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Homework Help: Why does an abelian group of order G have G conjugacy classes

  1. Oct 14, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi guys,

    The title pretty much says it. I need to explain why:
    (a) an abelian group of order |G| has precisely |G| conjugacy classes, and
    (b) why the irreducible representations of abelian groups are one-dimensional.

    Also in my description below, if I make any mathematical errors of any sort in my explanations please let me know because I really want to learn this!!!

    2. Relevant equations
    I dont think there are any

    3. The attempt at a solution
    Okay so this is one of those open-ended questions, so all I can do is present a few of my initial thoughts, but they arent really explanations. So what i know is this:

    - if we take the conjugate of any element [itex]g \in G[/itex] for an abelian group: [itex]g_{1}gg_{1}^{-1}[/itex], it is just equal to the element [itex]g[/itex] itself. So each element belongs to its conjugacy class; and because there are |G| elements, there must be |G| conjugacy classes.

    - for a finite group, the number of irreducible representations is less than or equal to the order of the group.

    - You can calculate the order of a finite group using the formula [itex]|G|=\sum_{R}d_{R}^{2}[/itex]; where [itex]R[/itex] is an index that runs through the list of irreducible representations, and [itex]d_{r}[/itex] is the dimension of the Rth representation. So if there is 1 representation for each group element (is this true?) then R must range from 1 to |G|; and thus [itex]d_{R} = d_{R}^{2} = 1[/itex]...so each representation is 1 dimensional.

    So can u guys tell me what I'm missing for either parts of the question, or if ive made any mistakes?

  2. jcsd
  3. Oct 15, 2013 #2


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    Science Advisor
    Homework Helper

    For the first part, I would state explicitly why ##g_1 g g_1^{-1} = g## (and maybe call the conjugating element something else, like ##h##, for clarity?) as this is the crucial point - any g1 commutes with g only because the group is abelian.

    For the second one, the number of irreducible representations of a group is equal to the number of conjugacy classes in the group. Do you have this result available? Because then (b) would follow quite straightforwardly from (a), I think.
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