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?
Best Answer
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 $gcd(m,n)=1$.