# Zariski dense subgroup of algebraic group [closed]

Let $G$ be a Zariski dense subgroup of algebraic group $H.$ Then is $H$ the Zariski closure of $G$?

Conversely, if $G$ is a subgroup of an algebraic group $H$ and if $H$ is the Zariski closure of $G,$ then is $G$ Zariski dense in $H$?

What are the equivalent conditions of being a Zariski dense subgroup of an algebraic group?

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## closed as off-topic by Venkataramana, Felipe Voloch, Peter Humphries, Daniel Loughran, Tom De MedtsJul 14 at 7:30

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• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Venkataramana, Felipe Voloch, Peter Humphries, Daniel Loughran, Tom De Medts
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Either I don't understand the question, or this is just a trivial consequence of the definition of "dense". – Martin Brandenburg Sep 27 '10 at 15:45
It seems like an elementary exercise in general topology, and it has nothing to do with Zariski topology or algebraic groups. – Qfwfq Sep 27 '10 at 15:49

I am also not sure, whether I understand the question properly; but it seems to me that one should distinguish carefully between the group scheme and its geometric points here.

To fix ideas, let $K$ be an algebraically closed field and $G/K$ an algebraic group. An algebraic subgroup of $G$ is by definition a closed subgroup scheme of $G$. Furthermore there is the abstract group $G(K)$, and we have the Zariski topology on $G(K)$. There is a bijection $H\mapsto H(K)$ from the set of smooth algebraic subgroups of $G$ to the set of closed subgroups of $G(K)$. (Cf. Milne's lecture notes on algebraic groups for example.)

If $\Gamma$ is a subgroup of $G(K)$, then there is a unique smooth algebraic subgroup $\Gamma^{zar}$ of $G$, such that $\Gamma^{zar}(K)=\overline{\Gamma}$. Some people call the algebraic group $\Gamma^{zar}$ the Zariski closure of $\Gamma$.

Assume now that $G$ is smooth. With the above terminology we see:

a) If $\Gamma$ is Zariski dense in $G(K)$, then $\Gamma^{zar}=G$.

b) If $H$ is a smooth algebraic subgroup of $G$ and $H(K)$ is Zariski dense in $G(K)$, then $H(K)=G(K)$ and consequently $H=G$.

And one cannot drop the smoothness assumption on $G$ here.

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