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Sep 19, 2019 at 8:37 comment added Ville Salo If I understand correctly, in your other answer, you find $a \in G$ such that a branch $B$ outside some separating set $A$, which we can take to be an $n$-ball, properly contains the set $aB$. Then $a$ indeed has infinite order, and we also obtain that $a$ is $n$-axial: any path from $1$ to $a^2$ in particular has to step into $aB$, since $a^2B$ is inside $aB$. Translate and apply the lemma that $(B_i : B_{i+1} : B_{i+2})$ implies $(B_i : B_j : B_k)$ for $i < j < k$.
Sep 19, 2019 at 7:01 comment added Ville Salo People are so fast, I had already basically written my answer after asking Cohen (I just had to add the axial corollary) and Cohen personally told me about this post, yet I was 15 minutes slower. If the cited answer works, it sounds much simpler than mine.
Sep 19, 2019 at 6:40 history answered AGenevois CC BY-SA 4.0