Given a matrix in GL_{2}(Fq) consider its characteristic polynomial
Exercise 2.1. If the roots (λ_{1},λ_{2}) are distinct in Fq then the matrix is conjugate to and to .
Exercise 2.3. If λ^{2}  (a + d)λ + (adbc) is irreducible in Fq[t], then the matrix is similar to .
To summarise, the conjugacy classes in GL_{2}(Fq) are as follows:
In all, there are
 (2.1) 
conjugacy classes. Detailed information about the conjugacy classes is collected in Table 1.

Let B be the subgroup of GL_{2}(Fq) consisting of invertible upper triangular matrices. Let N be the subgroup of upper triangular matrices with 1’s along the diagonal. Let T be the subgroup of invertible diagonal matrices.
Let w = .
Note that B is really a double coset B1B. So Proposition 2.5 really tells us that the double coset space B\GL_{2}(Fq)∕B has two elements and that {1,w} is a complete set of representatives for these double cosets.
Proof. A matrix lies in B if and only if c = 0. If c0, then
Given characters χ_{1} and χ_{2} of Fq^{*}, we get a character χ of T by
 (2.2) 
Let I(χ_{1},χ_{2}) be the representation of GL_{2}(Fq) induced from this character of B.
Proof. Let χ and μ be the characters of B obtained from the pairs χ_{1}, χ_{2} and μ_{1}, μ_{2} respectively as in (2.2). We regard χ and μ as onedimensional representations of B acting on the space C. We may identify Hom_{C}(C,C) with C as well. Then, using Mackey’s description of intertwiners (Theorem 1.12), we see that we must compute the dimension of the space of functions Δ : GL_{2}(Fq) → C such that
 (2.3) 
It follows from the Bruhat decomposition that Δ is completely determined by its values at 1 and w.
Taking g = 1 in (2.3), we see that for any t T,
Taking g = w in (2.3), we see that for any t T,
Therefore, if μ_{1}χ_{2} or μ_{2}χ_{1} then Δ(w) = 0. On the other hand, if μ_{1} = χ_{2} and μ_{2} = χ_{1}, let Δ_{w} be the function such that
An arbitrary functions satisfying (2.3) can be expressed as a linear combination of Δ_{1} and Δ_{w}, so we see that the dimension of the space of such functions must be e_{1} + e_{w}. □
Theorem 2.8. Let χ_{1}, χ_{2}, μ_{1} and μ_{2} be characters of Fq^{*}. Then I(χ_{1},χ_{2}) is an irreducible representation of degree q + 1 of GL_{2}(Fq) unless χ_{1} = χ_{2}, in which case it is a direct sum of two irreducible representations having degrees 1 and q. We have
 (2.4) 
or else
 (2.5) 
Proof. Apply Proposition 2.6 with χ_{1} = μ_{1} and χ_{2} = μ_{2}. We see that
Because the index of B in GL_{2}(Fq) is q + 1, the dimension of I(χ_{1},χ_{2}) is always q + 1. If χ_{1} = χ_{2}, the representation of GL_{2}(Fq) generated by the function f(g) = χ_{1}(det(g)) clearly satisfies f(bg) = χ(b)f(g) for all b B and g G. Therefore f I(χ_{1},χ_{2}). Moreover, (g ⋅f)(x) = χ_{1}(det(g))f. Therefore the onedimensional subspace spanned by f is invariant under the action of G, hence forms a onedimensional representation of G. The other component is therefore qdimensional.
If χ_{1}χ_{2} then, I(χ_{1},χ_{2}) is irreducible. By Proposition 2.6 there exists a nonzero element in Hom(I(χ_{1},χ_{2}),I(μ_{1},μ_{2})) if and only if χ_{1} = μ_{1} and χ_{2} = μ_{2} or χ_{1} = μ_{2} and χ_{2} = μ_{1}. By irreducibility, these homomorphisms must be isomorphisms. This proves the second part of the theorem. □
To summarise, in this section, we have constructed irreducible representations of GL_{2}(Fq) corresponding to characters χ = (χ_{1},χ_{2}) of T:
Recall from Schur theory, that the number of irreducible representations is the same as the number of conjugacy classes in a group. We have constructed
Recall that for a group of order n whose irreducible representations are π_{1},…,π_{r} of degrees d_{1},…,d_{r} respectively,
The sum of squares of degrees of the representations that we have constructed so far is
We will see in Section 3.4 that there are (q^{2}  q) irreducible representations of degree q  1 remaining. These will be constructed in Section 3.5.
