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If $g$ has infinite order in a hyperbolic group, the map $n\mapsto g^n$ is a quasi-isometric embedding of the integers. Equivalently $|g^n|\ge c|n|-C$ for some $c>0$. This is the homogeneous quasigeodesic theorem: local-to-global control of metric geodesics in a hyperbolic Cayley graph gives a quasigeodesic orbit for a suitable conjugate of a positive power; undoing conjugation and passing between consecutive powers changes only uniform constants. It implies $\tau (g)>0$. The theorem and its hypotheses are given in Löh, Theorem 6.5.9. It is not true for arbitrary finitely generated groups, as the Baumslag-Solitar group relation $ab{a}^{-1}=b^3$ shows.

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.

Let $r:G\to H$ be a retraction. Part (a) supplies a constant $C$ such thatThe finite set $r(S)$ generates $H$. PutIf a shortest $S$-word represents $h_1^{-1}{h}_2\in H$, applying the group homomorphism $r$ gives a $T$-word for the same element of length at most $D{d}_S(h_1,h_2)$. HenceThe inclusion is therefore bilipschitz and in particular a quasi-isometric embedding. Thus every finitely generated retract subgroup is quasi-isometrically embedded.

This is the Baumslag-Solitar group $G=\mathrm{BS}(1,2)$. Its defining relation gives, by mathematical induction,In the cyclic subgroup $H=⟨a⟩$ with generator $a$,whereas in $G$, with generators $a,b$,If inclusion were a $(\lambda ,\epsilon )$-quasi-isometric embedding, its lower bound would implyfor every $n$, which is impossible because an exponential function eventually dominates every linear function. Hence $H$ is not quasi-isometrically embedded; it is an exponentially distorted subgroup.

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.

Any two word metrics from finite generating sets on the same group are bilipschitz equivalent. Indeed, if $S,S^{\prime }$ are finite, let $L={\max }_{s\in S}|s{|}_{S^{\prime }}$; then $|g{|}_{S^{\prime }}\le L|g{|}_S$, and the reverse inequality follows symmetrically. Apply this once to the two finite generating sets of $G$ and once to those of $H$. Composing these bilipschitz identity maps with the inclusion changes only the multiplicative and additive constants in the quasi-isometric embedding inequalities. Thus being a quasi-isometrically embedded subgroup is independent of $S$ and $T$.

Let $h\in H$ and letbe a geodesic in the Cayley graph ${\mathrm{Cay}}_S(G)$. By $K$-quasiconvexity choose $h_i\in H$ with $d_S(x_i,h_i)\le K$, taking $h_0=1$ and $h_N=h$. ThenThe elements $h_i^{-1}{h}_{i+1}$ telescope to $h$, so the finite setgenerates $H$.

Moreover $|h{|}_U\le |h{|}_S$, while $|h{|}_S\le (2K+1)|h{|}_U$. Thus the inclusion $(H,d_U)\to (G,d_S)$ is a quasi-isometric embedding, and $H$ is quasi-isometrically embedded.

A finitely generated subgroup $H\le G$ is quasi-isometrically embedded when its inclusion, equipped with word metrics from finite generating sets, is a quasi-isometric embedding. This property is independent of those generating sets.

A subgroup $H\le G$ is a retract when there is a group homomorphism $r:G\to H$ whose restriction to $H$ is the identity. If both groups are finitely generated, the inclusion of $H$ is a quasi-isometric embedding: the inclusion and retraction are both Lipschitz in their word metrics.

For a finitely generated subgroup $H\le G$, subgroup distortion compares the intrinsic word length $|h{|}_H$ with the ambient word length $|h{|}_G$. The subgroup is undistorted exactly when its inclusion is a quasi-isometric embedding.

This is the subgroup word-metric instance of distortion.