Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 133 (Geometric Group Theory)

Write $\bar{a}=f(a)$, $\bar{t}=f(t)$, and $\bar{b}={\bar{a}}^{\bar{t}}={\bar{t}}^{-1}\bar{a}\bar{t}$. The elements $\bar{a}$ and $\bar{b}$ are conjugate group elements, so they have the same order of a group element, say $n$. The defining relation becomesIterating conjugation gives ${\bar{b}}^j\bar{a}{\bar{b}}^{-j}={\bar{a}}^{2^j}$. Since ${\bar{b}}^n=1$, taking $j=n$ yields $\bar{a}={\bar{a}}^{2^n}$, and henceThe stated consequence of Fermat's little theorem implies $n=1$. Thus $\bar{a}=1$, and the surjective group homomorphism $f$ shows that $Q$ is generated by $\bar{t}$. Therefore $Q$ is a cyclic group.

Introduce $b=t^{-1}at$. The presentation may be written as the HNN extensionof the Baumslag-Solitar group $\mathrm{BS}(1,2)=⟨a,b\mid ba{b}^{-1}=a^2⟩$, with stable letter $t$ identifying the infinite cyclic subgroups $⟨a⟩$ and $⟨b⟩$. Britton's lemma embeds the base group in the HNN extension. In particular, $a$ has infinite order, so $G$ is infinite. It is also nonabelian because $ba{b}^{-1}=a^2\ne a$.

Keep $b=t^{-1}at$. Repeatedly applying $ba{b}^{-1}=a^2$ givesIn the word metric on $G$ from the finite generating set $\{a,t\}$, the right-hand side therefore has length at most $6n+1$, because $b$ is represented by the word $t^{-1}at$. In the intrinsic word metric on the infinite cyclic group $⟨a⟩$ generated by $a$, however,A quasi-isometric embedding would bound the latter by an affine function of the former. The exponential sequence above violates every such bound, so the inclusion $⟨a⟩\hookrightarrow G$ is not a quasi-isometric embedding.

Direct multiplication gives $A^3=B^2=-I$, so $⟨A⟩$ has order $6$, $⟨B⟩$ has order $4$, and their intersection is $\{I,-I\}$ of order $2$. For the amalgamated free productthe Bass-Serre tree has verticesand edges $S{L}_2(\mathbb{Z})/\{\pm I\}$. Each edge $g\{\pm I\}$ joins $g⟨A⟩$ to $g⟨B⟩$. Thus it is an infinite bipartite tree in which the $⟨A⟩$-vertices have degree $3$ and the $⟨B⟩$-vertices have degree $2$:with a third branch leaving every ${\circ }_3$ vertex and the same pattern continuing on every branch.

Let $g$ be an isometry of the tree $T$. After subdividing edges if necessary, it has no edge inversion. If $g$ fixes a vertex, it is an elliptic isometry of a tree. Otherwise choose a vertex $v$ minimizing the positive integer $d(v,gv)$. The geodesic segments $g^n[v,gv]$ concatenate without backtracking: any backtracking would produce a vertex with smaller displacement. Their union over $n\in \mathbb{Z}$ is therefore a bi-infinite geodesic, the axis of a tree isometry, and $g$ translates it by $d(v,gv)$. This proves the elliptic-hyperbolic classification of an isometry of a tree.

An element of finite order cannot translate a line through a positive distance, since its powers would have unbounded displacement. Every finite-order element of $S{L}_2(\mathbb{Z})$ therefore fixes a vertex of its Bass-Serre tree. Vertex stabilizers are conjugates of $⟨A⟩$ and $⟨B⟩$, so the element is conjugate to a power of $A$ or a power of $B$.

Consider the principal congruence subgroupIt has finite index because $S{L}_2({\mathbb{F}}_3)$ is a finite group. It is torsion-free: by part (b), a finite-order element is conjugate to $A^i$ or $B^j$, while the reductions of $A$ and $B$ modulo $3$ still have orders $6$ and $4$, respectively. Hence no nonidentity power in either vertex group reduces to the identity.

It follows that $\Gamma (3)$ intersects every conjugate of the two vertex stabilizers trivially, so its action on the Bass-Serre tree is free. A group with a free group action on a tree is a free group. Consequently $S{L}_2(\mathbb{Z})$ contains the free subgroup $\Gamma (3)$ of finite index; equivalently, it is a virtually free group.

