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Thread: 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|=$\displaystyle p^3$, show G has $\displaystyle p^2 + p -1$ 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|=$\displaystyle p^3$, show G has $\displaystyle p^2 + p -1$ 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 $\displaystyle Z(G)$ be the center of $\displaystyle G$ and $\displaystyle C(g)$ the centralizer of $\displaystyle g \in G$ in $\displaystyle G.$ see that $\displaystyle |Z(G)|=p$ and also $\displaystyle |C(g)|=p^2,$ for all $\displaystyle g \notin Z(G),$ because $\displaystyle Z(G) \subset C(g).$ now apply the class equation. $\displaystyle \Box$
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  3. #3
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    So the class equation says:
    G=Z(G)+[G:C($\displaystyle x_i$)]

    meaning,
    $\displaystyle 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($\displaystyle x_i$)]

    meaning,
    $\displaystyle 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 $\displaystyle x_i$ in your sum in the right hand side are not in $\displaystyle Z(G)$ and thus, as i mentioned, $\displaystyle [G:C(x_i)]=\frac{|G|}{|C(x_i)|}=p.$ now suppose $\displaystyle k$ is the number of conjugacy classes. then:

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