Let $G$ be an affine group scheme over a characteristic zero field $k$ (the case I have in mind is $k=\mathbb{Q}$). Let $H$ be a subgroup of $G$ which satisfies $\mathrm{Lie}(H)=\mathrm{Lie}(G)$. Is then $H$ an open subgroup of $G$? I found this sort of implication in a literature, but I could not find the most precise statement in this direction. I appreciate to someone who is introducing a correct statement with a proof.
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4$\begingroup$ Do you assume $G$ is (at least locally) of finite type over $k$? If not then the assertion is false (consider the affine group scheme associated to an infinite torsion abelian group, for which the Lie algebra vanishes but $H=0$ is not open). If so then this is an application of dimension considerations due to Cartier's theorem according to which all group schemes (locally) of finite type over a field of characteristic 0 are smooth. It is harmless for the proof to extend scalars to $\overline{k}$ so that $k$ is algebraically closed (in case that is assumed in some references). Try Google. $\endgroup$ – nfdc23 Dec 21 '16 at 18:50

$\begingroup$ @nfdc23 I am really sorry, but could you elaborate a bit more? If I understood correctly, what I get from the equality $\mathrm{Lie}(H)=\mathrm{Lie}(G)$ is $\dim(H)=\dim(G)$. Does this imply the openess? $\endgroup$ – User0829 Dec 22 '16 at 13:42

2$\begingroup$ I will assume $G$ and $H$ are locally of finite type over $k$ (you still have not clarified your hypotheses). Then $G^0$ and $H^0$ are finite type, and a monic homomorphism between group schemes of finite type over a field is always a closed immersion (were you not aware of this?). Thus, $H^0$ is a closed subscheme of $G^0$, and by Cartier + Lie algebra equality their dimensions coincide, so they are equal. This gives the result (using $k$point translations with $k = \overline{k}$ to handle the rest). $\endgroup$ – nfdc23 Dec 23 '16 at 16:57
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Here is a related statement: let $G$ be an algebraic affine group scheme over a field. Assume $G$ is smooth and connected and $H$ a smooth subgroup. If $\mathrm{Lie}(H)=\mathrm{Lie}(G)$, then $H=G$.