It is a well-known result that, for any finitely-presented group $G$ and any integer $n \geq 4$, there exists a closed $n$-manifold whose fundamental group is isomorphic to $G$ (a sketch of proof can be found here). It is also well-known that such a result becomes false when $n \leq 3$. However, I wonder whether there exists an "elementary" way to notice that in dimension three. That is,
Question: Does there exist a "simple" argument proving that not all finitely-presented groups are fundamental groups of a closed 3-manifold?
Of course, "simple" is quite vague. Typically, I would like to avoid evolved cohomological arguments: a geometric proof would be perfect.
Every closed oriented 3-manifold admits a Heegaard splitting, from which it follows that its fundamental group admits a presentation with $g$ generators and $g$ relations for some $g$. (To get a corresponding condition for the not-necessarily-oriented case look at the orientation double cover.) Such a presentation is called balanced, and most groups don't admit a balanced presentation; for example, it turns out that $\mathbb{Z}^n$ admits a balanced presentation iff $n \le 3$, and in fact only $n = 0, 1, 3$ occur as $3$-manifold groups, corresponding to $S^3$, $S^1 \times S^2$ and $S^1 \times S^1 \times S^1$ respectively.
More generally you can classify all finitely generated abelian groups which admit balanced presentations and, among those, classify which ones appear as $3$-manifold groups; this is Proposition 4.7 in this survey.
Lots of things are known about $3$-manifold groups in general; see, for example, this survey.