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Math Help - Conjugacy classes - two problems.

  1. #1
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    Conjugacy classes - two problems.

    Hi, I was wondering if I could get some help with these questions?

    1.Let G be a non-abelian group of order 21. Show that if m is the number of conjugacy classes of size 3 and n is the number of conjugacy classes of size 7 then 3m+7n = 20. Deduce that m = n = 2.

    2. Let G be a group of order p^2 where p is a prime number. Show that all conjugacy classes in G have size 1 or p.


    I guess that I need to use The Class Equation or The Orbit-Stabilizer Theorem but I do not know how?

    Can anyone help me, please?
    Thanks!
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by leelooana View Post
    Hi, I was wondering if I could get some help with these questions?

    1.Let G be a non-abelian group of order 21. Show that if m is the number of conjugacy classes of size 3 and n is the number of conjugacy classes of size 7 then 3m+7n = 20. Deduce that m = n = 2.

    2. Let G be a group of order p^2 where p is a prime number. Show that all conjugacy classes in G have size 1 or p.


    I guess that I need to use The Class Equation or The Orbit-Stabilizer Theorem but I do not know how?

    Can anyone help me, please?
    Thanks!
    For the first question, prove that every non-abelian group of order pq has trivial center. (You may wish to first prove that if G/Z(G) is cyclic then G is abelian). You can then apply the class equation.

    For the second result, use the orbit-stabiliser theorem to investigate the action of conjugation. What is an element's orbit? What is the corresponding stabiliser? Is one of these a subgroup?

    Are you at uni in Scotland?
    Last edited by Swlabr; February 17th 2010 at 01:04 PM. Reason: Grammer...
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  3. #3
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    Thanks for Your reply! I just proved in previous problem that if p and q are distinct primes and G is non abelian then Z(G) is trivial but I did not notice the relation between these problems.

    so quick proof:
    From Lagrange Theorem |Z(G)| could be pq, p, q or 1. It is easy to prove that order of Z(G) can not be pq ( because in this case Z(G) = G therefore G is abelian what contradicts the initial claim ).
    Also can not be p because the quotient group G/Z(G) would be cyclic of order q but G/Z(G) cannot be cyclic unless G is abelian ( because G/Z(G) cyclic implies G abelian.)
    The same for q.

    So Z(G) must be trivial.

    And now I see that if |G| = 21 then |Z(G)| = 1 and from class equation I get that 3m+7n = 20 and must be n=m=2.


    But I still can not done the 2nd question. I know that :
    -the size of each orbit divides the group order
    -in each conjugacy classes all elements have the same order
    -in all abelian groups every conjugacy class is a set containing one element (if |G| = p^2 and p is prime then G is abelian)

    and I can infer that order of conjugacy classes of abelian group equals 1.

    how to show that conjugacy classes in G have size p?


    Quote Originally Posted by Swlabr View Post

    Are you at uni in Scotland?
    Yes
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by leelooana View Post
    Thanks for Your reply! I just proved in previous problem that if p and q are distinct primes and G is non abelian then Z(G) is trivial but I did not notice the relation between these problems.

    so quick proof:
    From Lagrange Theorem |Z(G)| could be pq, p, q or 1. It is easy to prove that order of Z(G) can not be pq ( because in this case Z(G) = G therefore G is abelian what contradicts the initial claim ).
    Also can not be p because the quotient group G/Z(G) would be cyclic of order q but G/Z(G) cannot be cyclic unless G is abelian ( because G/Z(G) cyclic implies G abelian.)
    The same for q.

    So Z(G) must be trivial.

    And now I see that if |G| = 21 then |Z(G)| = 1 and from class equation I get that 3m+7n = 20 and must be n=m=2.


    But I still can not done the 2nd question. I know that :
    -the size of each orbit divides the group order
    -in each conjugacy classes all elements have the same order
    -in all abelian groups every conjugacy class is a set containing one element (if |G| = p^2 and p is prime then G is abelian)

    and I can infer that order of conjugacy classes of abelian group equals 1.

    how to show that conjugacy classes in G have size p?
    Well, the conjugacy class of the identity is 1. Assume your group is non-abelian and take an element which is non-central. What is it's conjugacy class? It must have order p or p^2. What would happen if it's conjugacy class was p^2? What would this "mean"?

    Quote Originally Posted by leelooana View Post
    Yes
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