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

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Homework Help Overview

The discussion revolves around the cardinality of the centralizer of a normal subgroup K in a finite group G, specifically when |K|=p, where p is a prime number. The original poster is tasked with proving that |G/CG(K)| divides p-1.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to establish a relationship involving the cardinality of CG(K) and its implications for |G|. They express uncertainty about determining |CG(K)|. Other participants introduce the N/C-theorem and discuss the structure of Aut(Z_p), questioning the cardinality of this group and its implications.

Discussion Status

The discussion is active, with participants exploring different aspects of group theory related to the problem. Some guidance has been provided regarding the N/C-theorem and the nature of automorphisms, but there is no explicit consensus on the cardinality of Aut(Z_p) or its implications for the original problem.

Contextual Notes

Participants are navigating the complexities of group theory, particularly the properties of normal subgroups and automorphisms. There is an ongoing examination of definitions and assumptions related to the groups involved.

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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 [tex]N_G(H)/C_G(H)[/tex] 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 [tex]N_G(H)\rightarrow Aut(G)[/tex] to which we can apply the first isomorphism theorem.

Now, the only thing you still need to show is that [tex]|Aut(\mathbb{Z}_p)|=p-1[/tex]...
 
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...
 

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