I don't think that the following is known, but before going to other things, I would like to know what can be said about it. Thanks in advance for any relevant comment !

So here is the situation. Let $k'\subset k$ be a finite field extension. Take affine algebraic group schemes defined over $k'$ together with a central isogeny $\pi : \tilde{G}\rightarrow G$ (i.e. $\pi $ is surjective on underlying topological space, and is a finite flat morphism such that $ker$ $\pi\subset Z(G)$).

Given a subgroup $\Gamma \leq \tilde{G}(k)$, assume that $\Gamma _{down}=\pi_{k}(\Gamma )$ lies in fact already in $G(k')$. Can we conclude that $\pi_{k'}^{-1}(\Gamma_{down})$ is big enough compared to $\Gamma $ (more precisely, that its intersection with $\Gamma $ is of finite index in $\Gamma $ ?)

Ok, now more precision about assumption (but I would be interested to have counter-examples if not in that situation, especially for the first assumption) :

1. $k$ is a local field, and $k'$ a closed subfield
2. $\tilde{G}$ (resp. $G$) is absolutey simple simply connected (resp. adjoint)

Also, let me stress that I am mainly interetsed in the positive characteristic case, i.e. $k$ a finite extension of $\mathbb{F}_{p}$((T)), and that I do not assume that the extension is Galois (but I would be interested to know what Galois cohomology can bring to the matter, I'm not at all familiar with that theory).

EDIT (prompted by the answer of user 76758)

I'm very happy to have forgotten to mention the crucial assumption on $\Gamma $ since I got this very illuminating answers without it. But now, here it is : $\Gamma $ is in fact assumed to be an open compact subgroup of $\tilde{G}(k)$ (in the topology given by the local field $k$).

Also, the purpose of all that was to prove the following claim : if $\Gamma $ is an open compact subgroup of $\tilde{G}(k)$ avoiding the center (as above, $\tilde{G}$ absolutely simple simply connected), then any closed normal subgroup of $\Gamma $ is of finite index (I wonder whether this is a well-known fact or not, by the way).

Using a paper of Pink, this reduces either to prove the finite index condition mentionned above, or alternatively that the commutator of $\Gamma$ is open. I've seen some litterature about the oppeness of $[\Gamma , \Gamma ]$, but I have not really had the time to see whether that is completely settled or not.

I don't think that the following is known, but before going to other things, I would like to know what can be said about it. Thanks in advance for any relevant comment !

So here is the situation. Let $k'\subset k$ be a finite field extension. Take affine algebraic group schemes defined over $k'$ together with a central isogeny $\pi : \tilde{G}\rightarrow G$ (i.e. $\pi $ is surjective on underlying topological space, and is a finite flat morphism such that $ker$ $\pi\subset Z(G)$).

Given a subgroup $\Gamma \leq \tilde{G}(k)$, assume that $\Gamma _{down}=\pi_{k}(\Gamma )$ lies in fact already in $G(k')$. Can we conclude that $\pi_{k'}^{-1}(\Gamma_{down})$ is big enough compared to $\Gamma $ (more precisely, that its intersection with $\Gamma $ is of finite index in $\Gamma $ ?)

Ok, now more precision about assumption (but I would be interested to have counter-examples if not in that situation, especially for the first assumption) :

1. $k$ is a local field, and $k'$ a closed subfield
2. $\tilde{G}$ (resp. $G$) is absolutey simple simply connected (resp. adjoint)

Also, let me stress that I am mainly interetsed in the positive characteristic case, i.e. $k$ a finite extension of $\mathbb{F}_{p}$((T)), and that I do not assume that the extension is Galois (but I would be interested to know what Galois cohomology can bring to the matter, I'm not at all familiar with that theory).

EDIT (prompted by the answer of user 76758)

I'm very happy to have forgotten to mention the crucial assumption on $\Gamma $ since I got this very illuminating answers without it. But now, here it is : $\Gamma $ is in fact assumed to be an open compact subgroup of $\tilde{G}(k)$ (in the topology given by the local field $k$).

