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

Write the generators additively in the abelianization. The relators giveandAfter substitution, the last relation becomesThe commutator relators disappear automatically, sobecause $\gcd (5,4)=1$. Explicitly, an isomorphism to $\mathbb{Z}$ sends

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Letbe the orientation-preserving $(2,3,7)$ hyperbolic triangle group. DefineEvery defining relator of $G$ maps to the identity, and $A,B,C$ lie in the image, so this is a surjective group homomorphism. Since$\Delta (2,3,7)$ is a non-elementary Fuchsian group.

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The equality $x=a^2$ and the relator $[b,x]=1$ show that $x$ commutes with both $a$ and $b$. Since $c=(ab{)}^{-1}$, it also commutes with $c$, and then with $y=b^3$. Hence $x\in Z(G)$. Similarly, $y=b^3$ commutes with $b$ and, by $[a,y]=1$, with $a$; it therefore commutes with $c$ and $x$. Thus

Quotienting by $⟨x,y⟩$ givesA non-elementary Fuchsian group has trivial center: two hyperbolic elements with different pairs of boundary fixed points have only the identity in their common centralizer in ${\mathrm{PSL}}_2(\mathbb{R})$. Therefore the image in the quotient of every element of $Z(G)$ is trivial. It follows that

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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 $g\in G$,Thus $gL$ is another axis of $x$. Uniqueness gives $gL=L$, and hence the whole group $G$ preserves the line $L$.

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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 throughIn particular, the induced tree isometries satisfy

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. Moreoverso 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, soSince $a,b,c$ generate $G/⟨x,y⟩$, $G$ fixes a vertex of $T^{Z(G)}$. Thus the action of $G$ on $T$ is trivial in the tree-action sense.

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