Irreducible Characters of GL(2, q)

S S

(iv) Finally let g = 0 1

−r^q+1 r+r^q

!

. The eigenvalues of g are λ = r and λ = r^q. Since g ∼ r 0

0 r^q

!

,we have o(g) =o

r 0 0 r^q

!!

.So o(g) =lcm(o(r), o(r^q)).

Ifθ is a generator of the groupF^∗_q2 andr =θ^k for somek,such that q+ 1-k,then

o(g) = (q²−1)²

gcd(k, q²−1) gcd(kq, q²−1) gcd

q²−1

gcd(k,q²−1),_gcd(kq,q^q2−12−1)

.

Hence all elements of the group GL(2, q) have the stated orders, which completes the proof of the

Proposition.

Chapter 4 — GL(2, q) and Some of its Subgroups where, in Table 4.2,

• α, β∈F^∗_q, α6=β,

• s, t denote the integers for whichα=ε^s andβ =ε^t, ε is a generator of F^∗_q,

• r∈Fq²\Fq and r^q is excluded whenever r is included,

• in χ⁽¹⁾_k , χ⁽²⁾_k , k= 0,1,· · · , q−2,

• in χ⁽³⁾_k,l, 0≤k < l≤q−2,

• in χ⁽⁴⁾_k , q+ 1-k, k= 1,2,· · ·, q²−1 and kq is excluded whenever k is included,

• bis the homomorphism :F^∗_qd = hε_di −→C^∗ given by b(ε^j_d) =εb^j_d =e

2πj qd−1i

, for d= 1,2 and 0≤j≤q^d−2.

PROOF. The function det :GL(2, q)−→F^∗_q defines a group homomorphism. For k= 0,1,· · · , q−2, we set λ_k : GL(2, q) −→ U to be λ_k(g) := χ_k(det(g)), where χ_k is an irreducible character of F^∗_q. It is clear that the composition χ_k ◦det : GL(2, q) −→ U is a group homomorphism.

Thus it is an ordinary representation of degree 1 and consequently is an irreducible character of GL(2, q). We recall by Theorem 3.1.19 that the derived subgroup GL(n, q)⁰ is SL(n, q) ex- cept when n = q = 2. By Proposition 2.3.4, the number of linear characters of GL(2, q) is

|GL(2, q)|/|GL(2, q)⁰|=|GL(2, q)|/|SL(2, q)|=q−1.Thus, apart from the caseGL(2,2),theq−1 linear characters given by λk are all the linear characters of GL(2, q).In the case GL(2,2)∼=S3, we have the extra linear character corresponding to the sign of the permutations ofS3.

The next table shows the values of the linear characters λ_k on the conjugacy classes ofGL(2, q).

Table 4.3: Values of the linear characters on elements ofGL(2, q)

Class T_s⁽¹⁾ T_s⁽²⁾ T_s,t⁽³⁾ T_r⁽⁴⁾ λ_k χ²_k(α) χ²_k(α) χ_k(α)χ_k(β) χ_k(r^(q+1)) where 0≤k≤q−2.

In fact, theq−1 linear characters given by the powers of the determinants comprise all the linear characters of the groupGL(n, q),for anyn∈Nand any prime powerq,excepting the caseGL(2,2).

This will be proved in Theorem 5.6.3.

LetT be the torus inGL(2, q), which by Section 3.2 consists of the 2×2 diagonal matrices; that is T =

( a 0 0 d

!

|a, d∈F^∗q

) .

S S

We can see clearly thatT ∼=F^∗_q×F^∗_q and hence

Irr(T) ={χ_kχ_l|χ_k, χ_l∈Irr(F^∗q)}.

We recall by Definition 3.1.4 that the group SU U T(2, q) consists of the elements of the form 1 b

0 1

!

, b ∈ Fq. By Theorem 3.1.9, this group is normal in U T(2, q), where a typical element of U T(2, q) will have the form a b

0 d

!

, b ∈Fq, a, d∈ F^∗_q. The quotient U T(2, q)/SU U T(2, q) is isomorphic to T ∼= F^∗q×F^∗q. We will use the method of lifting of characters from the quotient of a group by a normal subgroup, to get characters of the main group as described in Section 2.4.

Hence Irr(T)⊆Irr(U T(2, q)),where χ_kχ_l

a b 0 d

!!

=χ_k(a)χ_l(d). (4.12)

The following table shows the values of these characters on classes ofU T(2, q),which they can be deduced easily from Table 4.1 [of course we need to check whether a class in GL(2, q) will remain as it is or will break into classes inU T(2, q)].The full character table ofU T(2, q) will be discussed in Section 4.6.

Table 4.4: Conjugacy classes and some irreducible characters ofU T(2, q)

Type T_k⁽¹⁾ T_k⁽²⁾ T_k,l⁽³⁾

Rep g α 0

0 α

! α 1 0 α

! α 0 0 β

!

