Number of elements of order $45$ in $C_{45} \times C_{15} \times C_{10}$ If I had to do this question, can I say:
$$C_{45} \times C_{15} \times C_{10} \cong C_{9} \times C_5 \times C_3 \times C_5 \times C_2 \times C_5$$
Now, as I want elements of order $45$, lets first look at $C_{45}$. Clearly there are some in here and so splitting it into its powers of primes, I get the number of elements as $\varphi(9) \times \varphi(5)$. Now, the orders that could give an order of $45$ in the other groups are arbitrary and so I would use the number $3,5,5,5$ and so the number of elements of order $45$ in my group would be
$$\varphi(9) \cdot \varphi(5) \cdot 3 \cdot 5^2 = 1800.$$
Is this correct?
 A: Let $G$ be an arbitrary abelian group. Then the following holds.
For each element $g$ of order $mn$ with $(m,n)=1$, there is a unique factorization $g=ab$ with $|a|=m$ and $|b|=n$. [1]
Thus, each element $g$ of order $45$ in $G=C_{45}\times C_{15}\times C_{10}$ defines a unique pair $(a,b)$ with $g=ab$ and $|a|=9$,$|b|=5$, and conversely. So we are really counting all pairs of elements of order $9$ and elements of order $5$.
Looking for order $9$ elements, we see they can come from a combination of the $C_9$ subgroup of $C_{45}$ and the $C_3$ subgroup of $C_{15}$. That is, we are counting order $9$ elements of $C_3\times C_9$. Notice that we cannot pair two elements of order $3$ to get an element of order $9$. It follows that we need to pair arbitrary elements of $C_3$ with elements of order $9$ in $C_9$. This gives $3\cdot\varphi(9)=18$ possibilities.
The subgroups of order $5^n$ are $C_5\subset C_{45}$, $C_5\subset C_{15}$, and $C_5\subset C_{10}$. So we count order $5$ elements in $C_5\times C_5\times C_5$. Well, all elements of this product but the identity have order $5$ in this elementary abelian group. Hence, we have $125-1=124$ elements of order $5$.
Therefore, the number of pairs $(a,b)$ of elements of $G$ with $|a|=9$, $|b|=5$ is the product $18\cdot 124=2232$.

[1] The first lemma of the introduction of Groups of Prime Power Order, Volume 1 by Yakov Berkovich is a stronger version of this statement.
A: Suppose $\,C_{45}=\langle a\rangle\;\;,\;C_{15}=\langle b\rangle\;\;,\;\;C_{10}=\langle c\rangle\,$  , and let us agree to write the elements of $\,G:=C_{45}\times C_{15}\times C_{10}\,$ as ordered thirds $\,(a^i,b^j,c^k)\,$
There are $\,\phi(45)=24\,$ elements of order $\,45\, $ in $\,C_{45}\,$ , so we get the following elements of order $\,45\,$ in $\,G$ :
$$\begin{align*}(1)&\;\text{The elements}&\,(a^i,1,1)&\,\,,\,\,(45,i)=1\\
(2)&\;\text{The elements}&(a^i,b^j,1)&\,\,,\,\,(45,i)=1&\,\,,\,0\leq j\leq 14\\(3)&\;\text{The elements}&\,(a^i,b^j,c^k)&\,\,,\,\,(45,i)=1&\,,\,0\leq j\leq 14&\,,\,k=2,4,6,8\\(4)&\;\text{The elements}&(a^i,1,c^k)&\,\,,\,\,(45,i)=1&\,,\,k=2,4,6,8\\(5)&\;\text{The elements}&\,(a^{5i},b^{3j},c^{2k})&\,,\,i=1,2,...,8&\,\,,\,j=0,1,2,3,4&\,,\,k=0,1,2,3,4\end{align*}$$
...and that's all I count: $\,24+360+1440+96+200=2,120\,$...?
