# Direct product of finite cyclic groups of coprime orders [duplicate]

The Question is this:

How many generators are there of the group $G\times H$, if $G$ and $H$ are cyclic groups of order $m$ and $n$, which are coprime?

Let's say that $G$ is generated by $g$, and $H$ by $h$. Here I already proved that $(g,h)$ is generator of $G\times H$ but I can't come up with another one. So does $G\times H$ really have any other generator?

Thanks!

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## marked as duplicate by YACP, vonbrand, Asaf Karagila, rschwieb, Dominic MichaelisApr 3 '13 at 10:51

Yes, the tuples $(g,h)$ - where $\langle g\rangle$ and $\langle h\rangle=H$ - exhaust precisely all generators for $G\times H$. For if $\pi_G,\pi_H$ are the coordinate projections $G\times H\to G,H$ respectively, then $\langle (a,b)\rangle=G\times H$ under $\pi_G$ yields $\langle a\rangle=G$, and similarly under $\pi_H$ yields $\langle b\rangle=H$. –  anon Apr 3 '13 at 3:22
(I post the above as an abstract, algebraic alternative to the other arguments given, which are counting in nature. Also I meant to type $\langle g\rangle=G$ instead of just $G$.) –  anon Apr 3 '13 at 3:29
@YACP: Let's not mark this as a duplicate of a question which is already a duplicate... –  TMM Apr 3 '13 at 10:12

A group $G\times H$ is cyclic if and only if, given $|G| = m$ and $|H| = n$, $\gcd(m,n) = 1$. And if $\gcd(m,n) = 1$, then the order of $G\times H$ is $\text{lcm}(|G||H|) = mn$

A cyclic group is by definition, generated by a single element, so if you've found that $(g, h)$ generates $G\times H$, then you've shown that $\langle (g, h)\rangle = G\times H$.

Indeed, any integer relatively prime to the modulus of a cyclic group will additively generate the cyclic group. That is, the number of elements generating the cyclic group $G = \phi(m)$, the number of elements generating the cyclic group $H = \phi(n)$. So the number of elements generating $$G\times H = \phi(m)\cdot\phi(n)$$

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(Note: "modulus" is not a term reserved for cyclic groups in general, as an isomorphism class of groups. Rather, it is reserved specifically for working with integers mod $n$.) –  anon Apr 3 '13 at 3:27
Thank you, anon! Corrected. –  amWhy Apr 3 '13 at 3:30
Thank you guys! That is really helpful. –  user68498 Apr 3 '13 at 3:45
You're welcome, user68498! Do you understand the Euler phi function, (aka Euler totient function) and what it counts? –  amWhy Apr 3 '13 at 3:47

Hint: If you've proven that $G\times H$ is cyclic, then it must be of order $mn$. What do the generators of a cyclic group of order $mn$ look like? Think of a generator with a relatively prime power to the order...

As a bonus, we can even say it has $\varphi(mn)=\varphi(m)\cdot\varphi(n)$ generators, where $\varphi$ is the Euler phi function.

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Hint: If a group of order $n$ has a generator then it has precisely $\phi(n)$ generators.

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