I am having trouble seeing why this statement is true: "If S admits a hyperbolic metric, then the centralizer of any non-trivial element of $\pi(S)$ is cyclic. In particular, $\pi(S)$ has trivial center".
This is on page 23 of Mapping Class Groups by Farb, Margalit.
I have seen some arguments online that do this explicitly by listing the possible elements of $\pi_1(S)$ as deck transformations on $H^2$, the covering space of $S$. However, the book I am using seems to have a different proof. The argument it gives is that if $a$ is centralized by $b$, then $a, b$ have the same fixed points, which I understand mostly. Then the authors claim that since the action of $\pi_1(S)$ on $S$ is discrete, then the centralizer of $a$ in $\pi_1(S)$ is infinite cyclic. I understand that the action is discrete but do not understand why this implies that the centralizer should be infinite cyclic...
Also, I understand that if $S$ had non-trivial center, then $\pi_1(S) = Center(\alpha) = \mathbb{Z}$. The book then claims that this implies that $S$ has infinite volume, which is a contradiction. But I do not understand why $\pi_1(S)= \mathbb{Z}$ implies that $S$ has infinite volume...