# |G mod Cg(K)| divides p-1

## 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

## 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

## Answers and Replies

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?

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...