Why is the fundamental group of a compact Riemann surface not free ? 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) . 
 A: I just wanted to point out that Vitali's comment, and Daniel Litt's elaboration, can be explained without any hyperbolic geometry.  As they point out, once we know that an orientable surface $M^g$ of genus $g>1$ has contractible universal cover, the desired result follow by computing $H_2 (M^g)$ (or $H^2 (M^g)$).  So, here are 3 other proofs that the universal cover $X$ of $M^g$ is contractible:


*

*$X$ is a 2-dimensional manifold, and it is non-compact because the fiber of the universal covering is the fundamental group, which is infinite (it has infinite abelianization).  Any  non-compact n-manifold has $H^i (M) = 0$ for $i>n-1$ (this is Proposition 3.29 in Hatcher).  So in our case, $H_i (X) = 0$ for $i>1$ and since $X$ is simply connected, $H_1 (X)=0$ also.  By the Hurewicz Theorem, all the homotopy groups of $X$ must be trivial as well.  Since $M^g$ is a CW complex, so is its universal cover, so Whitehead's theorem says $X$ is contractible.

*Hatcher Example B.14 proves this by describing $M^g$ in terms of a graph of groups in which the maps on the edges are injective.

*Topologically, the only non-compact simply connected surface is $R^2$ by the classification of (non-compact) surfaces.
A: By Baer's (I believe) theorem, the mapping class group of a closed surface is actually the outer automorphism group of the fundamental group. On the other hand, the action of this group on $H_1(S, \mathbb{Z}))$ is symplectic (preserves the symplectic form, and the image under the Torelli map is $Sp(2g, \mathbb{Z})).$ On the other hand, the action of the outer automorphism group of $F_{2g}$ on its abelianization is the whole $SL(2g, \mathbb{Z})$ of which $Sp(2g, \mathbb{Z})$ is a proper subgroup. So, since $Out(\pi_1(S_{g})) \neq Out(F_{2g})$ the groups themselves are not isomorphic.
A: You can use covering spaces instead of cohomology. 
(Edit: This proof does not use a presentation of the group, either -- just the weaker fact that its abelianization is free of rank $2g$.)
As Daniel pointed out in his comment, if the group were free it would have to be free on $2g$ generators. The genus $g$ surface has a $2$-sheeted covering space which is a genus $2g-1$ surface. Every index $2$ subgroup of a free group on $r$ generators is free on $2r-1$ generators, because it is the fundamental group of a $2$-sheeted covering space of a wedge of $r$ circles. So $2(2g-1)=2(2g)-1$, a contradiction.
A: 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.
A: There are many good answers already, but maybe you'll be interested in a counting argument.
Let $Q_8$ denote the quaternion group.  Consider the set
$$\mbox{Hom}(\pi_1(X),Q_8)$$
where $X$ is the closed surface of genus $g$.  Since we have a presentation for $\pi_1(X)$, we note that a homomorphism to $Q_8$ is given by a $2g$-tuple of elements in $Q_8$ satisfying some group word.  In particular, the cardinality of the set is finite.  Indeed, if the fundamental group were free, the cardinality would be a power of $8$.  However, combinatorics (or gentle representation theory) gives the cardinality of the set exactly:
$$|\mbox{Hom}(\pi_1(X),Q_8)| = 2^{6g-1}+2^{4g-1}$$
which (for $g>0$) is never a power of $2$, still less a power of $8$.
---Edit---
The "gentle representation theory" to which I refer can be found here: http://arxiv.org/abs/1102.4353
We show in section 4 that the cardinality of $\mbox{Hom}(\pi_1(X),G)$ can be written explicitly in terms of the dimensions of the irreducible representations of $G$.
A: I have a feeling that almost every field of pure mathematics has its "own" way to see that surface groups $\pi_g$ (of genus $g\ge 2$) are not free. The two arguments below are, of course, more complicated than the ones which are based on group theory, geometry, topology or homological algebra, but the idea is to see connection to other fields of mathematics. Here are two examples:


*

*Algebraic geometry (combination of abelian and nonabelian Hodge theory): If $\pi_g\cong F_r$, free group of rank $r$, then, by looking at the 1-st Betti numbers and using the usual Hodge theory we see that $r$ has to be even. On the other hand, by Narasimhan-Seshadri theorem, the moduli space ${\mathcal M}_g$ of semistable rank 2 holomorphic bundles (with fixed determinant) on genus $g$ surface is analytically isomorphic to $Hom(\pi_g, SU(2))/SU(2)=Hom(F_r, SU(2))/SU(2)$, which would have odd (real) dimension $3r-3$. Contradiction, since (being a complex-projective variety) ${\mathcal M}_g$ is even-dimensional. (This is, of course, an argument similar to, but more complicated, than Mohan's.) 

