This is a question based on Exercise 7.6.5 of Humphreys', "Linear Algebraic Groups". For a solution to that particular problem, see: A closed subset of an algebraic group which contains $e$ and is closed under taking products is a subgroup of $G$
The Question:
Is a closed, nonempty subset $H$ of a (linear)${}^\dagger$ algebraic group $G$ that is closed under taking products a subgroup of $G$?
The Details:
The first "closed" is closed under the Zariski topology; the second, closed as the binary operation of $G$ restricted to $H$. The "subgroup" is as a (linear) algebraic group.
Thoughts:
My guess is that, yes, $H$ would be a subgroup, in the same way that the two-step subgroup test works for abstract groups:
Theorem (Two Step Subgroup Test): Let $S$ be a nonempty subset of a group $T$. Then if:
- For all $x,y\in S$, $xy\in S$, and
- For all $x\in S$, $x^{-1}\in S$,
then $S\le T$.
My guess is that 2), in the setting of (linear) algebraic groups, is somehow replaced by Zariski closure.
One way to go about proving this is to analyse a proof of the inspiring exercise to see where $e$ is used, and then adjust where possible.
I have tried that with the proof here, the same as linked to above. But $e$ is not mentioned.
On the other hand, my intuition is trained so far mostly on abstract groups. So I could be wrong.
Further Context:
I am studying for a postgraduate research degree in linear algebraic groups.
Please help :)
$\dagger$: Humphreys states that, whenever "algebraic group" is mentioned in his book, he means, "linear algebraic group"; but I'm not sure if linearity is needed here . . .