= Solution
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 $r$ gives
$$
a^r b^s=c^{rs}b^s a^r\qquad(r,s\geq0).
$$
Therefore
$$
\boxed{a^k b^k a^{-k}b^{-k}=c^{k^2}\qquad(k\geq0).}
$$
This 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 matrices
$$
a\mapsto I+E_{12},\qquad b\mapsto I+E_{23},\qquad c\mapsto I+E_{13}.
$$
Their commutator is $I+E_{13}$, which commutes with both other matrices, so all defining relations hold. The image of $c^m$ is $I+mE_{13}$; 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 $\{a,b,c\}^{\pm1}$, the displayed commutator gives
$$
|c^{k^2}|_H\leq4k.
$$
For any finite generating set $T$ of $\langle c\rangle$, 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>, $B\geq1$. A word of length $m$ in $T$ has exponent at most $Bm$ in absolute value, so
$$
|c^{k^2}|_T\geq\frac{k^2}{B}.
$$
A <quasi-isometric embedding> of the inclusion would require $|c^{k^2}|_H\geq\lambda^{-1}|c^{k^2}|_T-\varepsilon$. Combining these bounds would give $4k\geq k^2/(B\lambda)-\varepsilon$ for every $k$, which is impossible. Thus \b[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>.
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