Which algebraic subvarieties of a group variety have a group structure?

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Let $G$ be an algebraic group. Given an algebraic subvariety $X\subseteq G$, is there any way simple criterion to determine whether $X$ has a group structure or not?

For example, when $X$ and $G$ are affine, then if $\mathcal{A}(G)$ and $\mathcal{A}(X)$ are the coordinate rings of $G$ and $X$, we know that $G$ being a group variety is equivalent to $\mathcal{A}(G)$ having a Hopf algebra structure. Since $\mathcal{A}(X)$ is a quotient of $\mathcal{A}(G)$, doesn't it immediately imply that $\mathcal{A}(X)$ also have a Hopf algebra structure?

An obvious immediate obstruction to $X$ having a group structure is if $X$ does not contain the identity. I assume this should somehow also be data one could read from the Hopf algebra structure.

Thanks in advance!

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As in the comment by LT1918, the issue is that not all ideals $I$ in a hopf algebra $A$ are Hopf ideals, namely the comultiplication on $A$, $\Delta : A \to A\otimes_k A$ ($k$ being the ground field), need not induce comultiplication on the quotient $$ \overline{\Delta} : A/I \to A/I \otimes_k A/I; $$ for instance, setting $G=\mathbb{G_a}/k$ so $A=k[X], \Delta:X \mapsto X\otimes 1 + 1\otimes X$, implies that for example the ideal $I= (X(X-1))\subseteq k[X]$ isn't Hopf, since the composition $$ \Delta : k[X] \to k[X]\otimes_k k[X] \xrightarrow{\pi\otimes_k\pi} k[X]/I \otimes_k k[X]/I $$ doesn't factor through $\pi$, as $$ \Delta(X(X-1)) = X^2\otimes 1 + 1\otimes X^2 +2X\otimes X - 1\otimes 1 - X\otimes 1 = (X^2 - X)\otimes 1 + 1\otimes (X^2 - 1) + 2X\otimes X \neq 0 \text{ in }k[X]/I\otimes k[X]/I. $$ This is clear because I chose an ideal which doesn't define a subgroup of $\mathbb{G}_a$ of course. The wiki article has more than what you'd need on Hopf ideals I think.

Unless I'm mistaken, I think your question is on possible group structures of a subvariety which aren't necessarily induced by that of $G$, which I think is about as general as asking which varieties are algebraic groups... I don't know whether this has a satisfying answer tbh.

Sorry for the hazy answer and I hope this helps! :)