# Show that the identity component $G^\circ\sqsubseteq G$ (i.e., is characteristic) in a linear algebraic group $G$.

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:

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.