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I know by Van Kampen's Theorem that we can obtain $\pi_1(S_1 \vee S_1) = \mathbb{Z} * \mathbb{Z}$, so I am wondering if we can construct a surface or 3-manifold whose fundamental group is $\mathbb{Z}_n * \mathbb{Z}_2$ or even $\mathbb{Z}_m \rtimes \mathbb{Z}_n$.

This might be an interesting problem because I have written semidirect product $\rtimes$ rather than the free product $*$. A torus knot $K$ is defined in Hatcher as the image of an embedding of a map $f : S^1 \to S^1 \times S^1 \to \mathbb{R}^3 \subset S^3$ given by $z \mapsto (z^m, z^n)$ then the fundamental group $\pi_1(\mathbb{R}^3 - K)$ is $Z_m \ast Z_n$ possibly up to some number-theoretic conditions. Hatcher doesn't quite give you the answer.

I think the semidirect product $\mathbb{Z}_m \rtimes \mathbb{Z}_n$ is unique. We have to specify $\mathbb{Z}_m \lhd G$ and then $G = \mathbb{Z}_m \ltimes \mathbb{Z}_n$.

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    $\begingroup$ This looks like a homework. Voted to close. $\endgroup$
    – user6976
    Commented Jul 28, 2018 at 15:34
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    $\begingroup$ @MarkSapir I've been out of school 5 years. $\endgroup$
    – john mangual
    Commented Jul 28, 2018 at 15:35
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    $\begingroup$ Regarding uniqueness, there is always the direct product $\mathbb Z/m\mathbb Z \times \mathbb Z/n\mathbb Z$, and sometimes that's the only one (if $m$ does not divide $\varphi(n)$). If we do have a non-Abelian semi-direct product, then it's unique for $n$ square-free at least, and I think in general. $\endgroup$
    – LSpice
    Commented Jul 28, 2018 at 15:35
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    $\begingroup$ Also, since the semi-direct product is a quotient of the amalgamated product, you should be able just to take a suitable cover of your $\mathbb R^3 \setminus K$, no? $\endgroup$
    – LSpice
    Commented Jul 28, 2018 at 15:36
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    $\begingroup$ In the reference you cite it is shown that $\pi_1({\mathbb R}^3-K)$ is the group generated by two elements $a$ and $b$ subject to the relation $a^m=b^n$. This group is torsionfree (as is true for all knots, not just torus knots) and it has ${\mathbb Z}_m*{\mathbb Z}_n$ as the quotient group when the center, which is the infinite cyclic group generated by the element $a^m=b^n$, is factored out. $\endgroup$
    – Allen Hatcher
    Commented Jul 28, 2018 at 22:21

3 Answers 3

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See pp. 449--457 of Peter Scott's article The geometries of 3-manifolds for a complete description of all 3-manifolds with finite fundamental group. The article is available on his website. There don't seem to be any with dihedral fundamental groups (see Allen Hatcher's comment below), but the fundamental groups of the prism manifolds are the binary dihedral groups, i.e. non-split central extensions of $D_{2n}\cong \mathbb{Z}/n\rtimes\mathbb{Z}/2$ by $\mathbb{Z}/2$.

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    $\begingroup$ The fundamental groups of prism manifolds are not the dihedral groups themselves but the {\it binary\/} dihedral groups $D^*_{2n}$ which are the preimages of the subgroups $D_{2n}\subset SO(3)$ under the projection $S^3\to SO(3)$. Thus $D^*_{2n}$ maps onto $D_{2n}$ with a ${\mathbb Z}/2$ kernel. $\endgroup$
    – Allen Hatcher
    Commented Jul 29, 2018 at 20:37
  • $\begingroup$ @AllenHatcher: Then I must have misunderstood the nomenclature. What should one call the quotients of the 3-sphere by the free dihedral actions that Scott describes on page 451 of his article? $\endgroup$
    – HJRW
    Commented Jul 30, 2018 at 8:58
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    $\begingroup$ This seems to be a mistake in Scott's paper. He says that certain subgroups of the unit quaternion group $S^3$ are either cyclic, dihedral, or generalized quaternion (also known as binary dihedral). From this he deduces in particular that dihedral groups act freely on $S^3$. However, $S^3$ has a unique element of order $2$, the quaternion $-1$, so $S^3$ cannot contain dihedral subgroups. On Scott's webpage there is a list of errata for this paper but it does not include this one. $\endgroup$
    – Allen Hatcher
    Commented Jul 30, 2018 at 21:29
  • $\begingroup$ @AllenHatcher: thanks for the useful clarification. I’ll correct the answer shortly. $\endgroup$
    – HJRW
    Commented Jul 31, 2018 at 10:23
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A connected sum of the appropriate lens space and $\mathbb{R}P^3$ will have fundamental group $\mathbb{Z}_m \ast \mathbb{Z}_2.$ Otherwise, the only abelian fundamental groups of $3$-manifolds are $\mathbb{Z},$ $\mathbb{Z}^3$ and $\mathbb{Z}/n \mathbb{Z}$ - see Stefan Friedl's notes (introduction to 3-manifolds and their fundamental group), so that rules out interesting direct products.

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This is an alternative to HJRW's answer, which also will rely on Perelman's affirmative resolution to the Poincare conjecture. In this case, we are using using it to say that 3-manifolds with finite fundamental group are covered by $S^3$. The problem can be reduced to this because a classification of 2-manifolds shows the only 2-manifold with finite order elements in $\pi_1$ is $RP^2$ and $\pi_1(RP^2) \cong \mathbb{Z}/2\mathbb{Z}$.

Scott's paper tells us that any manifold with finite fundamental group admits a homomorphism with non-trivial kernel onto the orientation subgroup of a spherical triangle group.

However, Thurston's book (see below) has a classification of Elliptic 3-manifolds (those covered by $S^3$) and your problem can be solved using two statements which are rather self-contained.

Exercise 4.4.3 says that the only order 2 element of O(n+1) that acts freely on $S^n$ is the antipodal map. So for you, $n$ must be odd in $\mathbb{Z}/n\mathbb{Z} \rtimes \mathbb{Z}/2\mathbb{Z}$. Then Proposition 4.4.4 shows that the only admissible semi-direct product in this case is in fact the direct product, since the antipodal map is just -Id in $O(4)$ which commutes with everything.

Thurston, William P., Three-dimensional geometry and topology. Vol. 1. Ed. by Silvio Levy, Princeton Mathematical Series. 35. Princeton, NJ: Princeton University Press. x, 311 p. (1997). ZBL0873.57001.

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