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Class equation and a group of order pq

  1. Dec 12, 2011 #1
    1. The problem statement, all variables and given/known data
    Use the class equation to show that greup G of order pq, with p and q prime, has an element of order p.


    2. Relevant equations
    |G| = [itex]\sum[/itex] |C(x)| (class equation)
    Z = center of the group G = {z[itex]\in[/itex]G: zy = yz for all y in G}
    |G| = |Z(x)||C(x)| (counting formula)


    3. The attempt at a solution
    So I was told that I should consider this problem in cases. First, assume that G is abelian and then assume G was non abelian.

    So when G is abelian, |Z|= pq so Z = G. Since G is finite and any finite abelian group is a product of its Sylow subgroups G = C[itex]_{p}[/itex] x C[itex]_{q}[/itex]. So we can then see that G would have an element of order p.

    Now, so far I have proved that if G is non abelian, our center has to be trivial i.e. |Z| = 1. Using this, the class equation becomes |G| = 1 + [itex]\sum[/itex] |C(x)| (the sum is for the remaining conjugacy classes). From here is where I get stuck...Since the order of G is pq, the sumation on the right hand side will have to sum to pq -1 (but I have no idea if that is relevant).

    Also, I tried assuming that there wasn't an element of order p. Then, every non trivial element x of G would have to be of order q (since it can't be of order 1, pq, or p and those are our only possible orders which divide the order of the group). From there, I can't quite see where to go from here. I know that every term on the right hand side must divide the order of the group, i.e. pq. So, since the order of the remaining conjugacy classes can't be 1 or pq, they have to be either of order p or order q. If the conjugacy class of x was order q, then the centralizer of any element x would have have to be of order p by the counting formula...We know the centralizer would have to contain the center and the element x of order q.

    I feel like I am so close but I am just not seeing it at this point. Help please.
     
  2. jcsd
  3. Dec 13, 2011 #2
    First, if G is non abelian, |Z|<pq. It is not necessary that |Z|=1.

    I think this is correct:
    If we assume that G does not have an element of order p, then |Z(x)|=q for all x not in the center, where Z(x) is the centralizer of x. Then |C(x)|=p for all x not in the center. Then the class equation implies that there must be p conjugacy classes of order 1, i.e., |Z|=p, a contradiction. So G has an element of order p.
     
  4. Dec 13, 2011 #3
    What you said is false.
    If G is a non abelian group of order pq where p and q are distinct primes, the center is trivial. I proved this is homework and I know this for a fact.

    Consider the quotient group G/Z. Suppose that Z is a non-trivial subgroup then its order is either p or q (because it can't be pq because then it would be abelian and can't be 1 because then it wouldn't be non trivial). So, the order of G/Z is either q or p. We know that every group of prime order is cyclic, so G/Z must be cyclic. But if G/Z is cyclic, then G is Abelian which is a contradiction.

    There's your proof for why the center must be trivial.
     
  5. Dec 13, 2011 #4
    My bad, you're totally correct. The reasoning for existence of an element of order p still goes through though. |Z|=p is still a contradiction.
     
  6. Dec 13, 2011 #5

    Dick

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    I think you already basically have it, don't you? If you assume there is no element of order p then all of the conjugacy classes have to have p elements, because as you said, a conjugacy class with q elements would give a centralizer subgroup with p elements. So now what does the class equation tell you?
     
    Last edited: Dec 13, 2011
  7. Dec 13, 2011 #6
    How does there not being an element of order p imply that the centralizer is of order q? If every non identity element isn't of order p, it obviously has to be of order q since it can't be of order 1 or pq. Then the centralizer of an element x would have to be at least size q (<x> must be contained in Z(x) ) but it can not be all of pq. Are you assuming that p<q so that you can say that because then <x> would be maximal?

    Also, I don't see how you are saying that the conjugacy classes being of order p implies that there must be p conjugacy classes of order 1? If the conjugacy class of every other element is of order p, doesn't the class equation imply that pq = 1 + np where n is some integer?
     
  8. Dec 13, 2011 #7

    Dick

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    I thought you had spelled this out in your original post. But if x is an element with a conjugacy class containing q elements, then doesn't the centralizer subgroup of x contain p elements? Contradicting the statement there is no element of order p? Maybe I was reading between the lines? And you are claiming I said some stuff I didn't say. But I do agree that finally the class equation would say pq=1+np. That doesn't work.
     
  9. Dec 13, 2011 #8
    Dick,
    I wasn't responding to your post. I get what you are saying and yes I had spelled out what you said in my original post. I have the problem solved now. I just didn't understand the logic of the other person who was posting.

    Thanks.
     
  10. Dec 13, 2011 #9

    Dick

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    Oh, ok. Sometimes good to quote the post you are responding to. Very welcome.
     
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