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$D_4$ here indicates the dihedral group of order 8. Does there exist any trivial example of topological spaces such that it has $D_4$ as it's fundamental group.

ChakSayantan
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    https://math.stackexchange.com/questions/36775/is-the-group-of-rational-numbers-the-fundamental-group-of-some-space – Eric Wofsey Mar 04 '18 at 20:29
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    What exactly are you looking for? Certainly $\pi_1 K(D_4, 1) = D_4$, and any finitely presented group is the fundamental group of some closed $4$-manifold (via killing the relations in $\pi_1((S^1 \times S^3)^{# n})$ with surgery). I don't know offhand of any of the canonical examples that have $D_4$ as its fundamental group, though. – anomaly Mar 04 '18 at 20:42
  • In addition to the methods described above, take the standard presentation of $D_4$: $<x,y| x^2=y^2=(xy)^4=1>$ and let $X$ be the corresponding "presentation complex", obtained by attaching three 2-dimensional cells to the bouquet of circles according to the above relators. Then $\pi_1(X)\cong D_4$. This method, of course, works for any group with the given presentation. Various versions of this question were asked many times at MSE. VTC as a duplicate. – Moishe Kohan Mar 05 '18 at 13:36
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    Possible duplicate of Every Group is a Fundamental Group See also: https://math.stackexchange.com/questions/2093715/a-topological-space-x-who-can-be-drawn-whose-fundamental-group-is-s-3?rq=1, – Moishe Kohan Mar 05 '18 at 13:39
  • In terms of manifolds I don't think we can go below dimension $4$. If there was a closed $3$-manifold with this fundamental group, then the universal cover would be $S^{3}$, by Perelman's theorem. We see from the list on page 13 of https://arxiv.org/pdf/1205.0202.pdf that $D_{4}$ is not the fundamental group of such a manifold. So I guess the best thing to hope for (in the world of manifolds) is a $4$-dimensional manifold with a particularly simple description. – Nick L Mar 05 '18 at 18:04

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