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Class equation

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data

    Given a class equation of a group H of 20=1+4+5+5+5, does H have a subgroup of order 5? Of order 4?

    3. The attempt at a solution

    Order 5 I can't get, but for order 4 I think I am correct in saying that H does have a subgroup of order 4 because each summand in the class equation is either |Z(H)| or is the index of the normalizer subgroup. So H has 3 subgroups of order 4.

    Is this enough of an answer for the subgroup of order 4 part?

    And any help on the order of 5 part?

    Thanks
     
  2. jcsd
  3. Nov 1, 2011 #2
    I think my answer for the order 4 part might not be enough. Is there a general method for this type of question?
     
  4. Nov 1, 2011 #3
    This question is from Artin (I remember doing this problem before!). But I've forgotten all of it. Anyway, let me see if I can help. The order of the group divided by the order of the conjugacy class of x in G gives you the order of the centralizer of x (see p. 196 of Artin), and the centralizer of an element is a subgroup of G. So isn't the answer yes to both? However, there's another part of the problem that asks if the subgroups are normal.
     
  5. Nov 1, 2011 #4
    Yeah it is from artin, but I don't have my text with me.
    So 20/4=5 so there is a subgroup of order 5, and because 20/5=4 there is a subgroup of order 4? Is this enough?
    Thanks
     
  6. Nov 1, 2011 #5
    I think so... maybe?

    Here's what Artin says:
    The stabilizer of an element x of G for the operation of conjugation is called the centralizer of x. It is often denoted by Z(x):
    Z(x) = {g in G : gxg^-1 = x}. The centralizer of x is the set of elements that commute with x.
    The orbit of x for conjugation is called the conjugacy class of x, and is often denoted by C(x). It consists of all of the conjugates gxg^-1:
    C(x) = {x' in G: x' = gxg^-1 for some g in G}.
    The counting formula tells us that
    |G| = |Z(x)||C(x)|
    |G| = |centralizer||conj. class|
    The conjugacy classes partition the group. This fact gives us the class equation of a finite group. If we number the conjugacy classes, writing them as C_1,...,C_k, the class equation reads
    |G| = |C_1| +...+ |C_k|.

    Again, I'm a bit rusty on this stuff, but I think you should be able to take it from here. Hope this helps!
     
  7. Nov 1, 2011 #6
  8. Nov 1, 2011 #7
    I think I understand what you mean. I looked at the link you posted. They're talking about normal subgroups: they're trying to figure out the possible orders of the normal subgroups. This is because all normal subgroups are unions of conjugacy classes. So that technique is useful for figuring out if our groups has a normal subgroup of a certain order. But just to find out if the group has a subgroup (not necessarily normal) of a certain order, I think the stuff I mentioned above is sufficient.
     
  9. Nov 1, 2011 #8
    oh ok, so what I said in my previous post means that a normal subgroup of order 5 exists but not a normal subgroup of order 4. However, both a subgroup of order 4 and 5 do exist?
     
  10. Nov 1, 2011 #9
    I think both a subgroup of order 4 and 5 do exist, based on the stuff from Artin that I posted. But you should try to pinpoint exactly why. There's more to it than 20/4 = 5. Like, what does all of this have to do with the centralizer of an element?

    As for the normal part: There is indeed no normal subgroup of order 4, for the reason you gave. But just because you can add the 1 and the 4 in the class equation to get 5 does not mean that there is a normal subgroup of order 5. In other words, every normal subgroup is the union of conjugacy classes, but it is not true that every union of conjugacy classes is a normal subgroup. In fact, I think you have to use the Sylow theorems in order to find out whether there is a normal subgroup of order 5 or not.
     
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