Number of elements of order 4 in $\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_2$

Question

I've read that number of elements of order 4 in $\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ is 8, but I can't figure it out where this number come from. I now that if we're searching number of elements of order 4 in $\mathbb{Z}_{16}$, we use Euler's phi function ($\phi(4)=2$). Do we use something similar here?

Answer 1

The order of an element in a direct product like this one is the l.c.m of the orders inside the respective coordinates. So an element of order $4$ in $\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ must be $1$ or $3$ in the first coordinate. In the second and third coordinate, you could have either $0$ or $1$. In each case you get elements of order $4$. There are $2 \cdot 2 \cdot 2 = 8$ such possibilities.

Answer 2

Suppose that $(a,b,c)$ is an element of order $4$. As $2b=2c=0$, $a$ has to be an element of order $4$. Therefore $a \in \{1,3\}$.

And then $b,c$ can be whatever you want. Therefore $8$ elements of order $4$.