Let G and H be affine algebraic groups defined over a field k of characteristic zero, with H a closed subgroup of G. Suppose they have the same k-points. Have they to be equal?
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$\begingroup$ Surely you want to require it be connected, or else examples from your other question will work as counterexamples. $\endgroup$– Charles SiegelCommented Jul 15, 2010 at 15:49
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No: Take $k$ to be the rational numbers and $G$ to be the group of third roots of unity. Then the only rational point in $G$ is $1$. Then take $H$ to be the component of the identity. This satisfies your conditions but $H \neq G$.
The problem here is that the groups are no connected.
edit: I notice now after posting my answer that David Speyer suggested the same example in a similar question...
answered Jul 15, 2010 at 15:52
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$\begingroup$ If the group is connected the answer is yes? $\endgroup$– AnaCommented Jul 15, 2010 at 17:16
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5$\begingroup$ Ana, as I explained in my comment to your previous question, in characteristic zero the rational points are always Zariski-dense in a connected linear algebraic group. (This is not obvious, and requires some serious input from the structure theory of such groups, both the reductive case and the general case.) Hence, the answer to your question is affirmative. $\endgroup$– BCnrdCommented Jul 15, 2010 at 17:29