Also, the purpose of all that was to prove the following claim : if $\Gamma $ is an open compact subgroup of $\tilde{G}(k)$ avoiding the center (as above, $\tilde{G}$ absolutely simple simply connected), then any closed normal subgroup of $\Gamma $ is of finite index. I finally found a Paper of Carl Riehm, "The congruence subgroup problem over local fields", which completely settles the question.

But still, I would be interesting to know what happens in the above situation when $\Gamma $ is open compact (and if the finite index fact is true in this case, do we need the compactness assumption ?).

I don't think that the following is known, but before going to other things, I would like to know what can be said about it. Thanks in advance for any relevant comment !

So here is the situation. Let $k'\subset k$ be a finite field extension. Take affine algebraic group schemes defined over $k'$ together with a central isogeny $\pi : \tilde{G}\rightarrow G$ (i.e. $\pi $ is surjective on underlying topological space, and is a finite flat morphism such that $ker$ $\pi\subset Z(G)$).

Given a subgroup $\Gamma \leq \tilde{G}(k)$, assume that $\Gamma _{down}=\pi_{k}(\Gamma )$ lies in fact already in $G(k')$. Can we conclude that $\pi_{k'}^{-1}(\Gamma_{down})$ is big enough compared to $\Gamma $ (more precisely, that its intersection with $\Gamma $ is of finite index in $\Gamma $ ?)

Ok, now more precision about assumption (but I would be interested to have counter-examples if not in that situation, especially for the first assumption) :

1. $k$ is a local field, and $k'$ a closed subfield
2. $\tilde{G}$ (resp. $G$) is absolutey simple simply connected (resp. adjoint)

Also, let me stress that I am mainly interetsed in the positive characteristic case, i.e. $k$ a finite extension of $\mathbb{F}_{p}$((T)), and that I do not assume that the extension is Galois (but I would be interested to know what Galois cohomology can bring to the matter, I'm not at all familiar with that theory).

I don't think that the following is known, but before going to other things, I would like to know what can be said about it. Thanks in advance for any relevant comment !

So here is the situation. Let $k'\subset k$ be a finite field extension. Take affine algebraic group schemes defined over $k'$ together with a central isogeny $\pi : \tilde{G}\rightarrow G$ (i.e. $\pi $ is surjective on underlying topological space, and is a finite flat morphism such that $ker$ $\pi\subset Z(G)$).

Given a subgroup $\Gamma \leq \tilde{G}(k)$, assume that $\Gamma _{down}=\pi_{k}(\Gamma )$ lies in fact already in $G(k')$. Can we conclude that $\pi_{k'}^{-1}(\Gamma_{down})$ is big enough compared to $\Gamma $ (more precisely, that its intersection with $\Gamma $ is of finite index in $\Gamma $ ?)

Ok, now more precision about assumption (but I would be interested to have counter-examples if not in that situation, especially for the first assumption) :

1. $k$ is a local field, and $k'$ a closed subfield
2. $\tilde{G}$ (resp. $G$) is absolutey simple simply connected (resp. adjoint)

Also, let me stress that I am mainly interetsed in the positive characteristic case, i.e. $k$ a finite extension of $\mathbb{F}_{p}$((T)), and that I do not assume that the extension is Galois (but I would be interested to know what Galois cohomology can bring to the matter, I'm not at all familiar with that theory).

EDIT (prompted by the answer of user 76758)

I'm very happy to have forgotten to mention the crucial assumption on $\Gamma $ since I got this very illuminating answers without it. But now, here it is : $\Gamma $ is in fact assumed to be an open compact subgroup of $\tilde{G}(k)$ (in the topology given by the local field $k$).

Also, the purpose of all that was to prove the following claim : if $\Gamma $ is an open compact subgroup of $\tilde{G}(k)$ avoiding the center (as above, $\tilde{G}$ absolutely simple simply connected), then any closed normal subgroup of $\Gamma $ is of finite index (I wonder whether this is a well-known fact or not, by the way).

Using a paper of Pink, this reduces either to prove the finite index condition mentionned above, or alternatively that the commutator of $\Gamma$ is open. I've seen some litterature about the oppeness of $[\Gamma , \Gamma ]$, but I have not really had the time to see whether that is completely settled or not.