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I saw this in a paper that I've been reading and I've been trying to figure out if this is true or not.

Let $G(F)$ be a affine, simple, connected, adjoint, algebraic group over a local field endowed with the topology induced by the local field. Suppose that $H$ is a Zariski-dense subgroup of $G(F)$ that is open with respect to this topology. Is $H$ necessarily finite index in $G$?

Can someone provide a hint?

Edit: The local field is non-archimedean.

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You say it's connected - then $H$ should be the same as $G$, since the union of all other cosets of $H$ form an open set disjoint from $H$. –  user27126 Jan 31 '13 at 5:32
    
$G$ is Zariski-connected, but it's definitely not connected with the topology induced by the absolute value on $F$. –  Josh Schwartz Jan 31 '13 at 14:34
    
Hint: Take the most obvious example of $G$ and $F$ you can think of, and then think what $H$ might be. –  David Loeffler Jan 31 '13 at 15:03
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The simplest example I can think of is $G = PSL_2(\mathbb{Q}_p)$ with $H = PSL_2(\mathbb{Z}_p)$. This seems like $H$ is not finite index in $G$, but my intuition with these groups is shaky at best. –  Josh Schwartz Jan 31 '13 at 16:22
    
Now I feel foolish, of course $H$ must be infinite index in $G$. Looking at a matrix realization for $PSL_2(\mathbb{Q}_p)$, the element $\mathrm{diag}(p^2,1,p^{-2})$ is an element of infinite order such that no power is in $PSL_2(\mathbb{Z}_p)$. Hence, the index cannot be finite. This is unfortunate because it means there's a big hole in this paper. Thank you for all your help! –  Josh Schwartz Jan 31 '13 at 18:11
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