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

1. Dec 1, 2011

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?

2. Dec 1, 2011

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.

3. Dec 1, 2011

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!

4. Dec 1, 2011

wisvuze

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

5. Dec 1, 2011

Deveno

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

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

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