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It is known that the fundamental group of a compact semisimple Lie group is finite. Can this fact be generalized to compact orbits of semisimple Lie groups? i.e. Let $G$ be a semisimple Lie group such that $G/H$ is compact where $H$ is a closed subgroup. Then is it true that $\pi_1(G/H)$ is finite?

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I must admit that I didn't know that the fundamental group of a compact semisimple Lie group is finite, but anyhow, the answer to your question is - no.

This is an immediate corollary of the fact (Borel) that every semisimple Lie group $G$ admits a cocompact lattice (i.e a discrete subgroup $\Gamma$ such that $G/\Gamma$ is compact). $\Gamma$ is closed (it's discrete), and also infinite (unless $G$ is compact). Furthermore $\Gamma = \pi_1(G/\Gamma)$ since $\Gamma$ acts on $G$ freely and properly discontinuously.

Yet why not keep it more concrete. Take the hyperbolic plain $X=\mathbb{H}^2$. It admits an octogonal regular tilling with degree 4 vertices (a fact which can be found in any textbook on hyperbolic geometry). The surface of genus 2: $S_2$ can be obtained by gluing the edges of the regular octagon in a particular way, and thus it inherits a hyperbolic Riemannian metric for which $X$ is the universal cover, and each tile is a fundamental domain. The fundamental group $\Gamma$ of $S_2$ acts freely and properly discontinuously on its universal cover $X$ by Deck isometrics. Thus $\Gamma$ is a discrete subgroup of the group of isometries of $X$ which is just $SL(2,\mathbb{R})$.

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