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

  • Thread starter chibulls59
  • Start date
  • #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


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
 

Answers and Replies

  • #2
22,089
3,297
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]...
 
  • #3
Since Aut(Z_p) is the set of maps that maps K-->K, wouldn't that mean that |Aut(Z_p)|=p?
 
  • #4
22,089
3,297
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...
 

Related Threads on |G mod Cg(K)| divides p-1

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
16
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
5
Views
1K
Top