Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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

2 Answers 2

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 $\lt 8$.

share|improve this answer

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 $\,G\,$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.