Let Aut(SL_{2}(Fq)) denote the group of all automorphisms of SL_{2}(Fq). GL_{2}(Fq) acts on SL_{2}(Fq) by conjugation. This gives rise to a homomorphism GL_{2}(Fq) → Aut(SL_{2}(Fq)). The kernel of this automorphism consists of scalar matrices in GL_{2}(Fq), and is therefore isomorphic to Fq^{*}. The image is therefore isomorphic to the group PGL_{2}(Fq), which is the quotient of GL_{2}(Fq) by the subgroup of invertible scalar matrices. The orbits of PGL_{2}(Fq) on SL_{2}(Fq) are precisely the conjugacy classes of GL_{2}(Fq) which are contained in SL_{2}(Fq) (note that SL_{2}(Fq) is a union of conjugacy classes of GL_{2}(Fq)).
On the other hand, the image of SL_{2}(Fq) in Aut(SL_{2}(Fq)) is PSL_{2}(Fq), the quotient of SL_{2}(Fq) by the subgroup {±1}. The conjugacy classes of SL_{2}(Fq) are precisely the PSL_{2}(Fq) orbits.
Now, PSL_{2}(Fq) is a subgroup of PGL_{2}(Fq) (when both groups are viewed as subgroups of Aut(SL_{2}(Fq))) of index two. Therefore, each conjugacy class of GL_{2}(Fq) whose elements lie in SL_{2}(Fq) is either a single conjugacy class in SL_{2}(Fq) or a union of two conjugacy classes in SL_{2}(Fq). If ε is an element of Fq^{*} which is not a square (since q is assumed to be odd, there are such elements), then the image of in Aut(SL_{2}(Fq)) does not lie in the image of SL_{2}(Fq).
Let σ SL_{2}(Fq). Whether or not the conjugacy class of σ in GL_{2}(Fq) splits or not can be determined by counting. The basic principle here is that the number of elements in an orbit for a group action is the index of the stabiliser of a point in the orbit.
With the above observations in mind, it is not difficult to prove that
Theorem 2.11. Let σ SL_{2}(Fq). Let Z denote the centraliser of σ in GL_{2}(Fq). Then [Z : Z ∩ SL_{2}(Fq)] is either q  1 or . In the former case, the conjugacy class of σ in GL_{2}(Fq) is a single conjugacy class in SL_{2}(Fq). In the latter case, the conjugacy class of σ in GL_{2}(Fq) is a union of two conjugacy classes in SL_{2}(Fq), represented by σ and σ1 respectively.
Exercise 2.12. Prove Theorem 2.11.
In all, there are
 (2.6) 
conjugacy classes.

Let B now consist of the upper triangular matrices in SL_{2}(Fq), N the upper triangular matrices with 1’s along the diagonal, and T the matrices in SL_{2}(Fq) which are diagonal. Note that the results of Exercise 2.4 are still valid, as is the Bruhat decomposition:
Taking μ = χ in Proposition 2.14 gives that
In this case, I(χ) contains the invariant one dimensional subspace of constant functions on G. Therefore I(χ) splits into a direct sum of two irreducible representations, the trivial representation and a representation of dimension q, which is called the Steinberg representation.
In this case it is necessary to make a closer analysis of End_{SL2(Fq)}(I(χ)). Let Δ_{1} denote the unique function SL_{2}(Fq) → C for which Δ_{1}(1) = 1, Δ_{1}(w) = 0, and Δ(b_{1}gb_{2}) = χ(b_{1})Δ(g)χ(b_{2}). Also let Δ_{w} denote the unique function SL_{2}(Fq) → C for which Δ_{w}(1) = 0, Δ_{w}(w) = 1, and Δ(b_{1}gb_{2}) = χ(b_{1})Δ(g)χ(b_{2}). These two functions form a basis of End_{SL2(Fq)}(I(χ)). Write I(χ) = ρ^{+} ⊕ ρ^{}, where ρ^{+} and ρ^{} are the two irreducible summands of of I(χ). The identity endomorphism in I(χ) can be written as a sum of two idempotents, coming from the identity endomorphisms of ρ^{+} and ρ^{}.
Δ_{1} * Δ_{1} = q(q  1)Δ_{1},  Δ_{1} * Δ_{w} = q(q  1)Δ_{w},  
Δ_{w} * Δ_{1} = q(q  1)Δ_{w},  Δ_{w} * Δ_{w} = q^{2}(q  1)χ(1)Δ_{1}. 
Exercise 2.17. Besides q1(q  1)1Δ_{1} and 0, show that the only idempotents in End_{SL2(Fq)}(I(χ)) are
Let ^{ε}I(χ) be the representation of SL_{2}(Fq) on the representation space of I(χ), but where the action of SL_{2}(Fq) is given by
Exercise 2.19. Show that _{1} = Δ_{1} and _{w} = Δ_{w}. Conclude that ^{ε}ρ^{+} = ρ^{} and ^{ε}ρ^{} = ρ^{+}.
Therefore, the two representations ρ^{+} and ρ^{} must have equal degrees. It follows that I(χ) is a sum of two irreducible representations, each of degree .
In this section, we have constructed