Calculating the order of an element in group theory Calculate the order of the elements
a.) $(4,9)$ in $\mathbb{Z_{18}} \times \mathbb{Z_{18}}$
b.) $(8,6,4)$ in $\mathbb{Z_{18}} \times \mathbb{Z_{9}} \times \mathbb{Z_{8}} $
I have the solutions given as
a.) $(4,9)^{18}= (0,0)$ so the order is 18
OR 
the orders of 4,9, in their respective groups are 9 and 2, so the lowest common multiple is 18
b.)$(8,6,4)^{18}=(0,0,0)$ so the order is $18$
OR
the orders of $8,6,4$ in their respective groups are $9,17$ and $6$ whose lowest common multiple is 18.
So as you can see I have two solutions for each question, but I do not understand how they have got, say $(8,6,4)^{18}=(0,0,0)$ so the order is $18$ or on the second styles of solution, how they have got the orders of $8,6,4$ in their respective groups as $9,17$ and $6$?
I understand how they have got lowest common multiple obviously but do not understand how they have found the orders of the given elements in ther respective groups. Could someone please explain so that i can start to understand this topic better. Many thanks for any help.
 A: The first style of solution is not good. In fact, in the first case, we can say $(x,y)^{18}=(0,0)$ for all $(x,y)\in\Bbb Z_{18}\times\Bbb Z_{18},$ but we certainly can't say that the order of every element of $\Bbb Z_{18}\times\Bbb Z_{18}$ is $18.$ The latter approach is the way to go.
Now, to find the order of an element $m=1,2,3,\dots,n-1$ of $\Bbb Z_n,$ you should start by finding the least common multiple of $m,n.$ That least common multiple, divided by $m,$ will be the order of $\Bbb m$ in $\Bbb Z_n.$ Alternately, divide $n$ by the greatest common divisor of $m,n.$ (It amounts to the same thing.) The order of $0$ in $\Bbb Z_n$ is $1.$
A: I think it is about additive order of given elements and
e.g. $(4,9)^{18}$ is meant as $18\cdot (4,9)=(18\cdot 4,18\cdot 9)$ which is indeed $(0,0)$ in $\Bbb Z_{18}\times\Bbb Z_{18}$. The point is that no smaller 'power' (rather to be called 'multiple' in the additive case) of $(4,9)$ will give $(0,0)$.
The (additive!) order of $4$ is $\displaystyle\frac{18}{\gcd(4,18)}=9$, that of $9$ is $2$ (there are only two multiples of $9$ in $\Bbb Z_{18}$, $9$ and $0$). So when the first coordinate $4+\dots+4$ becomes $0$ first, we will have $9$ pieces of $4$, but at the same time, as $9$ is odd, the second coordinate will be $9$. So, the order of $(4,9)$ in $\Bbb Z_{18}\times \Bbb Z_{18}$ is indeed $18$.
