I need to find the number of elements of order 9 and 6 in the group $C_4 \times C_9 \times C_3 \times C_3$. I have read the solution and they are supposed to be 54 elements of order 9 and 26 elements of order 6, but I am not sure how they got that number.
The order of an element $(w,x,y,z)$ is the least common multiple of the orders of $w$,$x$,$y$ and $z$. So an element of order 9 will have to be of the form $(1,x,y,z)$, where $x$ has order 9, while an element of order 6 is of the form $(w,x,y,z)$, where $w$ has order 2 and $x$, $y$ and $z$ all have order dividing 3 and at least one of them has order 3.
Does that help?