The Details:
In considering this question on the same isomorphism and trying to come up with an alternative proof of my own (than the one composed of the work in the question and @DerekHolt's comment), I got stuck.
I want to use the following presentation of $\Bbb Z\times \Bbb Z$,
$$\langle a,b\mid ab=ba \rangle,\tag{$\mathcal{P}$}$$
by killing some element of the presentation.
My guess is to let $c=ab$ then kill $c^3$ in $(\mathcal{P})$, since, say, $(1,0)\mapsto a$ & $(0,1)\mapsto b$ and $a$ & $b$ commute, some other pithy Tietze transformations might elicit an isomorphism of the quotient of $(\mathcal{P})$ by $\langle (3,3)\rangle$ with
$$\langle x,y\mid y^3, xy=yx\rangle,\tag{$\mathcal{Q}$}$$
a presentation of $\Bbb Z\times\Bbb Z_3.$
The Question:
Using presentations, prove $\frac{\Bbb{Z} \times \Bbb{Z}}{\langle(3,3)\rangle}\cong\Bbb{Z} \times \Bbb{Z_3}$.
Thoughts:
I really think I ought to be able to do this myself. I work with presentations an awful lot. However, it has taken me the better part of an hour to articulate my hunch.
Please help :)
You can introduce a new generator, $c=ab$, and then $b$ is redundant (since $b = a^{-1}c$). Then the subgroup we are quotienting by is just $\langle c^{3} \rangle$, so the quotient group becomes $\langle a,c \mid c^{3},\ ac = ca \rangle$ which is clearly isomorphic to the presentation $(\mathcal{Q})$.
In terms of the presentation (using Tietze transformations), we have $$\begin{align} (\mathcal{P}) &\to \langle a,b,c \mid ab=ba, c=ab\rangle \\ &\to \langle a,c \mid c=a^{-1} c a, c=c\rangle \\ &\to \langle a,c \mid ac=ca\rangle. \end{align}$$