Counting Elements of Order $2$ in a Direct Product of Cyclic Groups I am not sure if  I am oversimplifying this question or not, but here it is: 

Suppose we have the following direct product of groups: 
  $$G=\mathbb Z/60\mathbb Z \times \mathbb Z/45\mathbb Z \times \mathbb Z/12\mathbb Z \times \mathbb Z/36\mathbb Z$$
  How many elements of order $2$ are in $G$?

For an element in the direct product $G$ to have order 2, we must choose an element in $Z/60Z$, and element in $Z/45Z$,.... in $Z/36Z$ that each have order 2 (or 1). In $Z/60Z$, we have 0, 30 , in $Z/45Z$ we have 0, in $Z/12$ we have 0 and 6, and in $Z/36Z$ we have 0 and 18. 
This gives us 8 possibilities. 
Something feels like I am being insane...
 A: A way that, perhaps, makes it easier to see whether you have the right answer or not:
$$60=2^2\cdot3\cdot5\;,\;\;45=3^2\cdot5\;,\;\;36=2^2\cdot3^2\;,\;\;12=2^2\cdot3$$
Thus, for example, we could write (using the notation $\;C_m:=\Bbb Z_m\;$ and the fact that $\;(n,m)=1\implies C_n\times C_m=C_{nm}\;$) :
$$C_{60}=C_{12}\times C_5\;,\;\;C_{45}=C_9\times C_5\;,\;C_{12}=C_4\times C_3\;,\;\;\;etc.$$
So our group can be expressed as
$$G=\left(C_4\times C_4\times C_4\right)\times \left(C_3\times C_3\right)\times\left(C_9\times C_9\right)\times \left(C_5\times C_5\right) $$
the parentheses being used only for clarity. From this we can express an element in $\;G\;$ as $\;(a_1,a_2,...,a_9)\;$, with $\;a_i\;$ in the i-th factor above, and the product of two elements is coordinatewise, of course.
Now the elements of order $\;2\;$ (=:involutions): there is only one of these in each factor $\;C_4\;$ above (why?). Since the the product of two elements in a finite abelian group is the minimal common multiple of the orders of the two elements, we must have that if $\;(x_1,...,x_9)\;$ is such an element, then $\;x_i=1\;\;\forall\,i\ge4\;$ , and thus all depends on the first three factors there, and if $\;a,b,c\;$ are the corresponding generators of these three factors, then the number of involutions is the number of $\;3-\;$ary vectors $\;(x,y,z)\;$ with at least one of the components non trivial, meaning: or $\;x=a^2\;$ , or $\;y=b^2\;$ , or $\;z=c^2\;$ -- up to here, all the above is explanation of what is the actual calculation -- , and there are eight of these:
$$(a^2,b^2,c^2)\;,\;(a^2,b^2,1)\;,\;(a^2,1,c^2)$$
and the same as above with first component $\;1\;$ , plus the element $\;(a^2,1,1)\;$ . Total: seven involutions.
