To show N(H) is a subgroup take and show . This is fairy simple if you first note that if then .
The second part is a result of the counting formula which states that the size of the orbit is equal to the index of the stabilizer . and under the group action conjugation. The LHS is the orbit of H under conjugation, so these sets are equal in cardinality.
Plus there is a really easy bijection to establish this fact. by . Surjectivity is clear.
Injectivity and well-defined is also pretty obvious.