Openness of finite index subgroups of $\mathrm{GL}_n(\prod O_v)$ Let $K$ be a global field and set $O := \prod_{v\nmid \infty} O_v$ where $v$ runs over the finite places of $K$. Equip $\mathrm{GL}_n(O) = \prod_v \mathrm{GL}_n(O_v)$ with the product of the $v$-adic topologies and suppose that $H \le \mathrm{GL}_n(O)$ is a subgroup of finite index. 


*

*Is $H$ necessarily open? 

*Is $H$ necessarily a congruence subgroup? (I.e., does it contain $\mathrm{Ker}(\mathrm{GL}_n(O) \rightarrow \mathrm{GL}_n(O/NO))$ for some integer $N$?) 


A positive answer to 2. would imply a positive answer to 1. 
I will accept a complete answer to either 1. or 2. as an answer. Also, I would be especially happy if you discussed a generalization of 1. and 2. for arbitrary reductive group schemes (or even a more general class of group schemes) over the ring of integers of $K$.
 A: The answer to 1 is no. For example each $GL_n(O_v)$ has a continuous map to $\{ \pm 1\}$ so your product has a quotient isomorphic to $F_2^{\mathbf{N}}$. Zorn's lemma will give you a non continuous linear form on that, and the pullback of the quotient is a non closed subgroup of index $2$. 
Edit: the pullback of the kernel, not of the "quotient"
A: In view of Laurent's answer, one may ask more generally for a criterion on a connected reductive $K$-group $G$ to ensure that for all compact open subgroups $U$ of $G(\mathbf{A}_K^S)$ (for a finite set $S$ of places of $K$ containing the arch. places) the finite-index subgroups of $U$ are necessarily open.  If $G$ admits a non-trivial torus quotient then one typically gets an obstruction as in Laurent's answer.  So assume $G$ is semisimple.  Let $\widetilde{G} \rightarrow G$ be the simply connected central cover, with kernel $\mu$.  If $\mu \ne 1$ then elementary "spreading out" arguments and Lang's theorem (on vanishing of degree-1 cohomology of smooth connected groups over finite fields) show that $U$ again admits a massive commutative quotient (related to Galois cohomology of $\mu$ over dvr's, roughly speaking), so problems arise again.
But what if $G$ is simply connected (such as ${\rm{SL}}_n$, not ${\rm{GL}}_n$)?  Then there's no "obvious" commutative obstruction, and at least in characteristic 0 one might hope there is no obstruction.  It comes down to a question like the following:  is an index-N normal subgroup of $\prod_{p\nmid N} {\rm{SL}}_6(\mathbf{F}_p)$ necessarily open?  (It is easy to show that such a subgroup contains ${\rm{SL}}_6(\mathbf{F}_p)$ for all but finitely many $p$, since such groups are generated by $\mathbf{F}_p$-points of unipotent radicals, a method which generalizes to all simply connected semisimple groups over finite fields of size at least 4, but that doesn't give access to the direct product.)
