I need to find the 5 classes of the group D4, which basically is the symmetry group of the square, and it’s character table. Now, the problem is that the only method that I can think of is just finding all the group elementsm, their representations and take their traces (then, equal characters=same class), but I think there must be more efficient methods. I also don’t see how the character table works: do you have to use different representations and work out the traces for each of the classes, or is it something completely different?
That should be anyone in the title. Duh.
Since it’s only got 8 elements, I’d just go with the way you outlined. A pain in the ass, I know, but it works, and more importantly, you seem to understand why it works.
Mm, okay, thanks. But what about the character table? And is there a systematic way of generating the different irreps (5 in total, I think)?
I’m not sure. Representation theory is not one of those areas I’m really familiar with.
Can you talk to your professor? That might be the quickest way to get a definitive answer.
You might be interested in checking out the book Abstract Algebra by Dummit and Foote, it’s got some examples of this (including the group you’re considering, in particular).
As for the character table, there are two relations that are useful in dealing with finite groups: if you let each each representation (don’t forget the trivial one!) be written as a column, and each class be written as a row, then different rows are orthogonal to each other, and different columns are orthogonal to each other, with respect to a weight. The weights are given by the populations of each class when “dot-producting” columns together–dotting a column with itself should yield the order of the group. The rows are orthogonal without weights, but dotting a row with itself should yield the order of the group divided by the population of that class. One final thing–the sum of the squares of the dimensions of the representations will yield the order of the group.
These three facts will help in filling in blank spots in the table, but they won’t get you the whole way. You actually do have to write down a few representations and take their traces to get started, but at least you don’t have to do them all.
Also note that since the number of rows=number of columns, D4 has 5 irreducible representations.
–I just checked my homework that’s due next week, and I have to do this problem too! When I finish maybe I’ll be of more help. 
The last time I had to do this, I found that it was easier to work out the conjugacy classes first without thinking about the characters, rather than trying to deduce the classes from the characters. Then once you know the number of conjugacy classes, start working on the table like Don Roberto describes. (But then again you already know how many characters you’re looking for, so that advice is probably unnecessary.)
Finding the five conjugacy classes is trivial (I can do it in my head). One is the identity, one contains rotation by 180, one contains the other two rotations, one the diagonal reflections and the last the horizontal and vertical reflections. As for the character table, it is a long time since I have done that. One way is to form the group ring C(D4) and find the central idempotents, but there must be a better way. I guess there will be one 2 dimensional representation and four 1 dimensional ones.
This page might help ya out. Been a while since I’ve studied group theory though so not sure if this is what you’re lookin for…
Thanks for the help everyone. There are indeed 4 1-dimensional irreps which are all pretty simple, and one 2-dimensional irrep which I’m still working on. Shouldn’t be too difficult.