Solving Cyclic Group Questions: How Many Elements of Order What?

[jimmycricket]

I was hoping someone could check the following solutions to these 3 basic questions on cyclic groups and provide theorems to back them up.

1. How many elements of order 8 are there in $C_{45}$?

Solution: $\varphi(8)=4$

2. How many elements of order 2 are there in $C_{20}\times C_{30}$?

Solution: $C_{20}\times C_{30}\simeq C_2\times C_4\times C_3\times C_5\times C_5$
Since 2 is coprime to 3 and 5 the question is equivalent to how many elements are there in $C_2\times C_4$? Answer is $\varphi(2)\varphi(4)=3$

3. How many elements of order 35 are there in $C_{100}\times C_{42}$?

Solution: Since 2,4 and3 are coprime to 35 the question is equivalent to how many elements of order 35 are there in $C_{25}\times C_7?$
Answer is $\varphi(25)\varphi(7)=120$
Now this sounds a bit large to me. A peer of mine said the answer is $\varphi(7)\varphi(5)=24$. Can anyone explain the reasoning behind this?

 

[conquest]

Could you tell me what you mean by order of an element (and more importantly φ). Usually one would say, the order of an element a is the smallest positive integer n such that aⁿ=e. e denoting the unit of the group. In this case Lagrange's theorem

http://en.wikipedia.org/wiki/Lagrange's_theorem_(group_theory)

would contradict the answer of 1. Since 8 does not divide 45.

For 2. Although I have no idea what you mean by φ(2)φ(4), there are indeed 3 elements of order 2 in that product following the basic reasoning you set out.

For 3. the second answer seems correct, since I don't know what you mean by the φ I can't give a reasoning to it as such. Your reasoning seems sound up to the point where you use the φ. Maybe you should simply try thinking about the question in terms of arithmetic modulo the relevant numbers (25 and 7 in this problem). I don't understand how a function φ can exist such φ(25)φ(7) calculates the number of order 35 elements in anything since the expression φ(25)φ(7) would not refer to the group or the number 35 (the same would go for 2.).

I would be glad to give further comments after you give me the answer to the riddle that is φ!

 

[jimmycricket]

Sorry I should have explained that $\varphi(x)$ is Eulers' totient function i.e. the number of integers coprime to x. Thanks for your reply I have now resolved the issue. First restrict the direct product of cyclic groups to a direct product of only groups who share a factor with the order of the elements you are trying to find. Then find all the factors of the orders of the cyclic groups and find all pairs of these factors that have lowest common multiple equal to the order of the elements you are trying to find. For each of these pairs $(a,b)$ calculate $\varphi(a)\varphi(b)$ and then do the summation of these products for each pair.
So for 2: We have $C_2\times C_4$. Factoring the orders gives us 1,2 and 1,2,4 as factors of the orders. Now finding the pairs who have LCM=2 gives (1,2), (2,1), (2,2). Now $\varphi(1)\varphi(2)+\varphi(2)\varphi(1)+\varphi(2)\varphi(2)=1+1+1=3$ elements of order 2.