This writeup relates to group theory, a part of abstract algebra.

The normalizer of a subset H of a group G is the set N(H) = {all x in G | H^{x} = H}, where H^{x} is the conjugate of H by x (H^{x} = all y in G | xyx^{-1} is in H}.

It is easy to show that the normalizer of any subset H is a subgroup of the group G. It's also pretty obvious that if H is a subgroup of G, then N(H) is the largest subgroup of G in which H is normal.

**Important:** The number of distinct subsets in G that are conjugate to a given subset H is o(G)/o(N(H))^{*}.

* I don't see a reason to prove all of these theorems. The proofs are quick and straightforward. If this material is interesting to you, you should understand it well enough to prove them yourself!

Clearly the normalizer of any set with a single element a is just the set of all elements that commute with a (i.e. all elements b such that ab = ba). This follows immediately from the definition of the normalizer. If an element a belongs to the center of a group G (all elements in the group that commute with *every* element in G--the center is a subgroup of G), then N(a) = G. Moreover, the center Z_{G} = {all x in G | N(x) = G}.

Since the center of a group G and the normalizers of elements not in the center are subgroups of G, the following, called the conjugacy class equation, can be written.

o(G) = o(Z_{G}) + (sum over i) o(G)/o(N(x_{i})),

where x_{i} is an element of a conjugacy class (the set of all elements conjugate to a single other element) and i runs through all conjugacy classes.