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Let $X$ be a subset of $\mathbb R^3$ with its induced topology and let $a\in X$ be a point. Then the fundamental group $\pi_1(X,a)$ seems not to have elements of finite order (except the identity of course). Does anybody know an easy argument why this is the case?

The answer needs to use the fact that the ambient space is 3 dimensional since there are spaces with fundamental group of order two which can be embedded in $\mathbb R^4$.

Thanks for your help!

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    $\begingroup$ Apparently it's an open problem (or at least it was in 2013). $\endgroup$
    – Najib Idrissi
    Commented Apr 1, 2015 at 12:53
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    $\begingroup$ @ Francesco Polizzi: I did not assume that X is not simply connected although for simply connected X the statement that it does not contain elements of finite order (other than the identity) is not very interesting. $\endgroup$
    – Tom
    Commented Apr 1, 2015 at 12:59
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    $\begingroup$ mathoverflow.net/questions/4478/… $\endgroup$
    – Thomas Rot
    Commented Apr 1, 2015 at 12:59
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    $\begingroup$ @Najib Idrissi: Thanks for the link, I did not know that it was an open problem. I thought someone told me some years ago that this was a theorem but now I came across this question again and so I wanted to know how one proves it. $\endgroup$
    – Tom
    Commented Apr 1, 2015 at 13:00
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    $\begingroup$ If $X \subset \mathbb{R}^3$ is open and connected, then $\pi_1(X)$ is torsion free, see math.stackexchange.com/questions/680998/… $\endgroup$
    – Francesco Polizzi
    Commented Apr 1, 2015 at 14:05

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