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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...

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    a and b are hyperbolic isometries, and since they have the same fixed points, they have the same axis. Thus we can map the group they generate to R, using the translation length function. This is an injective homomorphism, so this group is a discrete subgroup of R.2012-10-08

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