Volume of a Riemannian manifold and its relation to fundamental group

I am reading a book (Mapping Class Group by Farb and Margalit) and it says (in a proof of one theorem):

If $S$ admits a hyperbolic metric (they define such a surface to be of finite area and complete) and we had $\pi_1(S)\cong \mathbb{Z}\$ then the surface would have an infinite volume which is a contradiciton. Hence $\pi_1(S)$ is NOT isomorphic to $\mathbb{Z}$.

Questions:

1. Why $\pi_1(S)\cong \mathbb{Z}\$ implies the volume of $S\$ is infinite?

2. Is there a general relation between the fundamental group and the volume of a surface?

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Hyperbolic surfaces are of the form $\mathbb{H}^2/\Gamma$, where $\Gamma$ is a Fuchsian group which is isomorphic to the fundamental group. If $\Gamma \cong \mathbb{Z}$, then it is generated by a single element $\gamma$. The order of $\gamma$ is infinite, so it cannot be elliptic. It is thus either parabolic (and hence conjugate to $\begin{pmatrix} 1 & \lambda \\ 0 & 1 \end{pmatrix}$ for some $\lambda$) or hyperbolic (and hence conjugate to $\begin{pmatrix} \lambda & 0 \\ 0 & 1/\lambda \end{pmatrix}$ for some $\lambda \neq 0$). It is an easy exercise to show that the quotient of $\mathbb{H}^2$ by the subgroup generated by $\begin{pmatrix} 1 & \lambda \\ 0 & 1 \end{pmatrix}$ has infinite volume, and similarly for $\begin{pmatrix} \lambda & 0 \\ 0 & 1/\lambda \end{pmatrix}$.

For compact surfaces with Riemannian metrics, the Gauss-Bonnet theorem determines the volume in terms of the Euler characteristic (and hence in terms of the fundamental group).

I recommend reading Katok's book "Fuchsian Groups" if you want to acquire comfort with the above types of arguments.

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@Sue Thanks for your help. Could you tell me about the relation between the fundamental group and Euler characteristic, please? I know how determine the volume from Gauss-Bonnet using the Euler characteristic. – JimWang Sep 11 '12 at 4:10
Read the textbooks of either Massey or Armstrong (or any other book that talks about the classification of surfaces and the fundamental group). If this isn't something that you're familiar with, you'll have serious trouble reading Farb-Margalit, so you're better off studying the foundations. – Sue Sep 11 '12 at 5:02

I will write what I understand so far:

The fundamental group of an orientable surface of genus $g$ with $b$ boundary components and $n$ punctures will be generated by $2g+b+n$ generators. The Euler characteristic in this case is $\chi=2-(2g+b+n)$ i.e $\chi=2-$ the number of generators. So we get $\chi=1.$ obviously $g=0$ and the either $b=1$ or $n=1$ i.e sphere with one boundary component OR sphere with a puncture .

If sphere with one boundary component: I do not know what to do because there is no guarantee that the boundary is totally geodesic.

If sphere with a puncture (this corresponds to the case of parabolic element), that is just the plane, and it has infinite area (No need for Gauss-Bonnet).