\displaystyle\Gamma_{1}:=\left\\{\begin{pmatrix}a&b\\\ c&d\end{pmatrix}\in
SL_{2}(\mathbb{Z}[1/p]):\,\begin{matrix}b,c\equiv 0\mod 3,5\\\ a,d\equiv\pm
1\mod 3\\\ a,d\equiv 1\mod 5\end{matrix}\right\\};$
$\displaystyle\Gamma_{2}:=\left\\{\begin{pmatrix}a&b\\\ c&d\end{pmatrix}\in
SL_{2}(\mathbb{Z}[1/p]):\,\begin{matrix}b,c\equiv 0\mod 3,5\\\ a,d\equiv 1\mod
3\\\ a,d\equiv\pm 1\mod 5\end{matrix}\right\\}.$

Indeed, just as in §3,
$\widehat{\Gamma_{1}}\cong\widehat{\Gamma(15)}\times\mathbb{Z}/2\mathbb{Z}\cong\widehat{\Gamma_{2}}$,
where $\Gamma(15)$ is the principle congruence subgroup of $\Gamma$ of level
15.
Let us stress out that it is still unknown whether or not
$SL_{2}(\mathbb{Z}[1/p])$ itself is profinitely rigid, and in fact, there are
some reasons to believe it is profinitely rigid (see for example [5, §4]). On
the other hand, increasing slightly the dimension, it has been shown that
$SL_{4}(\mathbb{Z}[1/p])$ is not profinitely rigid [5].
We would like to state a stronger version of Theorem 1.1 which includes the
above example. First, we need some further notations. Let $S\subseteq V(k)$ be
a finite set of places containing all the archimedean places. The ring of
$S$-integers of the number field $k$ is

$\mathcal{O}_{k,S}:=\\{x\in k:\,v(x)\geq 0\,\forall v\notin S\\}$

Let $\tilde{\mathbf{G}}$ be a connected, simply connected, absolutely almost
simple $k$-linear algebraic group with a fixed faithful $k$-representation
$\rho:\tilde{\mathbf{G}}\to GL(n_{\rho})$ . A subgroup
$\Gamma\subseteq\tilde{\mathbf{G}}(k)$ is called an $S$-arithmetic subgroup if
it is commensurable with $\tilde{\mathbf{G}}(\mathcal{O}_{S})$. As in §2,
there is a map from the profinite completion to the congruence completion,
denote its kernel by $C(\Gamma,S)$. The group $\Gamma$ is said to have the
congruence subgroup property (with respect to $S$) if $C(\Gamma,S)$ is a
finite group. Again, this is actually a property of the ambient group
$\tilde{\mathbf{G}}$, the field $k$ and the set $S$. The proofs given
throughout the paper, carry over to establish:

######  Theorem 6.1.

Let $n$ be a positive integer, $k$ a number field, $S$ a finite set of places
of $k$ containing all the archimedean places, $\tilde{\mathbf{G}}$ a
connected, simply connected and absolutely almost simple $k$-linear algebraic
group such that $\sum_{v\in S}\text{rank}_{k_{v}}\tilde{\mathbf{G}}\geq 2$ and
such that $\tilde{\mathbf{G}}(k)$ satisfies the congruence subgroup property
(with respect to $S$). Let $\Gamma\subseteq\tilde{\mathbf{G}}(k)$ be an
$S$-arithmetic subgroup. Then, unless $\tilde{\mathbf{G}}$ has type
$G_{2},F_{4}$ or $E_{8}$ and $k=\mathbb{Q}$, $\Gamma$ has infinitely many
sequences of pairwise non-isomorphic finite index subgroups
$\Gamma_{1},...,\Gamma_{n}$ with isomorphic profinite completions.

As in the main theorem, the exceptional cases are indeed exceptional.
Moreover, in these cases, if $\Gamma_{1}$ is an $S_{1}$-arithmetic subgroup
and $\Gamma_{2}$ is an $S_{2}$-arithmetic subgroup with
$\widehat{\Gamma_{1}}\cong\widehat{\Gamma_{2}}$ then $S_{1}=S_{2}$ and
$\Gamma_{1}\cong\Gamma_{2}$.

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