This is Exercise 7.6.6 of Humphreys', "Linear Algebraic Groups".
The Question:
Show that the identity component $G^\circ$ of a linear algebraic group $G$ is a characteristic subgroup.
The Details:
The identity component is the unique irreducible component of $G$ of the identity $e$.
A subgroup $H$ of $G$ is characteristic if it is stable under all automorphisms $\varphi\in{\rm Aut}(G)$; that is, $\varphi(H)\subseteq H$.
Thoughts:
I think we could make use of the following in Humphreys (paraphrased).
Proposition 7.3.1: Let $G$ be a linear algebraic group. Then $G^\circ$ is a normal subgroup of finite index in $G$, whose cosets are the connected as well as irreducible components of $G$.
It takes care of showing $G^\circ$ is a subgroup.
One idea I have is to hit $G^\circ$ with an arbitrary $\varphi\in{\rm Aut}(G)$. The normality might come into play but I think the main thing to exploit would be irreducibility.
Context:
For questions of mine involving irreducibility in the context of linear algebraic groups, see here; you will find the definition of irreducible I am familiar with in those. The best such question for our current purposes is: The components of a Noetherian space are its maximal irreducible closed subsets.
Two questions of mine on characteristic subgroups of abstract groups are:
- An abelian, characteristically simple group is divisible (supposedly).
- Let $G$ be f.g. with $H\le G$ s.t. $[G:H]<\infty$. Then $\exists K\le H$ with $K\sqsubseteq G$ and $[G:K]<\infty$.
Please help 🙂
Best Answer
If you already know $G^\circ\subset G$ is the unique irreducible component containing the identity $e$, then this is immediate.
Indeed, for any automorphism $\varphi$ of $G$, we have $\varphi(G^\circ)$ is also an irreducible component containing the identity $e$, so $\varphi(G^\circ)=G^\circ$ by uniqueness.