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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?

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No, for example you are not counting the element $(5,1,0)$ as an element of order $45$. – Cocopuffs Jan 14 '13 at 0:43
I get $2232$... – Alexander Gruber Jan 14 '13 at 1:02
$2232=3\cdot6\cdot124$ – peoplepower Jan 14 '13 at 2:37
up vote 2 down vote accepted

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.

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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: $\,24+360+1440+96+200=2,120\,$...?

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We also have $(a,b,1)$ with $|a|=9$ $|b|=5$. – peoplepower Jan 14 '13 at 2:51
True! Thanks, @peoplepower. I'll add this...and also others. – DonAntonio Jan 14 '13 at 2:52
I thought there was an overlap.. Notice that $(a^i,1,1)$ shows up in both (1) and (2). Alternatively, here is a way to combine the first four cases: $(a^i,b^j,c^k)$ with $(45,i)=1$ and $0\leq i<45$; $0\leq j<15$; $(5,k)=1$ and $0\leq k<10$. The fifth case actually finishes it, but the $b^{3j}$ is too restrictive. We can actually have $b^j$ with $0\leq j< 15$. – peoplepower Jan 14 '13 at 3:37
Yes, I see your point...that's what happens when I, whom suck big time at counting, decide to deal with this. I'll keep the answer so that anyone can see all. It is (midly) interesting to know whether there's some definite formula to calculate this mess. – DonAntonio Jan 14 '13 at 4:11

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