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I apologize if this question seems trivial or elementary. Is there any concrete topological space with divisible fundamental group? For example, is there any such a space the fundamental group in which is $(\mathbb{Q}, +)$? I know that from any presentation of a given group $G$, we can construct a 2-dimensional complex $C$ with $\pi_1(C)=G$. What I need is a more geometric object like a manifold or some understandable topological space.

Sh.M1972
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    Certainly to get $\mathbb{Q}$ as a fundamental group requires something a little exotic; for example, anything homotopic to a finite CW complex would never work, as its fundamental group would be finitely generated. – Simon Rose Oct 26 '15 at 08:24
  • See solenoid complements. – Douglas Zare Oct 26 '15 at 08:25
  • This is answered in the comments to the following question: http://mathoverflow.net/questions/192230/is-there-a-manifold-with-fundamental-group-mathbbq – S. Carnahan Oct 26 '15 at 08:49
  • thank you so much for the comments. Now I have enough sources for my work. – Sh.M1972 Oct 26 '15 at 08:57
  • From the description in those comments, I believe that they mean a solenoid complement. I'm having trouble finding the right reference. It doesn't seem to be in Bing's book "The Geometric Topology of $3$-Manifolds." – Douglas Zare Oct 26 '15 at 08:57
  • @Douglas Zare: I have one more question: Let $\Sigma\subseteq \mathbb{R}^3$ be a solenoid with $\pi_1(\Sigma^c)=\mathbb{Q}$. As you said, such a $\Sigma$ exists. Now suppose $x\in \Sigma^c$ is a base point and $p:[0, 1]\to \Sigma^c$ is a simple loop with the initial and terminal point $x$. As $\mathbb{Q}$ is divisible, there exists some loop $q:[0, 1]\to \Sigma^c$ such that $q^2$ is homotopic to $p$. I want to have an imagination of this $q$. – Sh.M1972 Oct 27 '15 at 06:28
  • @M.Shahryari: $p$ must be in some $C_n$, and therefore in some $C_{2t-1}$. That's the complement of a solid torus, so $p$ is a multiple of the meridian in that solid torus complement, $x_{2t-1}^a$. Then there is a multiple of the meridian of the next solid torus $q=x_{2t}^{at}$ so that $q^2=p$. – Douglas Zare Oct 27 '15 at 15:59

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