# Finding a generator for the external direct product of two finite cyclic groups

I am aware of how one can find a generator of a cyclic group; for example, it is easy to see that $$\langle 5 \rangle =(\mathbb{Z}/14\mathbb{Z}, +)$$ since,

$$5(0)=0 \in \mathbb{Z}/14\mathbb{Z}, \\ 5(1)=5 \in \mathbb{Z}/14\mathbb{Z}, \\ \vdots \\ 5(12)=60 \equiv 4 \: (mod \: 14) \in \mathbb{Z}/14\mathbb{Z}, \\ 5(13)=65 \equiv 9 \: (mod \: 14) \in \mathbb{Z}/14\mathbb{Z}$$.

However, I'm having more difficulty understanding how one can find an explicit generator for the external direct product of two finite cyclic groups (assuming the product group is also cyclic). Specifically, how do you go about finding a generator for the group ($$\mathbb{Z}/m \mathbb{Z} \: \times \: \mathbb{Z}/n \mathbb{Z}$$) where $$gcd(m, n)=1$$? At the link below, it states that one can find a generator for any given finite cyclic group by examining the row beneath the identity in the groups' Cayley Table: Determining whether a group is cyclic from its Cayley Table

Here there is a similar example with a product of two cyclic groups: https://proofwiki.org/wiki/Group_Direct_Product_of_Cyclic_Groups/Examples/C2_x_C3

In examples of the external direct product of two cyclic groups (specifically of the form $$\mathbb{Z}/m \mathbb{Z} \: \times \: \mathbb{Z}/n \mathbb{Z}$$) where we are given that the product is also cyclic, how does one go about finding a suitable generator?

If they are cyclic, $$(1,1)$$ should always generate it. You can tell if $$\mathbb{Z}_m \oplus \mathbb{Z}_n$$ is cyclic by just making sure $$gcd(m,n)=1$$.
• Thank you! Might I also ask: it's well known that $\mathbb{Z}/n\mathbb{Z}$ will have $\phi (n)$ generators (Euler's totient function); would I be correct in assuming that $\mathbb{Z}/m \mathbb{Z} \times \mathbb{Z}/n \mathbb{Z}$ has $\phi (mn)$ generators? – Luke Poeppel Oct 15 '19 at 1:48
• If you know $gcd(m,n)=1$, then $\mathbb{Z}_m \times \mathbb{Z}_n \equiv \mathbb{Z}_{mn}$. This is called the chinese remainder theorem. So then this is absolutely true. You should then be able to just pick any generator for the first group and any from the second group and it will generate the group. A very roundabout way to prove the totient function is multiplicative over things that are relatively prime. – CPM Oct 15 '19 at 1:54