Verifying Sylow-p Groups of Sp: A Homework Exercise

[TheForumLord]

Homework Statement

This question is about sylow-p groups of Sp.

I've proved these parts of the question:
A. Each sylow p-sbgrp is from order p and there are (p-2)! p-sylow sbgrps of Sp.
B. (p-1)! = -1 (mod p ) [Wilson Theorem]

I need your help in these two :
C. 1) Let G be a group of order n*p^(k) where gcd(n,p)=1 and k>=1.
Prove that the normalizers of the p-sylow sbgrps of G are conjugated

C. 2) Make use of C.1), and find the order of the normalizer of a sylow-p-sbgrp of Sp.

Homework Equations

The Attempt at a Solution

About c1:
I proved it using the sylow-theorem that says that two sylow sbgrps are conjugated... I'm not sure that is the thing with the order... Is it matter that the order is n*p^k? The argument is also true for an arbitrary group, no?

About c2:

I'm not that sure, but according to the theorem that says:
o(G)/o(N(P)) = Number of conjugates to P

We'll get that the order of the normalizer of N(P) (where P is a sylow-p-sbgrp) is p! (the order of Sp) divided by the number of conjugates to P, which are all the elements of the normalizers of the sylow-p-sbgrps...
Which means that it's p! / (p-2)!*p = (p-1)! / (p-2)! = p-1..




Verification and help are needed!

TNX everyone!

 

[mighty2000]

I can confirm that your proof for C.1) is correct. The Sylow theorem states that any two Sylow p-subgroups are conjugate, which means that their normalizers must also be conjugate. This is true for any group, not just Sp.

For C.2), your reasoning is also correct. The order of the normalizer of a Sylow p-subgroup will be equal to p! divided by the number of conjugates to P, which is equal to (p-1)! divided by the number of Sylow p-subgroups, which we already know to be (p-2)! according to A. Therefore, the order of the normalizer of a Sylow p-subgroup of Sp is (p-1)! / (p-2)! = p-1.

Great job on your proof and reasoning! Keep up the good work.