TITLE: Why $|G:Z(G)|$ is finite in this question?
QUESTION [5 upvotes]: Suppose $G$ is a group such that the order of any nontrivial element of $G$ is infinite. Prove that if $G$ has a cyclic subgroup with finite index then $G$ is cyclic.

my solution:suppose $|G:<x_0>|=n$ then $G=<x_0,g_1,...,g_n>$ .because $G=\dot{\cup}^{n}_{i=1}<x_0>g_i$,if I show that $|G:Z(G)|$ is finite,then by schur theorem ,we can have $G^{'}$ is finite,by hypothesis we should have $G^{'}=1$ and then $G$ is a abelian finitely generated group,so by fundamental theorem of abelian finitely generated group $G \cong \mathbb{Z} \times G_1 \times ... \times G_n $ where $G_i$ s are cyclic and are of order prime numbers,by the hypothesis we should have $G_i=1$ and then $G \cong \mathbb{Z}$ .
now my problem is why $|G:Z(G)|$ is finite and how should I show this?
it will be great if you help me with this,thanks.

REPLY [2 votes]: Let $N= \langle x \rangle$ be a normal infinite cyclic subgroup of $G$, and $g \in G$. Then $g^{-1}xg = x$ or $x^{-1}$.
Suppose that $g^{-1}xg = x^{-1}$. Since $|G:N|$ is finite, we have $g^n \in N$ for some $n>0$. Since $g^{-1}x^kg = x^{-k}$ for all $k \in {\mathbb Z}$, $g$ does not centralize any nontrivial element of $N$, and hence we must have $g^n=1$, contradicting $G$ being torsion free.
So $g^{-1}xg=x$ and $N \le Z(G)$.