A closed subset of an algebraic group which contains $e$ and is closed under taking products is a subgroup of $G$

Question

A closed subset of an algebraic group which contains $e$ and is closed under taking products is a subgroup of $G$.

Denote this set as $X$. If the condition of $X$ being closed is dropped, this statement does not hold. The set of nonzero integers in $\mathbb{G}_m$ over $\mathbb{C}$ is a counterexample.

It suffice to prove that for any $x \in X$, $x^{-1}X = X$. As $X$ is closed under taking products, it is clear that $X \subseteq x^{-1}X$. In order to prove the inverse inclusion, the closedness of $X$ (under Zariski topology) must be used.

Let $\phi: G \rightarrow G, y \mapsto x^{-1}y$ is a homoemorphism of $G$ as an algebraic variety. So $x^{-1}X = \phi(X)$ is a closed subset of $G$ containing $X$. But Why are they equal?

Thanks very much.

Answer 1

As David Loeffler points out, the (attempted) argument with connectedness is invalid. However, if $x \in X$, then $X \supset x X \supset x^2 X \supset \cdots \supset x^n X \supset \cdots$ is a decreasing chain of closed subsets of $G$, which by Noetherian induction must eventually stabilize. Thus in fact $X = x X$, as desired.

Answer 2

Consider a fixed $x\in X$ and the multiplication morphism $m_x:X\to X:y\mapsto xy$. It suffices to prove that $m_x$ is surjective, since then there will exist $y\in X$ with $xy=e$ and so $y=x^{-1}$ will belong to $X$.
And now watch out for the conclusion:

The endomorphism $m_x$ is injective, hence surjective because of the Theorem of Ax-Grothendieck (or here )