Identify $G$ with the finite-index subgroup $f(G)\le H$. If $S$ is a finite generating set for $H$ and $T$ is a finite set of right-coset representatives, Schreier's lemma gives a finite generating set for $f(G)$, and hence for $G$.

Equip both groups with word metrics from finite generating sets. The inclusion is Lipschitz because each generator of $G$ has bounded length in $H$. Conversely, rewriting a word in $S$ by tracking its cosets through the finite set $T$ expresses an element of $G$ as a word of length bounded linearly in its $H$-length. Finally, every element of $H$ lies within the maximum word length of an element of $T$ from $f(G)$. Thus the inclusion is a finite-index subgroup quasi-isometry, and composing it with the isomorphism $f:G\to f(G)$ proves that $G$ and $H$ are quasi-isometric.

Let $S$ be a finite generating set of $H$, choose one lift $\tilde{s}\in G$ for every $s\in S$, and put $K=\ker f$. The finite setgenerates $G$: lift a word representing $f(g)$ and observe that the discrepancy from $g$ lies in $K$.

Use this generating set for $G$. The quotient map does not increase word length, while lifting a shortest word in $H$ leaves only a final element of $K$, whose word length is at most one. HenceThe map is surjective, so it is a finite-kernel quotient quasi-isometry. Therefore $G$ is finitely generated and quasi-isometric to $H$.

The relations $ak{a}^{-1}k=1$ and $bk{b}^{-1}k=1$ say that conjugation by either $a$ or $b$ sends $k$ to $k^{-1}$. Thus $K=⟨k⟩$ is a normal subgroup of order at most $3$. Quotienting by it givesthe orientation-preserving hyperbolic triangle group $\Delta (4,4,4)$. This group acts properly discontinuously and cocompactly by isometries on the hyperbolic plane. The Milnor–Švarc lemma therefore makes $\Delta (4,4,4)$ quasi-isometric to $H^2$.

The quotient map $\Gamma \to \Gamma /K$ has finite kernel, so part (b), equivalently the finite-kernel quotient quasi-isometry, makes $\Gamma$ quasi-isometric to $\Delta (4,4,4)$. By transitivity of quasi-isometry,

Let $[u,v]$ be one side of the geodesic quadrilateral, and draw a diagonal from $u$ to the opposite vertex. A point $p\in [u,v]$ lies, by $\delta$-thinness of the first geodesic triangle, within $\delta$ either of an adjacent side or of the diagonal. In the latter case, $\delta$-thinness of the second triangle places the nearby point of the diagonal within another $\delta$ of one of the other two sides. The triangle inequality then places $p$ within $2\delta$ of the remaining three sides. The argument applies to every side, proving the geodesic quadrilateral in a hyperbolic metric space bound.

Let $p$ be the midpoint of a geodesic $[y_1,y_2]$. Since $Y$ is a convex subset of a geodesic metric space, $p\in Y$. In the geodesic triangle with vertices $x,y_1,y_2$, the point $p$ lies within $\delta$ of $[x,y_1]$ or $[x,y_2]$. By symmetry suppose $q\in [x,y_1]$ and $d(p,q)\le \delta$. Put $L=d(y_1,y_2)$. Thenand consequentlyBecause $y_1$ is a closest point of $Y$ to $x$ and $p\in Y$, $d(x,y_1)\le d(x,p)$. Combining the inequalities gives $L\le 4\delta$. This is the coarse uniqueness of a closest point in a hyperbolic metric space.

Let $p$ be the midpoint of a geodesic $[y_1,y_2]\subseteq Y$ and put $L=d(y_1,y_2)$. Apply the geodesic quadrilateral in a hyperbolic metric space bound to the quadrilateral with consecutive vertices $y_1,z_1,z_2,y_2$. The point $p$ is within $2\delta$ of one of the other three sides. It cannot be within $2\delta$ of $[z_1,z_2]\subseteq Z$, because every point of $Y$ has distance greater than $2\delta$ from every point of $Z$.

By symmetry there is therefore a point $q\in [y_1,z_1]$ with $d(p,q)\le 2\delta$. The triangle inequality givesand henceSince $p\in Y$ and $y_1$ is a closest point of $Y$ to $z_1$, we also have $d(z_1,y_1)\le d(z_1,p)$. Therefore $L\le 8\delta$, as required.