Let $T$ be the infinite trivalent tree. I want to show that if $\alpha,\beta,\alpha',\beta'$ are vertices of the tree such that the distances $d(\alpha,\beta)$ and $d(\alpha',\beta')$ are equal, then there is an automorphism $g$ of the tree such that $(\alpha^g,\beta^g)=(\alpha',\beta')$.
I think I can prove this, but since the tree is infinite, I want to make sure my argument is complete, and the terminology to use is unclear to me.
Proof: If $\alpha=x_0,x_1,\ldots,x_\ell=\beta$ and $\alpha'=x_0',x_1',\ldots,x_\ell'=\beta'$ are the unique shortest $\alpha-\beta$ and $\alpha'-\beta'$ paths, respectively, then let $g$ be the map $x_i \mapsto x_i'$. The two remaining neighbors of $x_0$ can be mapped to the two remaining neighbors of $x_0'$ in any arbitrary manner, and similarly for the two neighbors of $x_\ell$. Map the unique remaining neighbor of $x_i, i=1,\ldots,\ell-1$ to the unique remaining neighbor of $x_i'$. This map $g$ can be extended to an automorphism of the infinite tree (does this sentence require further justification?). QED.
I want to prove that the vertices of the tree can be partitioned into two subsets $\Delta$ and $\Delta'$ such that the distance between any pair of vertices in the same subset is even. I think this follows immediately from the fact that the tree is bipartite, and I wonder if further explanation is required just because the tree happens to be infinite. (This is exercise 1.5.15 from Dixon and Mortimer's Permutation Groups).
(While the above is my question, I would be glad to know if there are any important differences (in results or proofs) between the cases where the tree is finite versus infinite.)