TITLE: Fundamental group of 3-manifold with boundary
QUESTION [24 upvotes]: Is it true that any finitely presented group can be realized as fundamental group of compact 3-manifold with boundary?

REPLY [3 votes]: If a finite group $G$ is (isomorphic to) the fundamental group of a three-manifold, then $G$ embeds in $\mathrm{SO}(4)$.
To see this, suppose that $M$ is a compact, connected three-manifold, possibly with boundary, having $G = \pi_1(M)$ finite. If $M$ has two-sphere boundary components, we can cap them off without changing the fundamental group.  Since free products of non-trivial groups are always infinite, we deduce that $G$ is freely indecomposible.  Appealing to the Poincaré conjecture [solved by Perelman], we have that $M$ is irreducible: all two-spheres in $M$ bound three-balls.
If $M$ is non-orientable, then a theorem of Livesay implies that $M$ is homeomorphic to the real projective plane crossed with a unit interval.
Thus we reduce to the case where $M$ is compact, oriented, connected, irreducible, and has finite fundamental group.  All boundary components of $M$ are now oriented, and of genus at least one.  Applying "one-half lives, one-half dies" we find that $M$ has no boundary components.  Appealing to the elliptic part of the geometrisation conjecture [solved by Perelman] we find that the universal cover of $M$ is the three-sphere, and the deck group, and thus $G$, is (conjugate to a) subgroup of $\mathrm{SO}(4)$. QED
A quick google search finds a paper of Zimmermann giving a readable introduction to the finite subgroups of $\mathrm{SO}(4)$ - see section three of that paper.  One then has to determine which of these act freely.  Finally, there is another approach to this problem via spherical Seifert fibered spaces.
As a concrete example of a finite group that is not a three-manifold group, consider $\mathbb{Z}_p \times \mathbb{Z}_q$ where $p$ and $q$ are not coprime.  This is a isomorphic to a subgroup of $\mathrm{SO}(4)$, but it cannot act freely.  [See Theorem 9.14 in Hempel's book for a proof that, in this special case, avoids geometrisation.]