Why does an abelian group of order G have G conjugacy classes

[Dixanadu]

Homework Statement

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!

Homework Equations

I don't think there are any

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 $g \in G$ for an abelian group: $g_{1}gg_{1}^{-1}$, it is just equal to the element $g$ 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 $|G|=\sum_{R}d_{R}^{2}$; where $R$ is an index that runs through the list of irreducible representations, and $d_{r}$ 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 $d_{R} = d_{R}^{2} = 1$...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!

 

[CompuChip]

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.