# Conjugacy classes and irreducible representations of GL_2(q)

Let $q$ be a prime.

(a) Find representatives for the conjugacy classes of $GL_2(q)$.

(b) Show that $GL_2(q)$ has an irreducible complex representation of dimension $q$.

(c) Compute the character of this representation.

My Attempt: First, we know the group has $(q^2-1)(q^2-q)=q(q+1)(q-1)^2$ elements (just by making the matrices with two non-zero linearly independent vectors).

(a) The conjugacy classes are dependent on the eigenvalues of the matrices.

There are $(q-1)(q-2)/2$ conjugacy classes for matrices with two distinct eigenvalues that lie in the field. The representatives are $\pmatrix{a& 0\\0 &b}$ for $a>b$.

If the matrix has just one eigenvalue then it lies in the field and there are two possibilities, $q-1$ classes with representatives $\pmatrix{a& 0\\0 &a}$ and $q-1$ classes with representatives $\pmatrix{a& 1\\0 &a}$.

The final possibility is that the eigenvalues are conjugate and lie outside the field. There are $q(q-1)$ such classes with representatives $\pmatrix{0& a\bar{a}\\1 &a+\bar{a}}$.

Hence there are $q^2-1$ conjugacy classes.

(b) We know that there are $q^2-1$ irreducible representations. But this is where I am stuck. Perhaps I need to compute the number of one dimensional representations by looking at the abelianization and then make some sort of argument that the sum of the dimensions of the irreducible representations is equal to the order of the group which somehow forces there to be one of these dimensions equal to $q$. But I just can't see how to make that argument.

-
Use the fact that ${\rm GL}(2,q)$ has a 2-transitive permutation representation of degree $q+1$. –  Derek Holt Jun 7 '12 at 8:09
A good reference for this is Fulton and Harris. They have a very complete analysis of the irreps of $GL_2(\mathbb{F}_q).$ –  rotskoff Jun 7 '12 at 19:23