Find n such that the group of the n-th roots of unity has exactly 6 generators

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SUMMARY

The discussion centers on finding the integer \( n \) such that the group of the n-th roots of unity has exactly 6 generators. The solution involves the Euler's totient function \( \varphi(n) \), which counts the positive integers up to \( n \) that are relatively prime to \( n \). The equation to solve is \( \varphi(n) = 6 \). Through computation, it is established that \( n = 7 \) is the only solution, as \( \varphi(7) = 6 \).

PREREQUISITES
  • Understanding of cyclic groups and their properties
  • Familiarity with Euler's totient function \( \varphi(n) \)
  • Basic knowledge of number theory concepts
  • Ability to perform calculations involving integers and their properties
NEXT STEPS
  • Study the properties of cyclic groups in abstract algebra
  • Learn more about Euler's totient function and its applications
  • Explore the relationship between group theory and number theory
  • Investigate other values of \( n \) for which \( \varphi(n) \) yields specific integers
USEFUL FOR

Students of abstract algebra, mathematicians interested in group theory, and anyone studying number theory concepts related to generators of cyclic groups.

AutGuy98
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Hey guys,

Sorry that it's been a decent amount of time since my last posting on here. Just want to say upfront that I am extremely appreciative of all the support that you all have given me over my last three or four posts. Words cannot express it and I am more than grateful for it all. But, in light of that, I actually have some more questions for an exercise set that I have to do for one of my classes and I'm really unsure how to begin doing them. There are four of them and they all require proofs to some degree. Anyway, I was going to make one post and put all four parts of the same question in it (i.e. 2(a),2(b),2(c), and 2(d)), but was unsure whether or not it would be allowed here. So, for those reasons and to play it safe rather than try to do so, here is the fourth and final part that I've been having trouble with. Any help here is, once again, greatly appreciated and will leave me forever further in your gratitude.

Question: 2(d): "Find n such that the group of the n-th roots of unity has exactly 6 generators."

Again, I have no idea where to start with this, so any help is extremely gracious and appreciated.

P.S. If possible at all, I'd need help on these by tomorrow at 12:30 E.S.T., so please try to look this over at your earliest conveniences. Thank you all again for your help with everything already.
 
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Hi AutGuy98,

A cyclic group of order $n$ has $\varphi(n)$ generators, where $\varphi(n)$ is the number of positive integers not exceeding $n$ and relatively prime to $n$. The question is asking you to solve the equation $\varphi(n) = 6$ for $n$.
 
Euge said:
Hi AutGuy98,

A cyclic group of order $n$ has $\varphi(n)$ generators, where $\varphi(n)$ is the number of positive integers not exceeding $n$ and relatively prime to $n$. The question is asking you to solve the equation $\varphi(n) = 6$ for $n$.

Hi Euge,
The same situation applies here as was the case with 2(b) and 2(c). I only have a small amount of understanding when it comes to what is going on in my Elem. Abstract class, so if you would be so kind as to please just provide me with the solution whenever you are able, as I am still heavily pressed for time and honestly have no idea where to go with the advice you gave me, nor do I have that much time to figure it all out. Thank you again for your help with 2(a) and 2(b) though, that helped relieve the tension off of me a tremendous amount.
 
You can compute $\varphi(n)$ for small values of $n$: $\varphi(1) = 1$, $\varphi(2) = 2$, $\varphi(3) = 2$, $\varphi(4) = 2$, $\varphi(5) = 4$, $\varphi(6) = 2$, $\varphi(7) = \color{red}{6}$. Thus $n = 7$.
 

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