What is the Cardinality of CG(K) in G mod Cg(K) when |K| = p and p is prime?

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Homework Statement


Let G be a finite group and K a normal subgroup of G
If |K|=p where p is a prime
Prove that |G/CG(K)| divides p-1


Homework Equations





The Attempt at a Solution


I must show that |G| / |CG(K)| * something = p-1
I figured a good place to start would be to determine the cardinality of CG(K) which I'm having trouble with.
So I just started writing down things that would help
Since K normal in G => NG(K)=G
and since CG(K) is normal in NG(K) => CG(K) is normal in G
 
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Try to prove the following nice theorem:

N/C-theorem: If H is a subgroup of G, then N_G(H)/C_G(H) is isomorphic to a subgroup of Aut(G).

The proof of the theorem shouldn't be too difficult, as we can easily find a group homomorphism N_G(H)\rightarrow Aut(G) to which we can apply the first isomorphism theorem.

Now, the only thing you still need to show is that |Aut(\mathbb{Z}_p)|=p-1...
 
Since Aut(Z_p) is the set of maps that maps K-->K, wouldn't that mean that |Aut(Z_p)|=p?
 
chibulls59 said:
Since Aut(Z_p) is the set of maps that maps K-->K, wouldn't that mean that |Aut(Z_p)|=p?

I don't immediately see why this would be the case. Can you give me p elements in Aut(Z_p) then?

Recall that elements in Aut(Z_p) are not simply maps Z_p--> Z_p, but this maps must actually be bijections!

Anyways, you could start by actually writing down some elements in Z_p and maybe noticing a general pattern...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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