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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 $g\geq 2$ ?

Conclusion

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) .

5 added 31 characters in body

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 $g\geq 2$ ?

Conclusion
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 allows me to acknowledge acknowledges Vitali's contributionin his comment, which : Vitali was the first to sketch a solution to (in the questioncomments) .

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 $g\geq 2$ ?

Conclusion
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 .
I have chosen Daniel's answer because of its merit, but also because it allows me to acknowledge Vitali's contribution in his comment, which was the first to sketch a solution to the question.

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