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We've been learning about fundamental groups and various properties of $SU(2)$, and I want to improve my understanding by working on interesting problems I've come across. I have the following that I need help with:

Let $G$ be a finite subgroup of the group $SU(2)$. Why does there exist a compact, connected 3-dimensional manifold $M$ such that its fundamental group is isomorphic to $G$?

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Ask yourself what is $SU(2)$ as a topological space. –  Scott Carter May 23 '12 at 3:58
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up vote 8 down vote accepted

Let $G$ be a finite subgroup of $SU(2)$. Then $G$ acts by left translation on $SU(2)$ freely, so the quotient $SU(2)/G$ is a manifold. Moreover, since $SU(2)$ is simply connected, the map $SU(2)\to SU(2)/G$ exhibits $SU(2)$ as the universal covering space of $SU(2)/G$. In particular, this implies that $G$ is isomorphic to the fundamental group of $SU(2)/G$.

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FGITW: nicely done. +1 –  mixedmath May 23 '12 at 3:56
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