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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.

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.

Choose a finite generating set $T$ of $H$. A $d_T$-geodesic between two elements of $H$ maps under the inclusion to a uniform quasigeodesic in ${\mathrm{Cay}}_S(G)$ because $H$ is quasi-isometrically embedded. Since $G$ is a hyperbolic group, the Morse lemma for quasi-geodesics gives a constant $R$ such that this quasigeodesic and the ambient geodesic with the same endpoints have Hausdorff distance at most $R$. Every vertex of the former lies in $H$, so the latter lies in the closed $R$-neighborhood of $H$. Therefore $H$ is a quasiconvex subgroup.

A subgroup is quasiconvex when it is a quasiconvex subset of a Cayley graph. In a hyperbolic group, a finitely generated subgroup is quasiconvex exactly when it is quasi-isometrically embedded.

For a word metric from a finite generating set, the stable word length isSubadditivity and the Fekete lemma give the limit. Conjugating $g^n$ changes its length by at most a constant, so stable word length is conjugacy invariant. It satisfies ${\tau }_S(g^k)=|k|{\tau }_S(g)$. Finite-order elements have zero stable word length; in a hyperbolic group, every infinite-order element has positive stable word length.