Elements of order $6$ in an Abelian group of order $360$ Let $A$ be a finite abelian group of order $360$ which does not contain any elements of order $12$ or $18$. How many elements of order $6$ does $A$ contain?  I've got that $A$ is $C_2 \times C_6 \times C_{30}$ but not sure how to work out how many elements of order $6$ there are.  Any help appreciated!
 A: An element has exponent $6$ if and only if each component has exponent $6$. There are $2$ possibilities for the $C_2$ component, $6$ for the $C_6$ component, and $6$ for the $C_{30}$ component (since $C_{30}$ has a unique subgroup of order $6$. That gives $72$ elements of exponent $6$.
Of these elements, you want to subtract those of exponent $3$ and those of exponent $2$, and then add back the element of exponent $1$ (which you subtracted twice).
To get an element of exponent $3$ you need each component to be of exponent $3$: there is one choice for the $C_2$ component, three choices for the $C_6$ component, and three choices for the $C_{30}$ component. This gives you $9$ elements.
To get an element of exponent $2$, you need each component to be of exponent $2$; there are two choices for each of the components. This gives you $8$ elements.
So:
$$\begin{align*}
\#(\text{elements of order 6}) &= \#(\text{elements of exponent 6}) - \#(\text{elements of exponent 3})\\
&\qquad -\#(\text{elements of exponent 2}) + \#(\text{elements of exponent 1})\\
&= 72 - 9 - 8 + 1 = 56.
\end{align*}$$
A: Escentially, since your group is $G = H \times I \times J$ an element of order $6$ means that every component must be of order $6$, $3$, $2$, $1$, this can be proven by looking at the map $G/(H \times J)\to I$. Hence you get elements of the following form (note the numbers represent the orders of the elements): 
$$
(1,1,6)\; (1,6,1)\; (1,6,6)\; (1,2,6)\; (1,3,6)\; (1,2,3)\; (1,3,2)\; (1,6,3)\; (1,6,2)\;
(2,1,6)\; (2,6,1)\; (2,6,6)\; (2,2,6)\; (2,3,6)\; (2,2,3)\; (2,3,2)\; (2,6,3)\; (2,6,2)\; (2,3,1)\; (2,1,3).
$$ 
And you basically take how many elements of each order there is in each subgroup (i.e., $(1,6,1)$ is $1\times 2\times 1=2$), and just multiply them for each set. 
