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