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) . 



Why is the fundamental group of a compact Riemann surface not free ?
As per Theo's request, I'm posting this as an answer, though it's largely an expansion on Vitali's comment. Let $F\_n$ be the free group on $n$ letters; $K(F\_n, 1)$ is a wedge of $n$ circles and so has vanishing cohomology in degrees $>1$. On the other hand, if $X$ is a compact Riemann surface of genus $g>1$, $X$ is a $K(\pi\_1(X), 1)$ as its universal cover is the upper-half plane, which is contractible. But then $$H^2(\pi\_1(X), \mathbb{Z})=H^2\_{sing}(X, \mathbb{Z})=\mathbb{Z},$$ which is non-zero. In particular, $\pi\_1(X)$ has non-vanishing cohomology in degree $2$ and so is not free, as Vitali says.


