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Def. Let $G$ be an algebraic group and let $P \subset G$ be a closed subgroup. Then $P$ is said to be parabolic if $G/P$ is complete.

Apparently, we can prove

Lemma. $P \subset G$ is parabolic $\Leftrightarrow$ $(P \cap G^0) \subset G^0$ is parabolic.

so that we needn't assume $G$ to be connected in the definition of "parabolic".

However, I seem to be missing one or another insight. I know that $G^0$ is a subgroup of finite index, and I suspect that $G^0/(P\cap G^0)$ is an irreducible component of $G/P$. Could someone give a hint for how to give a rigorous proof?

(One can draw a diagram like the one used in the nine lemma, but strictly speaking this only holds in an Abelian category or in the category of groups. Of course $0 \to P \to G \to G/P \to 0$ isn't an exact sequence in either category.)

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    I'm familiar with a result that says that $P$ is parabolic in $G$ if and only if $P^0$ is parabolic in $G^0$. But I am not sure whether it is always the case that $(P \cap G^0) = P^0$. – 2017-02-25

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