A hyperbolic <isometry of a tree> has a unique invariant <axis of a tree isometry>, on which it acts by a nonzero translation. Let $L$ be the axis of $x$. By part (c), $x$ is central, so for every $g\in G$,
$$
x(gL)=g(xL)=gL.
$$
Thus $gL$ is another axis of $x$. Uniqueness gives $gL=L$, and hence the whole group $G$ preserves the line $L$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 133 / 3 / d / Solution

{title2=$\ell(g)$}
{wiki}

The translation length of an isometry $g$ of a metric space is $\ell(g)=\inf_x d(x,gx)$. For a hyperbolic tree isometry, it is attained exactly on the <axis of a tree isometry>.

Translation length

{title2=$A_g$}
{wiki=Bass–Serre_theory}

The axis of a hyperbolic tree isometry $g$ is its unique invariant line. The isometry acts on this line by translation through its translation length.


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 Incoming links: Axis of a tree isometry