Present the group $\mathbb{Z}_{m} \times \mathbb{Z}_{n}$, where $\gcd(m, n) > 1$.

I have few questions regarding these type of tasks. As far as I know (still beginner) I need to find elements that generate this group and then find "equations" which will give us enough information so we can form multiplication table for that group. In all groups of this type $\mathbb{Z}_{m} \times \mathbb{Z}_{n}$, $(1,1)$ is generator, and its order is ${\rm lcm}(m,n)$, but I don't know if that is enough to present group.

How can I be sure, when I get few "equations" that I am done?

Any hint is helpful.

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    $\begingroup$ $(1, 1)$ is never a generator when $\operatorname{gcd}(m, n) > 1$. Consider $\mathbb Z_4 \times \mathbb Z_2$. $\langle (1, 1) \rangle$ is a proper subgroup in this example. $\endgroup$ – Ayman Hourieh Aug 25 at 18:36
  • $\begingroup$ $\mathbb{Z}*\mathbb{Z} = <a,b>, \\ \mathbb{Z}\times\mathbb{Z} = <a,b>/<ab=ba>, \\ \mathbb{Z}_m\times\mathbb{Z}_n = <a,b>/<ab=ba, a^m=1, b^n=1>$? $\endgroup$ – dcolazin Aug 25 at 18:46
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    $\begingroup$ Here: $$\Bbb Z_m\times\Bbb Z_n\cong\langle a, b\mid a^m, b^n, ab=ba\rangle.$$ $\endgroup$ – Shaun Aug 25 at 18:49
  • $\begingroup$ To answer your question, a good rule of thumb is to show, however possible it may be, that one can derive each element of the group using both the proposed generators & relations. You're entering into combinatorial-group-theory here, though, where there is many a problem that is undecidable (in the sense that there is no algorithm that will halt in finite time with an answer one way or the other). $\endgroup$ – Shaun Aug 25 at 19:21
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    $\begingroup$ Why are so many people answering the question in comments? $\endgroup$ – Derek Holt Aug 25 at 20:27

To make my comments into an answer . . .

A presentation for $\Bbb Z_m\times\Bbb Z_n$ is

$$\langle a,b\mid a^m, b^n, ab=ba\rangle.$$

One can think of $a$ as $([1]_m, [0]_n)$ and $b$ as $([0]_m, [1]_n)$.

A good rule of thumb for determining whether you're done is to show, whenever possible, that any element of the group can be derived from the candidate generators & relators, and that no other elements can be made; however, it might not be possible to define an algorithm here, since this is , a place where many things are undecidable (in the sense that no Turing machine exists such that, given a presentation, it will halt in finite time whether or not you're done - but there's so many such results that it's difficult to keep track, so don't take my word for it here; this is why I was hesitant about answering rather than commenting).

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    $\begingroup$ Incidentally this presentation is valid whether or not $\gcd(m,n) = 1$, but I guess the point is that if $\gcd(m,n)=1$, then the group is cyclic so there is a shorter presentation with a single geenrator and a single relation. $\endgroup$ – Derek Holt Aug 26 at 8:51
  • $\begingroup$ Sorry if this is random. But why is my post unclear? $\endgroup$ – Arbuja Aug 31 at 12:52

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