List the elements of $\mathbb{Z}_2 \times\mathbb{Z}_4$. Find the order of each of the elements.
I know the elements of $\mathbb{Z}_2 \times\mathbb{Z}_4$ are $$\{(0,0), (0,1), (0,2), (0,3), (1,0), (1,1), (1,2), (1,3)\},$$ but I'm not sure how to find the order of each element.
Do we have to find the identity? Where $x^n=e$ and $|x|=n$?
Everywhere that I have looked just shows the answer but not how to find the solution, I'm conflicted.
Hint $:$ $|(a,b)| = \text {lcm} (|a|,|b|).$
Note that $\Bbb Z_2 \times \Bbb Z_4$ is an additive group where the addition is defined component wise. Let's say $(a,b),(c,d) \in \Bbb Z_2 \times \Bbb Z_4$ then $(a,b) + (c,d) = (a +_2 b , c +_4 d),$ where $+_2$ and $+_4$ are respective additions in $\Bbb Z_2$ and $\Bbb Z_4.$ The order of an element $(a,b) \in \Bbb Z_2 \times \Bbb Z_4$ is the least positive integer $n$ for which $n(a,b) = (na,nb) = (0,0).$ But that implies $na=0$ in $\Bbb Z_2$ and $nb=0$ in $\Bbb Z_4.$ So $|a| \mid n$ and $|b| \mid n$ and hence $\text {lcm} (|a|,|b|) \mid n = |(a,b)|.$
Now let $\text {lcm} (|a|,|b|) = m.$ Then $ma = mb = 0.$ So $m(a,b) = (ma,mb) = (0,0) \implies |(a,b)| \mid m = \text {lcm} (|a|,|b|).$ So we have proved what we have claimed i.e. $|(a,b)| = \text {lcm} (|a|,|b|).$