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Why does an abelian group of order G have G conjugacy classes
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[QUOTE="Dixanadu, post: 4538302, member: 451060"] [h2]Homework Statement [/h2] 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! [h2]Homework Equations[/h2] I don't think there are any[h2]The Attempt at a Solution[/h2] 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 I've made any mistakes? Thanks! [/QUOTE]
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Why does an abelian group of order G have G conjugacy classes
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