# 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!

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Your title has a typo. Please check for typos in what is the most visible part of your question! – Mariano Suárez-Alvarez May 22 '12 at 21:04

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*}

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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.

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Moreover, the other answer is wrong by simply looking at the group ((Z2,+)X(Z3,+)) which has two elements of order 6 (1,2) and (1,1) but the components have no elements of order 6 – Pioter May 23 '12 at 2:30