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

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?

• $(1, 1)$ 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. – Ayman Hourieh Aug 25 at 18:36
• $\mathbb{Z}*\mathbb{Z} = <a,b>, \\ \mathbb{Z}\times\mathbb{Z} = <a,b>/<ab=ba>, \\ \mathbb{Z}_m\times\mathbb{Z}_n = <a,b>/<ab=ba, a^m=1, b^n=1>$? – dcolazin Aug 25 at 18:46
• Here: $$\Bbb Z_m\times\Bbb Z_n\cong\langle a, b\mid a^m, b^n, ab=ba\rangle.$$ – Shaun Aug 25 at 18:49
• 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). – Shaun Aug 25 at 19:21
• Why are so many people answering the question in comments? – Derek Holt Aug 25 at 20:27

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 $$([0]_m, [1]_n)$$.
• 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. – Derek Holt Aug 26 at 8:51