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

  1. Dec 1, 2011 #1
    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. jcsd
  3. Dec 1, 2011 #2


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    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:

    [tex]|G| = \sum_{[x]} [G:N(x)][/tex], 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:

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

    where again, we only sum over distinct conjugacy classes.
  4. Dec 1, 2011 #3
    Isn't N(x) (as you've defined it) actually the centralizer of x?

    Edit: I see, the centralizer is the normalizer of [itex] \{x\}[/itex]. Sorry!
  5. Dec 1, 2011 #4
    the normalizer is the centralizer if you take S to be just an element i.e. x^-1 S x = S
  6. Dec 1, 2011 #5


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    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.
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