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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.

The subgroup of $\tilde{G}_g$ generated by $t$ is contained in the center and the quotient is $G_g$. It is an easy exercise to Below I will show that this subgroup is infinite cyclic. We thus have a central extension

EDIT

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 $\psi : Here's another group-theoretic proof which \tilde{G}_g \rightarrow H$ 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 more "combinatorial 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 theory"to itself are isomorphisms. Using A simple-minded cancellation-based proof (using Nielsen reductiontheory) can be found in Proposition 2.7 of Lyndon and Schupp's book "Combinatorial group theory". Alternatively, one Malcev proved that all residually finite groups are Hopfian (this can show also be found in Lyndon and Schupp), and there are many proofs that free groups are residually finite; see the answers to the question http://mathoverflow.net/questions/20471/why-are-free-groups-residually-finite/20472

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(for a nice exposition of this, see Proposition 2.7 in Lyndon and Schupp's book "Combinatorial group theory"; it is not a particularly hard result). 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.

EDIT 2 : I thought I'd give the details of my "easy exercise" above since I realized that it is a little harder than I thought. Recall that this claims that $t \in \tilde{G}_g$ generates an infinite cyclic subgroup.

Let $N_{2g}$ be the free $2$-step nilpotent group on $2g$ generators $\alpha_1,\beta_1,\ldots,\alpha_g,\beta_g$. Letting $H = \mathbb{Z}^{2g}$, this fits into a central extension$$1 \longrightarrow \wedge^2 H \longrightarrow N_{2g} \longrightarrow H \longrightarrow 1.$$The generators for $N_{2g}$ correspond to a basis $[\alpha_1],[\beta_1],\ldots,[\alpha_g],[\beta_g]$ for $H$ such that $\alpha_i$ and $\beta_i$ project to $[\alpha_i]$ and $[\beta_i]$ in $H$. Moreover, we have relations of the form$$[x,y] = [x] \wedge [y]$$for $x,y \in \{\alpha_1,\beta_1,\ldots,\alpha_g,\beta_g\}$.Terry Tao's blog post http://terrytao.wordpress.com/2009/12/21/the-free-nilpotent-group/ has a nice description/construction of this group.

Anyway, by examining the above presentation of $\tilde{G}_{2g}$ one can see that there is a homomorphism $\psi : \tilde{G}_{2g} \rightarrow N_{2g}$ with $\psi(a_i)=\alpha_i$ and $\psi(b_i)=\beta_i$. This homomorphism takes $t$ to the element$$[\alpha_1] \wedge [\beta_1] + \cdots + [\alpha_g] \wedge [\beta_g] \in \wedge^2 H \subset N_{2g}.$$Since this element generates an infinite cyclic subgroup of $N_{2g}$, so does $t$.

 
 
 
 
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EDIT 2 : I thought I'd give the details of my "easy exercise" above since I realized that it is a little harder than I thought. Recall that this claims that $t \in \tilde{G}_g$ generates an infinite cyclic subgroup.

Let $N_{2g}$ be the free $2$-step nilpotent group on $2g$ generators $\alpha_1,\beta_1,\ldots,\alpha_g,\beta_g$. Letting $H = \mathbb{Z}^{2g}$, this fits into a central extension$$1 \longrightarrow \wedge^2 H \longrightarrow N_{2g} \longrightarrow H \longrightarrow 1.$$The generators for $N_{2g}$ correspond to a basis $[\alpha_1],[\beta_1],\ldots,[\alpha_g],[\beta_g]$ for $H$ such that $\alpha_i$ and $\beta_i$ project to $[\alpha_i]$ and $[\beta_i]$ in $H$. Moreover, we have relations of the form$$[x,y] = [x] \wedge [y]$$for $x,y \in \{\alpha_1,\beta_1,\ldots,\alpha_g,\beta_g\}$.Terry Tao's blog post http://terrytao.wordpress.com/2009/12/21/the-free-nilpotent-group/ has a nice description/construction of this group.

Anyway, by examining the above presentation of $\tilde{G}_{2g}$ one can see that there is a homomorphism $\psi : \tilde{G}_{2g} \rightarrow N_{2g}$ with $\psi(a_i)=\alpha_i$ and $\psi(b_i)=\beta_i$. This homomorphism takes $t$ to the element$$[\alpha_1] \wedge [\beta_1] + \cdots + [\alpha_g] \wedge [\beta_g] \in \wedge^2 H \subset N_{2g}.$$Since this element generates an infinite cyclic subgroup of $N_{2g}$, so does $t$.

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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 $G_g$. It is an easy exercise to 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.

EDIT : Here's another group-theoretic proof which is more "combinatorial group theory". Using Nielsen reduction theory, one can show 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 (for a nice exposition of this, see Proposition 2.7 in Lyndon and Schupp's book "Combinatorial group theory"; it is not a particularly hard result). 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.

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