2

I am trying to show that the $\phi(\phi(n))$ is equal to the number of generators of $\mathbb{Z}_n^*$ for all $n$ such that a generator exists.

Generators only exist for cyclic groups, and so I know that $n$ must be equal to $1,2,4,p^k$, or $2p^k$ where $p$ is an odd prime and $k>0$.

I also know that the number of elements of order $n$ (the number of generators) will be equal to $\phi(n)$

From here though I am unsure where to go. Any help would be greatly appreciated.

TNoms
  • 97
  • Well, you know that if $n$ is of the type you mention, then $\mathbb{Z}_n^*$ is cyclic of order $\phi(n)$, so you just need to show that a cyclic group of order $k$ has exactly $\phi(k)$ generators. – Arturo Magidin Apr 04 '21 at 06:09
  • And I would do this by showing that there will be one generator for every number between $1$ and $k-1$ that is relatively prime to $k$? – TNoms Apr 04 '21 at 06:22
  • That would be one way to try to do it, yes. – Arturo Magidin Apr 04 '21 at 06:23
  • Is it a good way to do it, or is there something "better" perhaps? – TNoms Apr 04 '21 at 06:24
  • 1
    The best way for one to do it is the way in which one succeeds at doing it. – Arturo Magidin Apr 04 '21 at 06:24
  • What should be better than to use that exactly the numbers coprime to $\phi(n)$ do the job ? It is clear that $\phi(\phi(n))$ such numbers exist. – Peter Apr 04 '21 at 07:30
  • Sorry Peter, that fact is not especially clear to me at this stage. – TNoms Apr 04 '21 at 07:38

0 Answers0