Order of the product of two cyclic groups

Question

Let $G_1,G_2$ be two finite cyclic groups with orders $\text{ord}(G_1)=n_1, \text{ord}(G_2)=n_2$ and $G:=G_1\times G_2$. Why is $\text{ord}(G)=n_1\cdot n_2$ if $G$ is cyclic?

I know that there are $a_1,a_2 \in G_i$ with $G=\langle a_1,a_2\rangle $, but which theorem says that $\text{ord}(a_1,a_2)=\text{ord}(a_1)\cdot \text{ord}(a_2)=n_1\cdot n_2$?

Answer 1

$|G_1 \times G_2| = |G_1| \times |G_2| = n_1 n_2$, which follows from the formula for the cardinality of the cartesian product of sets. Also, $G$ is only cyclic if $\gcd(n_1,n_2) = 1$, see here