Let $G=C_{19}\rtimes C_9=\langle a,b\ |\ a^{19}=b^9=1, a^b=a^7\rangle$ be a nonabelian group of order $171$. Is there a (compact) 3-manifold $M$ with $\pi_1(M)\cong G$?
Thanks for any help!
Let $G=C_{19}\rtimes C_9=\langle a,b\ |\ a^{19}=b^9=1, a^b=a^7\rangle$ be a nonabelian group of order $171$. Is there a (compact) 3-manifold $M$ with $\pi_1(M)\cong G$?
Thanks for any help!
If $\pi_1(M) = G$, then the universal cover of $M$ is a compact (because $G$ is finite) simply connected $3$-manifold, hence (by Poincare) a $3$-sphere, on which $G$ acts. The spherical space form conjecture then implies that $G$ is a group of isometeries of $S^3$, i.e. that $G$ is a subgroup of $O(4)$. (See also this wikipedia entry.) Since $G$ has odd order, in fact $G$ is a subgroup of $SO(4)$, or of $PSO(4)$. I don't think that your group $G$ embeds into $PSO(4)$ (use the isomorphism $PSO(4) \cong PSO(3)\times PSO(3)$ and the known list of finite subgroups of $PSO(3)$), and so I think the answer is no.
(I don't know enough about the area to know if this kind of question can be answered without appealing to the full machinery of Poincare/geometrization.)