Prove $\mathbb{Z} \times \mathbb{Z} / \left\langle (6,9)\right\rangle$ has an element of order 3 I am looking at this released exam question number two which states:

Let $G$ be the group $\mathbb{Z} \times \mathbb{Z}$, let $a = (6,9) \in G$. Prove that $G/\langle a \rangle$ has an element of order 3.

I am not sure what is meant by the group $\mathbb{Z} \times \mathbb{Z}$, is it implied to be under addition or multiplication?
I have a feeling that the answer is to choose:
$$(2,3)$$
$$(2,3)^2=(4,6)$$
$$(2,3)^3=(6,9) \stackrel{?}{=} (0,0)$$
But: 


*

*that would be assuming under addition, however I cannot think of a situation where multiplication would work.

*Doesn't $G/ \langle a \rangle$ mean that there is no $(6,9)$ in this set and $(2,3)^3$ would be undefined?

 A: The Cartesian product of groups follows the natural definition. In this case, we have $(z_1,z_2) + (y_1,y_2) = (z_1+y_1,z_2+y_2)$. Multiplication here is not a group operation as we are dealing with $\mathbb{Z}$, which has no multiplicative inverses for any elements that are not 1 or -1. 
Recall $\left\langle a\right\rangle = \{ ...(-6,-9),(0,0),(6,9),(12,18),(18,27),... \} = \{(z6,z9) | z \in \mathbb{Z}\} $, i.e. integer multiples of $(6,9)$. (Here, multiplication stands in only for repeated addition and notional convenience.) 
The quotient group $G/\left\langle a\right\rangle$ follows the normal definition for quotient groups, i.e. we associate the element $(1,2)$ with $(7,11)$, $(13,20)$, $(-5,-7)$... i.e., every element $(z_1,z_2)$ represents the equivalence class $(z_1+k6,z_2+k9)$ for any $k \in \mathbb{Z}$. In this case, $\mathbb{Z}\times \mathbb{Z} / \left\langle (6,9)\right\rangle = \mathbb{Z}_6 \times \mathbb{Z}_9$, so we must show some element of $\mathbb{Z}_6 \times \mathbb{Z}_9$ has an element of order 3. You are right in your choice of $(2,3)$, as $\overline{(2,3)} + \overline{(2,3)} + \overline{(2,3)} = \overline{(6,9)} = \overline{(0,0)}$, so $\overline{(2,3)}$ has order 3.
