How to prove that a Group is cyclic?

Question

I have to prove that $\mathbb{Z}_2\times \mathbb{Z}_3$ is cyclic by finding a generator. Using the usual operation on product group, addition, I found that $$ (1,1)+(1,1)+(1,1)+(1,1)+(1,1)+(1,1)=(0,0). $$ Hence we have that $(1,1)$ has order $6$. How to finish this argument? Could be enough my arguments such that $\mathbb{Z}_2\times \mathbb{Z}_3$ to be cyclic? What is the point with the generator?

Answer 1

Let $k(1,1)=(0,0),$ where $k\in\mathbb N$.

Thus, from the first coordinate $k$ is divisible by $2$ and from the second coordinate $k$ is divisible by $3$.

Can you end it now?