Would someone explain the class equation and its corrolaries for me?

[AdrianZ]

I've read about the class equation on Herstein's abstract algebra but haven't understood it well. Would someone here explain the class equation to me and give a short proof for it and then tell me where it can be used?

 

[Deveno]

ok, i'll start with the highest-level explanation i can, and if we need to, we'll work our way down.

first of all, we can define an equivalence relation on G, by x~y iff x = gxg^-1 for some g in G.

an equivalence relation partitions a set into disjoint subsets, the equivalence classes. so it is natural to ask what is the equivalence class of x, [x], under ~?

well, clearly it is the set of all conjugates of x, {gxg^-1: g in G}.

now, consider the subgroup N(x) = {g in G: gx = xg}, that is, every element of G that commutes with x. if n is in N(x), then nxn^-1 = xnn^-1 = x.

so every element of N(x) just gives x upon conjugation.

what we want to show is that there is a bijection between the set of all conjugates of x (that is, the equivalence class [x]) and the set of left cosets gN(x).

suppose two conjugates of x are the same:

so gxg^-1 = hxh^-1. then,

x = g^-1hxh^-1g = (g^-1h)x(g^-1h)^-1, that is:

(g^-1h)x = x(g^-1h).

this means that g^-1h is in N(x), so gN(x) = hN(x).

this means that the map gN(x) → gxg^-1 is injective, provided it is well-defined.

well suppose h is in gN(x), so h = gn, for some n in N(x).

then hxh^-1 = (gn)x(gn)^-1 = g(nxn^-1)g^-1

= gxg^-1 (since n is in N(x), and thus commutes with x).

so our map is indeed well-defined, it only depends on the coset gN(x), and not on the element we pick from it to represent it.

clearly this map is surjective, too, since any conjugate gxg^-1 of x has the pre-image the coset of N(x) which contains g.

so gN(x) → gxg^-1 is indeed a bijection between the (left) cosets of N(x) and the conjugates of x in [x].

but we can count the number of cosets of a subgroup rather easily, it is the index of the subgroup in G.

thus: |[x]| = [G:N(x)], there are exactly as many conjugates of x, as there are cosets of the normalizer of x.

now |G| is just the sum of the sizes of these equivalence classes, so we have:

$$|G| = \sum_{[x]} [G:N(x)]$$, where we count each equivalence class only once.

this is almost the class equation. one more wrinkle to work out. if x is in the center of G, so that xg = gx, for every g in G, it's normalizer N(x) is all of G. so in this case, [G:N(x)] = [G:G] = 1.

rather than sum these all "one at a time", we just count them "all at once" by finding the order of the center. this gives us:

$$|G| = |Z(G)| + \sum_{x \not \in Z(G)} [G:N(x)]$$

where again, we only sum over distinct conjugacy classes.

 

[spamiam]

Isn't N(x) (as you've defined it) actually the centralizer of x?

Edit: I see, the centralizer is the normalizer of $\{x\}$. Sorry!

 

[wisvuze]

the normalizer is the centralizer if you take S to be just an element i.e. x^-1 S x = S

 

[Deveno]

for a set S, with |S| > 1, the centralizer of S, C_G(S) = {g in G: gx = xg, for all x in S}, whereas the normalizer of S, N_G(S) = {g in G: gSg^-1 = S}.

but if |S| = 1, that is: S = {x}, then

gxg^-1 = x is equivalent to gx = xg.