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This is an elementary question that probably admits an elementary counterexample, but ...

Let $G$ be a compact Lie group and $G_0$ its identity component. One then has a short exact sequence $$ 1 \to G_0 \to G \to \pi_0(G) \to 1; $$ the last object is the component group.

Does the component group embed as a subgroup?

jdc
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    See http://math.stackexchange.com/questions/941066/lie-group-decomposition-as-semidirect-product-of-connected-and-discrete-groups – Hanno Jan 23 '15 at 08:29
  • I should have thought to search for "semidirect." Thanks. – jdc Jan 23 '15 at 08:45

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