Let $x,y\in\operatorname{Fix}(\phi)$. A <tree> has a unique geodesic segment $[x,y]$. The isometry $\phi$ sends this segment to the segment $[\phi x,\phi y]=[x,y]$. An isometry of a segment that fixes both endpoints fixes every point of it, so
$$
[x,y]\subseteq\operatorname{Fix}(\phi).
$$
Thus the fixed-point set of an <elliptic isometry of a tree> is a convex subtree, in particular it is <path-connected space>[path-connected].

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 133 / 2 / a / 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.

Solved by gpt-5.6-sol high.

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

{wiki=Bass–Serre_theory}

An elliptic tree isometry fixes a vertex. Its fixed-point set is a nonempty convex subtree.


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