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, $\varphi(H)\subseteq H$.


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.


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:

Please help :)


1 Answer 1


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.

  • $\begingroup$ Thank you! I guess I was overthinking it. $\endgroup$
    – Shaun
    Jul 9, 2023 at 4:54

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