Suppose $G$ is a reductive group that is not necessarily connected and $Z \subset G$ is a central subgroup. Suppose $G^0$ is the identity component of $G$. Is it true that $G/G^0Z= \pi_0(G/Z)$? I can see there is a surjection $G \twoheadrightarrow \pi_0(G/Z)$ which induces a surjection $G/G^0Z \twoheadrightarrow \pi_0(G/Z)$ but I don't see why this has to be injective.
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2$\begingroup$ $G/G^0Z$ is a finite group and every homomorphism from $G$ to a finite group that vanishes on $Z$ is trivial on $G^0$ and on $Z$ and hence factors through $G/G^0Z$. Hence $G/G^0Z$ is the initial object in the category of such quotients of $G$, and hence you have an isomorphism. This works for an arbitrary algebraic group $G$ and normal subgroup $Z$. (I'm working in the absolute world, i.e. over algebraically closed fields. If your question concerns rationality aspects, it should be restated.) $\endgroup$– YCorCommented Feb 17, 2019 at 22:52
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$\begingroup$ This is very elegant. I wish I had thought of it ;). Thanks for explaining the argument. $\endgroup$– AlexanderCommented Feb 18, 2019 at 6:28
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$\begingroup$ @Alexander: It would help to motivate your question if you gave some typical examples of non-connected reductive groups. $\endgroup$– Jim HumphreysCommented Feb 26, 2019 at 21:59
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