Hg toolkit infinite hyperbolic groups have infinite order elements

Let $k$ be the positive finite number of cone types. The group is not contained in its finite closed radius-$k$ set, by F2 and infinitude. Choose $g$ with $\mathrm{\mid }g\mathrm{\mid }=m>k$ and a shortest word path $e=g_0,g_1,\dots ,g_m=g$. Prefixes of a shortest word are shortest, so $\mathrm{\mid }{g}_i\mathrm{\mid }=i$ and $\mathrm{\mid }{g}_i^{-1}{g}_j\mathrm{\mid }=j-i$ for $i<j$; shortening either prefix or intervening subword would shorten the whole word. Among the $m+1$ cone types of these vertices, two are equal. Fix $i<j$ with $\mathcal{C}(g_i)=\mathcal{C}(g_j)$, and put $h=g_i^{-1}{g}_j$, $l=j-i>0$. Then $\mathrm{\mid }h\mathrm{\mid }=l$ and $\mathrm{\mid }{g}_{i}h\mathrm{\mid }=i+l$.

We prove by induction that $\mathrm{\mid }{h}^n\mathrm{\mid }=nl$ and $\mathrm{\mid }{g}_i{h}^n\mathrm{\mid }=i+nl$ for every integer $n\ge 0$. For $n=0$ this is the identity, and for $n=1$ it is step 1.1. Suppose the assertion holds for $n$. Then $h^n\in \mathcal{C}(g_i)=\mathcal{C}(g_{i}h)$, so $\mathrm{\mid }{g}_i{h}^{n+1}\mathrm{\mid }=\mathrm{\mid }{g}_{i}h\mathrm{\mid }+\mathrm{\mid }{h}^n\mathrm{\mid }=i+(n+1)l\mathrm{.}$ The inequalities $\mathrm{\mid }{g}_i{h}^{n+1}\mathrm{\mid }\le i+\mathrm{\mid }{h}^{n+1}\mathrm{\mid }\le i+(n+1)l$ force $\mathrm{\mid }{h}^{n+1}\mathrm{\mid }=(n+1)l$. This completes the induction.

For every $n>0$, step 2.1 gives $\mathrm{\mid }{h}^n\mathrm{\mid }=nl>0$, hence $h^n\mathrm{\ne }e$. Thus $h$ has infinite order. In the hyperbolic case F1 supplies the required finite cone hypothesis, proving the first assertion. The proof uses only a single long word and finitely many cone comparisons, not an infinite ray selection or AC.