Abelian subgroups of hyperbolic groups are virtually cyclic

If $A$ contains an element $g$ of infinite order, then $A\subseteq C_G(g)$. By [L1], the cyclic subgroup $⟨g⟩$ has finite index in $C_G(g)$. Thus $A\cap ⟨g⟩$ has finite index in $A$ and is cyclic as a subgroup of $⟨g⟩$. Hence $A$ is virtually cyclic.

If every element of $A$ has finite order, each finitely generated subgroup of $A$ is finite: for generators of orders $n_1,\dots ,n_k$, commutativity makes it a quotient of the finite group ${\prod }_i\mathbb{Z}\mathrm{/}{n}_i\mathbb{Z}$. By [L2] it has at most $B$ elements. Were $A$ to contain $B+1$ distinct elements, their finitely generated subgroup would contradict this bound. So $A$ is finite, hence virtually cyclic. The two cases prove the claim.