Which algebraic subvarieties of a group variety have a group structure? 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!
 A: 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! :)
