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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.