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Jul 27 at 4:14 comment added LSpice @Maxime, assuming that a reductive group is required to have finitely many components, this follows from the existence of a finite subgroup meeting every component, which can be conjugated into your favourite maximal compact subgroup to show that every extension of a finite group by a connected, reductive complex group can be "realised compactly".
Sep 26, 2013 at 4:49 comment added Maxime Do you happen to know a source for the bijection between (possibly disconnected) complex reductive groups and (possibly disconnected) compact Lie groups?
Oct 18, 2010 at 18:59 history answered Ben Webster CC BY-SA 2.5