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The following quote is from An Introduction to Lie Group and Lie Algebra by Alexander Kirillov. By "discrete" he means that $|G/G^0|$ is finite. However, according to Introduction to Smooth Manifolds by J. Lee, a topological manifold has countable (not finite in general) components. So, I'd like to know what condition makes lie group to have a finite components.

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    $\begingroup$ It isn't. ${}{}$ $\endgroup$ – Mariano Suárez-Álvarez Oct 4 '15 at 3:20
  • $\begingroup$ For example, one wants that a closed subgroup of a Lie group itself, but that is false if we insisted in that the number of components be finite, as the group of integers inside the additive group of real number shows. $\endgroup$ – Mariano Suárez-Álvarez Oct 4 '15 at 3:23
  • $\begingroup$ Thanks for your answer! That solved my question. $\endgroup$ – Math.StackExchange Oct 4 '15 at 3:24

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