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

  • Thread starter Dixanadu
  • Start date
  • #1
254
2

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 dont 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 [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!
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
4,302
47
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.
 

Related Threads on Why does an abelian group of order G have G conjugacy classes

  • Last Post
Replies
1
Views
2K
Replies
4
Views
3K
Replies
0
Views
946
Replies
10
Views
6K
  • Last Post
Replies
2
Views
407
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
0
Views
2K
Top