Find a surface or 3-manifold whose fundamental group is $(\mathbb{Z}/n\mathbb{Z}) \rtimes (\mathbb{Z}/2\mathbb{Z})$ 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 $G = \mathbb{Z}_m \ltimes \mathbb{Z}_n$.
 A: 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 $\mathbb{Z}/2$.
A: 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.
A: 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.
