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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 :)

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    $\begingroup$ What about sending $\overline{(1,0)} \to x$ (rather $\overline{a} \to x$) $\overline{(1,1)} \to y$ (or rather $\overline{ab} \to y$)? You just need to show this is an isomorphism from your quotient group to $\mathbb{Z}\times\mathbb{Z}_{3}$. $\endgroup$
    – xxxxxxxxx
    Sep 8, 2019 at 22:29
  • $\begingroup$ This is probably a stupid question, @MorganRodgers, but what do you mean by, say, $\overline{a}$? It reminds me of tranversals à la Magnus et al., but I doubt that's what you mean. $\endgroup$
    – Shaun
    Sep 8, 2019 at 22:41
  • $\begingroup$ I've corrected the notation to be clearer, @MorganRodgers; I still don't know what you mean by $\overline{a}$ and $\overline{ab}$. The overline is used differently in Magnus et al.'s book "Combinatorial Group Theory: [. . .]". $\endgroup$
    – Shaun
    Sep 9, 2019 at 1:05
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    $\begingroup$ I mean if $g$ is in the group, $\overline{g}$ is the corresponding element in the quotient group (ie the coset $g + \langle (3,3)\rangle$) $\endgroup$
    – xxxxxxxxx
    Sep 9, 2019 at 4:14
  • $\begingroup$ No, $(\mathcal{P})$ and $(\mathcal{Q})$ are presentations. I'm using the tag{$\mathcal{X}$) code for each, @MorganRodgers. It's a standard tool here. $\endgroup$
    – Shaun
    Sep 9, 2019 at 8:17

1 Answer 1

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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 $$\begin{align} (\mathcal{P}) &\to \langle a,b,c \mid ab=ba, c=ab\rangle \\ &\to \langle a,c \mid c=a^{-1} c a, c=c\rangle \\ &\to \langle a,c \mid ac=ca\rangle. \end{align}$$

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