Why is the fundamental group of a compact manifold finitely presented?
[Math] Fundamental group of a compact manifold
at.algebraic-topologyfundamental-groupsmooth-manifolds
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A couple of extra points.
Any compact 3-manifold with boundary $M$ can be doubled to give a closed 3-manifold $D$. 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 Baumslag--Solitar groups that Autumn mentions above) applies just as well to compact 3-manifolds as closed 3-manifolds.
Other classes of poison subgroups can be constructed from cohomological conditions. The Kneser--Milnor Theorem implies that any closed, irreducible 3-manifold 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 3-manifold (and hence of a compact 3-manifold, by the previous paragraph).
EDIT:
Oh, and yet another source of poison subgroups comes from Scott's theorem that 3-manifold 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.
No, the answer is negative in general (if you require $M$ to be compact). $M$ comes with a map $M \to BG$ that is, by definition, $n+1$-connected (iso on $\pi_i$ for $i=0,...,n$, epi on $\pi_{n+1}$). You can turn it into a weak equivalence by attaching cells of dimension $\geq n+1$. From that you see, that there is a model for $BG$ having finite $n$-skeleton. This is a special property of a group that is called $F_n$ (for more information, see http://berstein.wordpress.com/2011/03/16/morse-theory-finiteness-properties-and-bieri-stallings-groups/). Finitely presented groups are $F_2$ and you find that a necessary condition on your $G$ is that it is of type $F_n$. The are concrete examples of groups that are $F_i$ but not $F_{i+1}$ for each $i$, which are discussed in same blog post (on page 423 in Hatcher's AT, you find the same examples in a slightly different context).
On the other hand, let $G$ be $F_n$ and let $K$ be the $n$-skeleton of $BG$; a finite complex. Then I claim there is a closed manifold $M$ with the desired properties. $M$ can be chosen of arbitrary dimension $d \geq 4,2n+1$ and to be stably parallelizable. Start with a sphere $S^d \to K$ and do surgery on $S^d$ to get rid of the homotopy groups in low dimensions. The precise formulation is for example Proposition 4 in Kreck's paper "Surgery and duality".
So we can say that a necessary and sufficient condition is that $G$ is of type $F_n$. Caveat: I might have confused $n$ and $n+1$ at various places.
If you want to have $dim M \leq 2n$, you meet a new obstruction enforced by Poincare duality and things become really difficult.
Best Answer
Every compact manifold has the homotopy type of a finite CW-complex, and a finite CW-complex has finitely presented fundamental group by van Kampen's theorem.