Let $G=\mathbb{Z_{10}}\times \mathbb{Z_{15}}$. Then find the number of elements.

Question

Let $G=\mathbb{Z_{10}}\times \mathbb{Z_{15}}$. Then,

$G$ contains exactly one element of order $2$.
$G$ contains exactly $5$ element of order $3$.
$G$ contains exactly $24$ element of order $5$.
$G$ contains exactly $24$ element of order $10$.

I know how to calculate number of elements when external direct product of groups is given. But in this case I don't know how to proceed. Please give me some hints. Thanks.

Please specify what you mean by an external direct product, and why this isn't one. — Daron, Nov 01 '15 at 14:43
Sorry actually I got caught in the notations, this is indeed the external direct product. — Kushal Bhuyan, Nov 01 '15 at 14:48
It might be useful to exploit that this is isomorphic to, e.g., $\Bbb Z_2 \times \Bbb Z_3 \times \Bbb Z_5 \times \Bbb Z_5$. — Travis Willse, Nov 01 '15 at 15:00

score 2 · Accepted Answer · answered Nov 01 '15 at 14:46

Hint: If the element $a \in \mathbb Z_{10}$ has order $A$ and the element $b \in \mathbb Z_{15}$ has order $B$ then the order of the element $(a,b) \in \mathbb Z_{10} \times \mathbb Z_{15}$ is the least-common-multiple of $A$ and $B$.

So in order that $(a,b)\in \mathbb Z_{10} \times \mathbb Z_{15}$ has order $2$ we must have $A,B \le 2$. But if $A=B=1$ then $(a,b) = (0,0)$ has order $1$. So either $\{A,B\} = \{2,1\}$ or $\{A,B\} = \{2\}$. Now look at each of $\mathbb Z_{10} $ and $\mathbb Z_{15}$ to determine what $a$ and $b$ can be.

Let $G=\mathbb{Z_{10}}\times \mathbb{Z_{15}}$. Then find the number of elements.

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