# Calculate the number of elements of order $24$ in $C_{24} \times C_{40}$

This question came up today in my exam, I said:

Splitting my group up will give me $C_8 \times C_3 \times C_8 \times C_5$.

We also have $24 = 8 \cdot 3$, so I want element of $8$ and $3$ in the group I've split up. So the number of elements of $24$ is going to be

$$\varphi(8) \times \varphi(3) \times \varphi(8) = 32$$

Is this correct?

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Not quite, as the element need not have order $8$ in each of those two factors, just in one of them. – Tobias Kildetoft Jan 15 '13 at 14:11
Ohh yeah! So it would be $\varphi(8) \cdot \varphi(3) \cdot 8 = 64$? – Kaish Jan 15 '13 at 14:18

Hint: Look at the $C_8\times C_8$ part. The bad elements of this are the elements $(a,b)$ such that $a$ and $b$ each have order $\lt 8$.

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Write

$$G:=C_{24}\times C_{40}=C_{120}\times C_8:=\left\{(a^i,b^j)\;\;;\;\;0\leq i\leq 119\,\,,\,\,0\leq j\leq 7\right\}$$

There are $\,\phi(24)=8\,$ elements of order $\,24\,$ in $\,C_{120}\,$ , and since $\,8\mid 24\,$ we get the following $\,64\,$ elements:

$$\left\{(a^{5i},b^j)\;\;;\;(24,i)=1\,\,,\,\,0\leq j\leq 7\right\}$$

But we also have all the elements of order $\,3\,$ with all the ones of order $\,8\,$:

$$\left\{(a^{40i},b^j)\;\;;\;\;i=1,2\,\,,\,\,0\leq j\leq 7\right\}$$

So, if I'm not wrong, there are $\,64+16=80\,$ elements of order $\,24\,$ in $\,G\,$

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