ag.algebraic geometry - Subgroup Groups and Coordinate Algebra Subalgebras

It may be possible in rare circumstances, but the natural thing is that $mathcal{O}(H)$ is a quotient of $mathcal{O}(G)$, not a subalgebra. The quotient map is dual to the inclusion $Hto G$.

I guess in the case that $G$ is a direct product $H times K$, you get

$$mathcal{O}(G) simeq mathcal{O}(H) otimes mathcal{O}(K),$$
and in this case you have what you want. But in general, the arrow should go the other direction.

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