Mobius Inversion, finite subgroup

[mathsss2]

The parts of this problem form a proof of the fact that if $$G$$ is a finite subgroup of $$F^*$$, where $$F$$ is a field (even if $$F$$ is infinite), then $$G$$ cyclic. Assume $$|G|=n$$.
(a) If $$d$$ divides $$n$$, show $$x^d-1$$ divides $$x^n-1$$ in $$F[x]$$, and explain why $$x^d-1$$ has $$d$$ distinct roots in $$G$$.
(b) For any $$k$$ let $$\psi(k)$$ be the number of elements of $$G$$ having order $$k$$. Explain why $$\sum_{c|d}\psi(c)=d=\sum_{c|d}\phi(c)$$, where $$\phi$$ is the Euler $$\phi$$-function.
(c) Use Mobius Inversion to conclude $$\psi(n)=\phi(n)$$. Why does this tell us $$G$$ is cyclic?

Here is what I have so far:

a) The first part is trivial [$$y - 1$$ divides $$y^k - 1$$, there values of $$y$$ and $$k$$ will give us our result - I need to show this]. If $$|G| = n$$ then the roots of $$x^n - 1$$ are precisely the elements of $$G$$, which are distinct. (Note how strong this result is. Specifying the order of $$G$$ specifies $$G$$ uniquely.) Since the roots of $$x^d - 1$$ are a subset of the roots of $$x^n - 1$$, they must also consist of $$d$$ distinct elements of $$G$$.

b) $$\sum_{c | d} \psi(c)$$ is the number of elements having order dividing $$d$$, and those are precisely the roots of $$x^d - 1$$. On the other hand, $$x^d - 1 = \prod_{c | d} \Phi_c(x)$$ where $$\text{deg } \Phi_c = \phi(c)$$.

c) The first part is trivial [we use Möbius inversion to uniquely extract the value of $$\psi$$ from the equation. But the equation is the same for $$\phi$$]. Since $$\psi(n) > 0$$, $$G$$ has elements of order exactly $$n$$.

I am still confused on all of the parts. I am not sure if I am doing this right. Any ideas/comments would be great.

 

[rscosa]

Hi!
I can not help you with the part of $\psi$ function but yes with the other part.
It is well known that
$$x^{n}-1=\prod_{d|n}\Phi_d(x),$$ where $$\Phi_d$$ is the dth-ciclotomic polynomial which has degree $$\phi(d)$$, in other words, the euler function of d, so know (a) should be striagtforward. Because if d divides n, then each divisor of d is a divisor of n.
(b) Is a fact of degrees, so is by definition.
I suggest to read the info about ciclotomic polynomial at wikipedia or www.mathworld.com