*Geometric analysis: Suppose that $\pi_g\cong F_r$. Realize this isomorphism $\rho$ by a harmonic map $h$ from $S_g$ (genus $g$ compact Riemann surface) to a metric graph $\Gamma_r$ (the rose with $r$ leaves and unit edges). Preimages of generic points in $\Gamma_r$ under $h$ will be compact 1-dimensional submanifolds. By the maximum principle (look at the lifted harmonic map from the universal cover of $S_g$ to the tree), components of these submanifolds cannot bound disks in $S_g$, hence, they are not nul-homotopic. Hence, $\rho$ cannot be injective. 
So, are there proofs (even difficult ones) which make essential use of other fields of mathematics, e.g.:
a. Number theory (algebraic or analytic number theory, or arithmetic algebraic geometry)?
b. Measure theory? (This might be difficult since $\pi_g$ and $F_r$ are "measure-equivalent".) 
c. Probability? (An argument using random walk on graphs maybe?) 
d. Dynamical systems/ergodic theory? 
e. Functional analysis? (Maybe infinite-dimensional unitary representations of $\pi_g$ and $F_r$?) 
f. Logic? 
g. Commutative algebra?
Update: Here is a proof by the commutative algebra. Consider the affine schemes $S=Hom(\pi_g, GL(2))$ and $S'=Hom(F_r, GL(2))\cong GL(2)^r$. The latter, being an open subscheme of the affine space, has the same dimension of Zariski tangent space at every point. Consider $S$, let $R$ be its coordinate ring and $m_1, m_2\subset R$ be the ideals corresponding to the points $\rho_1$ (the trivial representation) and $\rho_2$, the representation which sends all but one standard generators to $1$ and the remaining generator to any noncentral matrix in $GL(2)$. Then (by a reasonably simple computation) $d_1=dim(R/m_1)=8g$ while $d_2=dim(R/m_2)=8g-2$. Hence, $d_1>d_2$ and dimensions of Zariski tangent spaces to $S$ at $\rho_1, \rho_2$ are different. In particular, the schemes $S, S'$ (equivalently, their rings) cannot be isomorphic and $\pi_g$ cannot be isomorphic to $F_r$.      
A: Since the abelianization of $\pi_1(\Sigma_g)$ is $\mathbb{Z}^{2g}$, if this group were free it would be free of rank $2g$. Since free groups are Hopfian, any $2g$ generators for $F_{2g}$ are free generators. The universal property of free groups then gives
$$\text{Hom}(\pi_1(\Sigma_g),G)\simeq \text{Hom}(F_{2g},G)\simeq G^{2g}$$ for any group $G$, i.e. every collection of $2g$ elements $a_1,b_1,\ldots,a_g,b_g$ in any group $G$ would have to satisfy $[a_1,b_1]\cdots[a_g,b_g]=1$. This would imply that every torsion-free group is abelian, since for any $x$ and $y$ in $G$ setting $a_i\mapsto x$ and $b_i\mapsto y$ gives $[x,y]^g=1$.
This is patently absurd. For explicit counterexamples, send $a_i$ and $b_i$ to reflections across two parallel lines, or reflections across two lines meeting at an irrational angle, or to the matrices $\begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix}$ and $\begin{bmatrix} 1 & 0\\ 1 & 1\end{bmatrix}$.
A: Original proof. How about a surface group is 1-ended since its universal cover is the hyperbolic plane and the non-abelian free group has infinitely many ends since the universal cover of the wedge of circles is a tree. 
Second proof. Since Misha has suggested a proof from each area here is a proof from universal algebra. Let V be the variety of extensions of elementary abelian 2-groups by elementary abelian 2-groups. The word problem for the relatively free group on X in this variety is well-known. A word is trivial if and ony if it is trivial in $(\mathbb Z/2)^X$ and the loop it labels in the Cayley graph of $(\mathbb Z/2)^X$ with respect to $X$ traverses each geometric edge an even number of times. 
If the surface group were free it would have to be free on $2g$-generators because of the abelianization. So if we factor by the verbal subgroup associated to V we would get a free group in this variety on 2-generators. But the group in V on 2g-generators with the surface defining relation is a proper quotient of the relatively free group because the product of commutators in question uses each edge of the Cayley graph of $(\mathbb Z/2)^{2g}$ exactly once. Since these are finite groups, a proper quotient is not free. 
A: *

