How many elements of order $4$ are there in $\mathbb{Z}/2$ × $\mathbb{Z}/4$ × $\mathbb{ Z}/6$?

Question

So my thinking is that obviously $\mathbb{Z}/2$ and $\mathbb{Z}/6$ don't have elements of order $4$ (Lagrange's theorem) but I can look at the order of $2$ for both these groups?

Answer 1

Do it directly.

Take $x=(a,b,c) \in \mathbb Z/2 \times \mathbb Z/4 \times \mathbb Z/6 $.

Then $4x=0$ iff $4a \equiv 0 \bmod 2$, $4b \equiv 0 \bmod 4$, $4c \equiv 0 \bmod 6$.

This places no restriction on $a$ or $b$ but requires $c \equiv 0,3 \bmod 6$.

Now, both $0$ and $3$ have order $2 \bmod 6$. So, for $x$ to have order $4$, we need $b$ to have order $4$. Therefore, we have $8$ elements of order $4$: $a=0,1$; $b=1,3$; $c=0,3$.

Answer 2

$\mathbb{Z}/4\mathbb{Z}$ has $2$ elements of order $4$ and hence the group has exactly $2 \cdot 2 \cdot 2=8$ elements of order $2$, since $\mathbb{Z}/6 \mathbb{Z}$ has two elements of order dividing $2$ and so does $\mathbb{Z}/2\mathbb{Z}$.