If $H\subset{G}$ is a topologically closed subgroup of a compact, connected and semisimple Lie group G, then is $H$ also semisimple? If yes, I need some references where this is stated.
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1Let $H$ be a Cartan subgroup of $G$. Then $H$ is abelian. – Dietrich Burde Mar 12 '17 at 15:58
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Yes, by definition. But, how is this related to my question? Or is it too obvious for you and you are just fooling around? – user3257624 Mar 12 '17 at 16:02
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I thought that abelian Lie groups are not semisimple? Why do you say fooling? – Dietrich Burde Mar 12 '17 at 16:03
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No. In fact every compact semisimple Lie group has at least one non-semisimple closed subgroup: a maximal torus. For a specific example, take the diagonal subgroup of $\mathrm{SU}(2)$.
(A torus is not semisimple since it is abelian, and hence it's Lie algebra has non-trivial solvable ideals.)
Spenser
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