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Is it true that the class of isomorphism classes of fundamental groups of Lie groups coincides with the class of isomorphism classes of finitely generated abelian groups?

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  • $\begingroup$ The fundamental groups of Lie groups are all abelian. So if you restrict to compact ones, they are all finitely generated abelian groups. Not sure what your question is asking about.... $\endgroup$ – Bombyx mori Mar 9 '18 at 19:36
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    $\begingroup$ I know that they are abelian, but my question is about their presentations... $\endgroup$ – RRR Mar 9 '18 at 19:38
  • $\begingroup$ For a finitely generated group, both "the number of generators" and "the number of relations" are not defined, simply because "generators" and "relations" are not defined. You maybe want to define something related to presentations. Such as the existence of presentations with certain properties. $\endgroup$ – YCor Mar 9 '18 at 19:38
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    $\begingroup$ Next, since the class of isomorphism classes of $\pi_1$ of Lie groups coincides with the class of isomorphism classes of finitely generated abelian groups, there is no point to incorporate Lie groups in the question, which (once more carefully stated) is a question about group presentations of finitely generated abelian groups. $\endgroup$ – YCor Mar 9 '18 at 19:41
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    $\begingroup$ @YCor Your answer provides examples that I am thinking about, but actually your answer answers much more general question. $\endgroup$ – RRR Mar 9 '18 at 19:52
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Yes, these two classes coincide. One direction splits into two: for a Lie group $G$,

  1. the fundamental group is abelian: this is already answered here this MO question, and
  2. the fundamental group is finitely generated. This is equivalent (passing to the universal covering of the unit component) to the property that every discrete central subgroup of a connected Lie group is finitely generated, and this is answered here.

Conversely any finitely generated abelian group is fundamental group of a Lie group. By the classification of finitely generated abelian groups, it is enough to consider cyclic groups (and then use products). Indeed $\mathbf{Z}$ is isomorphic to the fundamental group of the circle group $\mathbf{R}/\mathbf{Z}$, and $\mathbf{Z}/n\mathbf{Z}$ is isomorphic to the fundamental group of $\mathrm{PSL}_n(\mathbf{C})$.

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