TITLE: Why is the fundamental group of a compact Riemann surface not free ?
QUESTION [148 upvotes]: Consider a compact Riemann surface $X$ of genus $g$.
It is well-known that its fundamental group $\pi_1(X)$ is the free group  on the generators $a_1,b_1,...,a_g,b_g$ divided out by the normal subgroup generated by  the single relator $[a_1,b_1]\cdot \ldots\cdot [a_g,b_g]$.
(This has of course nothing to do with the complex structure of $X$, but may be  computed by considering the underlying topological manifold as a cell complex.)
This group is trivial for $g=0$ and free abelian on two generators for $g=1$. 
For $g\geq 2$, however, I had always taken for granted  that it is not free  but I have just realized that I cannot prove that. 
So, although I guess   the answer is no, I'll ask my official question in an open way : Is   $\pi_1(X)$ free for $g\geq 2$ ? 
Edit: Users have now brilliantly solved the problem in multiple ways.
Non-freeness is definitely established, with 12 proofs plus sketches of proofs in the comments!
It is clearly impossible to select in a reasonable way an answer for "acceptance" among all these great answers .
Since the software forces me to make only one choice, I have chosen Daniel's answer because of its merit, but also because it acknowledges Vitali's  contribution: Vitali was  the first to sketch a solution (in the comments) .

REPLY [3 votes]: Free groups are residually finite.
Finitely generated and residually finite implies Hopfian.

So $F\{a_1,b_1,...,a_g,b_g\}$ is Hopfian so it is not isomorphic to the quotient group by any nontrivial normal subgroup, hence
$$F\{a_1,b_1,...,a_g,b_g\}\not\cong \pi_1(X)=\dfrac{F\{a_1,b_1,...,a_g,b_g\}}{\langle[a_1,b_1]\cdot...\cdot[a_g,b_g]\rangle}\quad (1)$$ 
But taking the epimorphism $\pi_1(X)\to\pi_1(X)_{ab}\cong \mathbb{Z}^{2g}$ we see that $\pi_1(X)$ has least set of generators of size $2g$ and from $(1)$ it isn't isomorphic to the only free group in $2g$ generators.