1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?

    Thanks!
     
  2. jcsd
  3. Oct 15, 2013 #2

    CompuChip

    User Avatar
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted