Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 133 / 1

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.