Finite subgroups of a hyperbolic group have uniformly bounded order

Let $F\le G$ be finite and put $M=F\cdot e$, a finite set of vertices of $X$. For a vertex $x$ define $R(x)={\max }_{m\in M}d(x,m)$. The nonempty set of integer values $R(x)$ has a least value $R$, attained at some vertex $x$. Left translation by each $f\in F$ preserves $M$ and distances, so $R(fx)=R(x)=R$. Thus the center set $C=\{v\in G:R(v)=R\}$ is $F$-invariant and contains the orbit $Fx$.

Let $x,y\in C$, write $D=d(x,y)$, and let $z$ be the midpoint of a geodesic $[x,y]$. For any $m\in M$, slimness of the triangle with vertices $x,y,m$ gives a point $p$ on $[x,m]$ or $[y,m]$ with $d(z,p)\le \delta$. In the first case, $d(x,p)\ge D\mathrm{/}2-\delta$ and $d(x,m)\le R$, so $d(z,m)\le R-D\mathrm{/}2+2\delta$; the second case is symmetric. Choose a vertex $v$ of the edge containing $z$, with $d(v,z)\le 1\mathrm{/}2$. Then $R(v)\le R-D\mathrm{/}2+2\delta +1\mathrm{/}2$. Minimality of $R$ forces $D\le 4\delta +1$. Hence every two vertices of $C$, and in particular of $Fx$, are at distance at most $4\delta +1$.

The map $f\mapsto fx$ is injective by [L2]. The orbit $Fx$ lies in the vertex ball about $x$ of radius $N=\lceil 4\delta +1\rceil$, whose cardinality is the fixed finite number $B=\mathrm{\mid }{B}_X(e,N)\cap G\mathrm{\mid }$ by Cayley vertex transitivity. Thus $\mathrm{\mid }F\mathrm{\mid }=\mathrm{\mid }Fx\mathrm{\mid }\le B$ for every finite $F\le G$. Taking $B_S=B$ proves the assertion for the given $S$.