Solution
= Solution
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 $L$.