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Math Help - Finding the number of conjugacy classes

  1. #1
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    Finding the number of conjugacy classes

    Given a group G, non-abelian, with |G|= p^3, show G has p^2 + p -1 distinct conjugacy classes.

    Any help would be great -- I'm sure there's a useful theorem or fact I'm missing out on.

    thanks!
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  2. #2
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    Quote Originally Posted by kimberu View Post
    Given a group G, non-abelian, with |G|= p^3, show G has p^2 + p -1 distinct conjugacy classes.

    Any help would be great -- I'm sure there's a useful theorem or fact I'm missing out on.

    thanks!
    let Z(G) be the center of G and C(g) the centralizer of g \in G in G. see that |Z(G)|=p and also |C(g)|=p^2, for all g \notin Z(G), because Z(G) \subset C(g). now apply the class equation. \Box
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  3. #3
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    So the class equation says:
    G=Z(G)+[G:C( x_i)]

    meaning,
    p^3=p + this sum, but what exactly is the sum? Is it just p, m times? (sorry, I've never used this equation before. thanks so much for the help!)
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  4. #4
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    Quote Originally Posted by kimberu View Post
    So the class equation says:
    G=Z(G)+[G:C( x_i)]

    meaning,
    p^3=p + this sum, but what exactly is the sum? Is it just p, m times? (sorry, I've never used this equation before. thanks so much for the help!)
    the x_i in your sum in the right hand side are not in Z(G) and thus, as i mentioned, [G:C(x_i)]=\frac{|G|}{|C(x_i)|}=p. now suppose k is the number of conjugacy classes. then:

    p^3=p + (k - |Z(G)|)p=p+(k-p)p, which will give us the result.
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