Showing that the multiplicative group of roots of $x^n - 1$ in $\overline{F}$ has $\phi(n)$ distinct generators

Question

Basically the question is how can it be shown that the n-th cyclotomic extension, which is a cyclic group under multiplication, has exactly $\phi(n)$ distinct elements, where $\phi$ is Euler's totient function.

I kind of have an idea why, however I am struggling to put it in nice mathematical language. I know that this extension is in general isomorphic to $\mathbb{Z}/n\mathbb{Z}^{x}$, and the generators of this group are numbers that are coprime to $n$, and for any given $n$ in the naturals, there are exactly $\phi(n)$ such numbers. That's my sketch, however I'm afraid it's too native.

Answer 1

In a cyclic group $\langle g \rangle$ of order $n$, the order of $g^k$ is $\dfrac{n}{\gcd(k,n)}$. (*)

In particular, the generators are $g^k$ with $\gcd(k,n)=1$ and so there are $\phi(n)$ of them.

(*) Let $d=\gcd(k,n)$. Then $d=ku+nv$ and so $x^d=(x^k)^u$ and this implies $\langle g^d \rangle=\langle g^k \rangle$.

Answer 2

The Roots of Unity under Multiplication form Cyclic Group of order $n$. If $g$ is a generator, then all generators are given by $g^k$ with $gcd(k,n)=1$.

The group $(\mathbb{Z}/n\mathbb{Z})^{\times}$ has order $\phi(n)$ and hence has $\phi(\phi(n))$ generators. It is the Galois group of the cyclotomic extension $\mathbb{Q}(\zeta_n)$ over $\mathbb{Q}$, where $\zeta_n$ is an $n$-th primitive root of unity.