1. The problem statement, all variables and given/known data 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. 2. Relevant equations 3. 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!