= Solution
Let $T^{Z(G)}$ be the common fixed subtree of the centre, which is nonempty by hypothesis and is $G$-invariant because $Z(G)$ is central. Restrict the action to this subtree. There $x$ and $y$ act pointwise trivially, so the action factors through
$$
G/\langle x,y\rangle\cong\Delta(2,3,7).
$$
In particular, the induced tree isometries satisfy
$$
a^2=b^3=c^7=1,\qquad abc=1.
$$
Assume, as usual for a combinatorial tree action, that edge inversions have been removed by barycentric subdivision. The finite-order elements $a,b,c$ are then <elliptic isometry of a tree>[elliptic]. Moreover
$$
ab=c^{-1},\qquad bc=a^{-1},\qquad ca=b^{-1},
$$
so each pairwise product is elliptic. <Serre lemma for tree actions> implies that the fixed subtrees of each pair intersect. Convex subtrees of a tree have the <Helly property>, so
$$
\operatorname{Fix}(a)\cap\operatorname{Fix}(b)\cap\operatorname{Fix}(c)\ne\varnothing.
$$
Since $a,b,c$ generate $G/\langle x,y\rangle$, $G$ fixes a vertex of $T^{Z(G)}$. Thus the action of $G$ on $T$ is trivial in the tree-action sense.
Back to article page
