Find an element such that its order equals the cardinality of the group of cartesian product

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I've been stuck on this question for quite some time and can't seem to be able to solve it (it's in my algorithms course).

Consider the group: $$G = (\Bbb Z_6 \times \Bbb Z_8, \oplus)$$

where: $$(a, b) \oplus (c, d) = (a+c \bmod 6,b + d\bmod 8)$$

Find an element $a \in G$ such that: $${\rm ord}(a) = |G|$$

I have been able to find the identity element which I think is $(6,8)$, but I still don't understand how to calculate the order of an element in $G$.

Say I take an element like $(2,4)$, its order, if I'm not mistaken, should be the smallest integer $m$ such that $a^m=e$ How do I calculate $a^m$? I tried $(a,b)\oplus(a,b)$ but can never get $(6,8)$ (the identity element).

I can't find what I'm doing wrong, and I'm obviously doing something wrong.

Thank you.

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If $G_1$ and $G_2$ are two groups and $G= G_1 \times G_2$ is the usual Cartesian product between them ( i.e $ (a,b) * (c,d) = (a\cdot c, b \cdot d)$, where the first $\cdot$ is operation in $G_1$ and the second in $G_2$), then order of any element $(a,b)$ in $G$ is least common multiple of order of $a$ in $G_1$ and order of $b$ in $G_2$. In your case, order of element in $Z_6$ is a factor $x$ of 6 and order of any element in $Z_8$ is a factor $y$ of 8. For any such $x$ and $y$, least common multiple would be 24. But the cardinality of your group is 48. Hence no such element exists.