Do Euler's Totient Function and Relative Primality Determine Group Generators?

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Homework Statement


Does Euler's totient function tell me how many elements are in my group?

And once I know how many elements are in my group. the generators are the ones that are relatively prime with the number of element in my group.
Are my statements correct.
 
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cragar said:

Homework Statement


Does Euler's totient function tell me how many elements are in my group?

And once I know how many elements are in my group. the generators are the ones that are relatively prime with the number of element in my group.
Are my statements correct.

Could you please clarify a bit? What group exactly are you talking about? Are you talking about the invertible elements of \mathbb{Z}_n??
 
I am talking about the elements in Z(star)n . so the elements in my group have no common factors other than 1 with n,
 
Yes, the number of elements in 2n is indeed phi(n). With phi the Euler totient function.
But your statement of the generators is not true. It's very difficult to find generators for such a groups...
 
thanks for your help, my book is not very clear on how to find generators. Should I try to read another book on finding generators or make a Cayley table.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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