Is it true that any finitely presented group can be realized as fundamental group of compact 3manifold with boundary?

15$\begingroup$ There's no need to apologize for a stupid question, and this isn't one. $\endgroup$ – Tom Church Nov 19 '09 at 17:17
A couple of extra points.
Any compact 3manifold with boundary $M$ can be doubled to give a closed 3manifold $D$. As $M$ is a retract of $D$, it follows that $\pi_1(M)$ injects into $\pi_1(D)$. Therefore, any "poison subgroup" (such as the BaumslagSolitar groups that Richard mentions above) applies just as well to compact 3manifolds as closed 3manifolds.
Other classes of poison subgroups can be constructed from cohomological conditions. The KneserMilnor Theorem implies that any closed, irreducible 3manifold with infinite fundamental group is aspherical. It follows that any freely indecomposable infinite group with cohomologial dimension greater than 3 cannot be a subgroup of a closed 3manifold (and hence of a compact 3manifold, by the previous paragraph).
EDIT:
Oh, and yet another source of poison subgroups comes from Scott's theorem that 3manifold groups are coherent, meaning that every finitely generated subgroup is finitely presented. This rules out subgroups like $F\times F$ (where $F$ is a free group), which is not coherent.

1$\begingroup$ Do you know if there is an algorithm to decide if a group of 3mnfldwithbry is trivial? $\endgroup$ – Anton Petrunin Nov 20 '09 at 18:15

2$\begingroup$ There is an algorithm to determine if $\pi_1(M)$ is trivial for compact $M$: If the boundary has a component that is not a sphere, then $M$ will have nontrivial homology. If the boundary is a union of spheres, then cap them off with balls to get a closed manifold $N$. Then, by the Poincare conjecture, which we now know, the question is whether or not $N$ is the threesphere. You can use Rubinstein's algorithm to recognize the threesphere to do this. $\endgroup$ – Autumn Kent Nov 20 '09 at 20:25

$\begingroup$ A less intelligent but more simpleminded algorithm is just to apply geometrization and deduce that the word problem is (uniformly) solvable in 3manifold groups. If you know how to solve the word problem then it's easy to tell if a group is trivial. $\endgroup$ – HJRW Nov 21 '09 at 17:51

$\begingroup$ Could you explain a little bit how $M$ turns out to be a retract of its double $D$? $\endgroup$ – Maharana Dec 27 '09 at 7:05

$\begingroup$ Maharana, sure. D = M<sub>1</sub> U M<sub>2</sub>, where the M<sub>i</sub> are copies of M. Now the identifications M<sub>i</sub>>M agree on their boundaries, so extend to a map D>M. This is the retraction. $\endgroup$ – HJRW Dec 27 '09 at 15:57
No. The Baumslag solitar groups $\langle a, b  ab^m a^{1} = b^n \rangle$ are not $3$manifold groups when $m \neq n$.
See
Heil, Wolfgang H. Some finitely presented non$3$manifold groups. Proc. Amer. Math. Soc. 53 (1975), no. 2, 497500.
(See also Peter Shalen, ThreeManifolds and BaumslagSolitar groups. Topology Appl. 110 (2001), 113118)

$\begingroup$ Are you sure that manifolds there can haave boundary? $\endgroup$ – Anton Petrunin Nov 19 '09 at 17:22

1$\begingroup$ Yes. These groups are never the fundamental group of any 3manifold. $\endgroup$ – Autumn Kent Nov 19 '09 at 17:27
I recently heard of a result due to Aitchison and Reeves which shows that any finitely presented group arises as the fundamental group of a 3dimensional orbifold (where fundamental group means the topological and not the orbifold fundamental group). In fact, they say that the orbifold can be taken to be the quotient of a closed oriented hyperbolic 3manifold by an isometric involution with isolated fixed points, all modelled on $x\mapsto x$.
(I'm certainly no expert on this topic, just passing on what I heard.)

3$\begingroup$ I asked Aitchison, he only can make it to be fundamental group of $M^3/\mathbb Z_2$ where $M^3$ is closed orientable 3manifold and $\mathbb Z_2$ acts on $M$ with isolated fixed points. The question if $M$ can be made hyperbolic and $\mathbb Z_2$ action isometric is not yet resolved. $\endgroup$ – Anton Petrunin Mar 22 '11 at 2:36

1$\begingroup$ Ah, that's good to know  I must have misunderstood, although I thought I was quite insistent on knowing about the hyperbolic case. Still, that's quite a while ago now, probably I've misremembered things. $\endgroup$ – Joel Fine Mar 24 '11 at 19:18