Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 133 / 3

Use the fact that infinite cyclic subgroups of hyperbolic groups are undistorted: if $z$ has infinite order in a hyperbolic group, then its stable word lengthis strictly positive. The limit exists by subadditivity and the Fekete lemma. It is invariant under conjugation, because conjugating $z^m$ changes its length by at most twice the conjugator's length. It also satisfies $\tau (z^k)=|k|\tau (z)$.

In the given Baumslag-Solitar group, $b$ has infinite order. To verify this without presuming the presentation's normal form, let $a$ act on the real line by $t\mapsto 3t$ and $b$ by $t\mapsto t+1$. These affine bijections satisfy $ab{a}^{-1}=b^3$, and every nonzero power of $b$ is a nonidentity translation. Thus the element $b$ in the presented group cannot have finite order.

If an injective group homomorphism to $G$ existed, its images $z$ of $b$ and $u$ of $a$ would satisfy $uz{u}^{-1}=z^3$, with $z$ still of infinite order. Consequentlyforcing $\tau (z)=0$, a contradiction. ThereforeEquivalently, the relation gives $a^{k}b{a}^{-k}=b^{3^k}$, so powers of $z$ with exponent $3^k$ would have length at most $2k|u|+|z|$. This contradicts the linear lower bound for an undistorted infinite cyclic subgroup. This alternative also makes the exponential compression obstruction explicit.

The two commutation relations make $c$ central, and the remaining relation is $ab=cba$. Repeatedly move $a$ past $b$, keeping the central powers in front. Induction first on $s$ and then on $r$ givesThereforeThis is the commutator identity in the integer Heisenberg group. The order of the factors fixes the positive sign of the exponent.

To check the claimed infinite order, map the generators to matricesTheir commutator is $I+E_{13}$, which commutes with both other matrices, so all defining relations hold. The image of $c^m$ is $I+m{E}_{13}$; hence $c$ has infinite order. This realizes the presentation in the Integer Heisenberg group and is sufficient to certify the cyclic subgroup, without needing to prove faithfulness of the whole representation.

With the ambient word metric from $\{a,b,c{\}}^{\pm 1}$, the displayed commutator givesFor any finite generating set $T$ of $⟨c⟩$, write its elements as powers of $c$ and set $B=\max \{|j|:c^j\in T\}$, with inverses included. Since $T$ generates an infinite cyclic group, $B\ge 1$. A word of length $m$ in $T$ has exponent at most $Bm$ in absolute value, soA quasi-isometric embedding of the inclusion would require $|c^{k^2}{|}_H\ge {\lambda }^{-1}|c^{k^2}{|}_T-\epsilon$. Combining these bounds would give $4k\ge k^2/(B\lambda )-\epsilon$ for every $k$, which is impossible. Thus the inclusion of the central infinite cyclic subgroup is not a quasi-isometric embedding. Changing the ambient finite generating set only changes multiplicative constants, by equivalence of finite word metrics. This is the quadratic distortion of the Heisenberg center.