*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.
A: FINAL EDIT : This edit cleans up the first proof (and simplifies it -- there are no longer any references to the free nilpotent group) and adds some remarks to the second proof following the discussion in the comments.

PROOF 1.
Here's a low-tech way to see that a surface group is not free (though cohomology is secretly lurking in the background).  Let $G_g = \langle a_1,b_1,\ldots,a_g,b_g\ |\ [a_1,b_1]\cdots [a_g,b_g]=1 \rangle$ be the surface group.  Form the group
$$\tilde{G}_g = \langle a_1,b_1,\ldots,a_g,b_g,t\ |\ [a_1,b_1]\cdots [a_g,b_g]=t, [a_i,t]=1, [b_i,t]=1\ \text{for all $1 \leq i \leq g$} \rangle$$
The subgroup of $\tilde{G}_g$ generated by $t$ is contained in the center and the quotient is $G_g$.  Below I will show that this subgroup is infinite cyclic.  We thus have a central extension
$$1 \longrightarrow \mathbb{Z} \longrightarrow \tilde{G}_g \longrightarrow G_g \longrightarrow 1.$$
If $G_g$ were free, then this would split as a direct product.  However, since $t$ becomes zero when we abelianize $\tilde{G}$, there is no splitting homomorphism $\tilde{G}_g \rightarrow \mathbb{Z}$.  Thus $G_g$ cannot be free.
It remains to show that the subgroup generated by $t$ is infinite cyclic.  Let $H$ be the $3$-dimensional Heisenberg group, ie the group of upper-triangular $3 \times 3$ integer matrices with $1$'s on the diagonal.  As is well-known, $H$ has a presentation
$$H = \langle x,y,z\ |\ [x,y]=z, [x,z]=1, [y,z]=1 \rangle.$$
Examining the presentations, there is a homomorphism $\psi : \tilde{G}_g \rightarrow H$ with 
$$\psi(a_1) = x \quad \text{and} \quad \psi(b_1) = y \quad \text{and} \quad \psi(t) = z$$
and
$$\psi(a_i) = \psi(b_i) = 1 \quad \quad (2 \leq i \leq g)$$
Since $z$ generates an infinite cyclic subgroup of $H$ (as a matrix, $z$ is the matrix with $1$'s on the diagonal and at position $(1,3)$ and $0$'s elsewhere), it follows that $t$ generates an infinite cyclic subgroup of $\tilde{G}_g$. 

PROOF 2.
It is known that free groups are Hopfian, i.e. that all surjections from a free group to itself are isomorphisms.  A simple-minded cancellation-based proof (using Nielsen reduction) can be found in Proposition 2.7 of Lyndon and Schupp's book "Combinatorial group theory".  Alternatively, Malcev proved that all residually finite groups are Hopfian (this can also be found in Lyndon and Schupp), and there are many proofs that free groups are residually finite; see the answers to the question Why are free groups residually finite?
This implies that if $F$ is a free group on $n$ generators and $S$ is a generating set for $F$ which has $n$ elements, then $S$ is a free generating set.  But this implies the result -- letting $G_g$ be the surface group as above, by abelianizing we see that if $G_g$ were a free group, then it would be free on $2g$ generators.  But $a_1,b_1,\ldots,a_g,b_g$ is a generating set of size $2g$ which is not free since it satisfies a relation.  Thus $G_g$ is not free.
A: Here is a Stallings-style argument. Suppose that $S$ is a closed connected surface.  Let $G = \pi_1(S)$.  Suppose $T$ is a graph and $v \in T$ a vertex.  Let $F = \pi_1(T,v)$ and suppose that $\phi \colon G \to F$ is any homomorphism. 

