This is discussed in the standard textbooks on algebraic topology.
Pick a presentation of the group $G = \langle g_1,g_2,…,g_n|r_1,r_2,…r_m \rangle$
where $g_i$ are generators and $r_j$ are relations. Then we have a wedge of $n$ circles and attach two-cells to the wedge sum according to the relations $r_j$. Denote the final space $X$. Then van Kampen says $\pi_1(X)=G$.
While usually $X$ is not a manifold, it is well-known that every finitely generated group $G$ can be realized as the fundamental group of some 4-manifold $X$. Can someone sketch the proof? Also, if $X$ could not
be some manifold of dimension $<4$, what is the obstruction?
[Math] finite generated group realized as fundamental group of manifolds
at.algebraic-topologygt.geometric-topology
Best Answer
Theorem. Every finitely presentable group is the fundamental group of a closed 4-manifold.
Sketch proof. Let $\langle a_1,\ldots,a_m\mid r_1,\ldots, r_n\rangle$ be a presentation. By van Kampen, the connected sum of $m$ copies of $S^1\times S^3$ has fundamental group isomorphic to the free group on $a_1,\ldots, a_m$. Now we can quotient by each relation $r_j$ as follows. Realise $r_j$ as a simple loop. A tubular neighbourhood of this looks like $S^1\times D^3$. Do surgery and replace this tubular neighbourhood with $S^2\times D^2$. This kills $r_j$. QED
There are many restrictions on 3-manifold groups. One of the simplest arises from the existence of Heegaard splittings. It follows easily that if $M$ is a closed 3-manifold then $\pi_1(M)$ has a balanced presentation, meaning that $n\leq m$.
Other obstructions to being a 3-manifold group were discussed in this MO question.