I don't think I believe that this is true. If G is abelian, then the normalizer of any subgroup is equal to G.
Let be a group of order where is prime, and let be a proper subgroup of . Prove that where is the normalizer of .
I have not yet studied theorems on p - groups. All I have done is theorems and concepts of subgroups, cosets and the Lagrange's Theorem. Can this question be solved using these concepts?