Theorem: If $\phi$ is injective then $S$ is the two-sphere.

Here's the proof. Let $e$ be any edge of $T$ and let $p \in e$ be the midpoint.   The given generating set $\{a_i,b_i\}$ for $G$ gives a one-skeleton for $S$, with one vertex; call the vertex $u$.  Let $Q$ be the single two-cell remaining in $S$.  
We may now define a map $f$ from the one-skeleton of $S$ to $T$ by mapping $u$ to $v$ and by mapping the edges of $Q$ to paths in $T$ as instructed by $\phi$.  Note that the relation in $G$ is killed by $\phi$ -- so, thinking of the image as a word $w$, we find $w$ is a completely reducible word.  The reduction of $w$ tells us how to extend $f$ to the two-cell $Q$.  (Draw $Q$.  Subdivide and label the edges of $Q$ by their images.  The first reduction of $w$ gives a triangle cutting off a pair of these new, smaller edges. Etc.)  
It follows that $f$ induces the homomorphism $\phi$. (It does the correct thing to the generators and to the relator.)  Now consider $C = f^{-1}(p)$.  By construction, $C$ is a collection of circles in $S$.  If any component of $C$ is non-trivial in $G$, then $\phi$ is not injective, a contradiction.  From the Jordon curve theorem deduce that all components of $C$ bound disks.  Thus we may homotope $f$ so that $p$ is not in the image.  Thus we may homotope $f$ so that the interior of $e$ is not in the image.  Thus we may reduce the number of edges in $T$.  We now induct downwards.  In the base case, where $T = v$, we find that $f$ is the constant map, so $G$ is the trivial group, so $S$ is the two-sphere. 
A: Riffing off of Ben Steinberg's answer, which amounts to the statement that a tree is not quasi-isometric to the hyperbolic plane, here's a proof which doesn't require knowing anything about ends.
Tile $H^2$ by regular $2g$-gons which are fundamental domains for the action of $\pi_1 X$. The boundaries of these $2g$-gons form a Cayley graph $\Gamma$ of $\pi_1 X$. If $\pi_1 X$ were a free group then $\Gamma$ would be quasi-isometric to a tree, and so there would exist constants $C \ge 0$ and $s \in (0,1)$ such that every closed edge path $\gamma$ in $\Gamma$ can be written as a concatenation $\gamma = \gamma_1 * \gamma_2$ so that $Length(\gamma_1), Length(\gamma_2) \ge s Length(\gamma)$, and the initial and terminal endpoints of $\gamma_1$ have distance $\le C$ in $\Gamma$. But for sufficiently large $r > 0$, choosing $\Gamma$ to be a closed edge path that stays within a uniform distance of the radius $r$ circle in $H^2$, we get a contradiction.
A: There are already many excellent answers, and I'm adding another more out of  solidarity than anything else. Unlike the other answers, this one is perversely complicated, although
 perhaps it  addresses the question raised in Misha's answer about the existence of an arithmetic proof. I might also add my own question that I have wondered about:
Question: There are many known obstructions for  group to be Kahler. Which of these
extend to tame fundamental groups?  

Lemma: Let $X$ be a smooth projective variety defined over an algebraically closed field.
  Let $\ell$ be a prime different from the characteristic, then the pro-$\ell$ part of
  the etale fundamental group $\pi_1^{et}(X)$ is not the pro-$\ell$ completion of a free group.

Sketch. (This is just a translation of the argument indicated in Mohan's comment to the original question.) Suppose it was,  then after passing to  a suitable etale cover, we get
a new variety $Y$ with 
$$b_1(Y) := \dim H^1(Y_{et},\mathbb{Z}_\ell)= \dim Hom(\pi^{et}_1(Y),\mathbb{Z}_\ell)$$
odd. This is impossible, because by the hard Lefschetz theorem (Deligne) $H^1$ has a nondegenerate symplectic structure.