No. of CC q−1 q−1 (q−1)(q−2)

|C_g| 1 q−1 q

|C_{U T(2,q)}(g)| q(q−1)² q(q−1) (q−1)² χ_kχ_l χ_k(α)χ_l(α) χ_k(α)χ_l(α) χ_k(α)χ_l(β)

Now, we use these linear characters of U T(2, q) as a base for Frobenius method of induction of characters to get characters of GL(2, q). Thus let χ_k,l = χ_kχ_l↑^GL(2,q)_{U T(2,q)}. We consider the following two cases

Case I: Suppose that χ_k6=χ_l.Then we have

Chapter 4 — GL(2, q) and Some of its Subgroups

(i) If g is in a class of the first type; that is g=αI₂ for someα∈F^∗_q,then χk,l(g) = |C_GL(2,q)(g)|

|C_{U T(2,q)}(g)|χk(α)χl(α) = (q+ 1)χk(α)χl(α).

(ii) Ifg is in a class of the second type; that is g= α 1 0 α

!

for someα∈F^∗_q,then

χk,l(g) = |C_GL(2,q)(g)|

|C_{U T(2,q)}(g)|χk(α)χl(α) = q(q−1)

q(q−1)χk(α)χl(α) =χk(α)χl(α).

(iii) If g is in a class of the third type; that isg= α 0 0 β

!

for some α, β ∈F^∗q, α6=β.Then we can check that the conjugacy class represented by g⁰ = β 0

0 α

!

which is conjugate to g in GL(2, q) is no longer conjugate tog inU T(2, q).In this case we have

χ_k,l(g) = |C_GL(2,q)(g)|

χ_k(α)χ_l(β)

|C_{U T(2,q)}(g)|+ χ_k(β)χ_l(α)

|C_{U T(2,q)}(g⁰)|

= (q−1)²

χk(α)χl(β)

(q−1)² +χk(β)χl(α) (q−1)²

=χ_k(α)χ_l(β) +χ_k(β)χ_l(α).

(iv) Ifg is in a class of the fourth type; that isg= 0 1

−r^q+1 r+r^q

!

for somer∈F^∗_q2\F^∗_q,then χk,l(g) = 0,since there is no intersection between a class of this type and the groupU T(2, q).

Note that an element of the fourth type has eigenvalues r andr^q which are inF^∗_q2\F^∗q,while the eigenvalues of any element ofU T(2, q) are inF^∗q.

Let us now check the irreducibility of the above characters.

hχ_k,l, χk,li = 1

|GL(2, q)|

X

g∈GL(2,q)

χk,l(g)χk,l(g)

= (q−1)(q+ 1)²

q(q−1)²(q+ 1)χk(α)χl(α)χk(α)χl(α) + (q−1)(q²−1)

q(q−1)²(q+ 1)χk(α)χl(α)χk(α)χl(α) + (q−1)(q−2)q(q+ 1)

2q(q−1)²(q+ 1) (χ_k(α)χ_l(β) +χ_k(β)χ_l(α))(χ_k(α)χ_l(β) +χ_k(β)χ_l(α))

= (q+ 1) q(q−1)+1

q (4.13)

+ 1

2(q−1)² X

α6=β

(χ_k(α)χ_l(β) +χ_k(β)χ_l(α))(χ_k(α)χ_l(β) +χ_k(β)χ_l(α)).

In the last term of (4.13), we have divided by 2 because interchangingαwithβ in the main diagonal of an element in a class of typeT⁽³⁾,does give the same conjugacy class.

S S

To evaluate the last sum of the right hand side of equation (4.13), we will use the abelian group T ∼=F^∗_q×F^∗_q of order (q−1)² and hence of indexq(q+ 1) in GL(2, q).Let tα,β = α 0

0 β

! be an element of T. With χkχl be an irreducible character of T, defined as before and for fixed k and l, the functionξ :T −→GL(2,C) given by

ξ(t_α,β) = χk(α)χl(β) 0 0 χ_k(β)χ_l(α)

!

is a representation of T. Thenξ =ξ1⊕ξ2,whereξ1, ξ2:T −→GL(1,C) given by ξ₁(t_α,β) =χ_k(α)χ_l(β), ξ₂(t_α,β) =χ_k(β)χ_l(α).

It is clear that ξ1, ξ2 ∈Irr(T).Also

χ_ξ(t_α,β) =χ_k(α)χ_l(β) +χ_k(β)χ_l(α).

Thushχ_ξ, χ_ξi= 2 and hence 2 =hχ_ξ, χ_ξi = 1

|T| X

g∈T

χ_ξ(g)χ_ξ(g) = 1 (q−1)²

X

α,β∈F^∗q

χ_ξ(t_α,β)χ_ξ(t_α,β)

= 1

(q−1)² X

α=β

χ_ξ(t_α,α)χ_ξ(t_α,α) + 1 (q−1)²

X

α6=β

χ_ξ(t_α,β)χ_ξ(t_α,β)

= 1

(q−1)²



 X

α∈F^∗q

(χ_k(α)χ_l(α) +χ_k(α)χ_l(α))(χ_k(α)χ_l(α) +χ_k(α)χ_l(α))





+ 1

(q−1)²



 X

α6=β

(χk(α)χl(β) +χk(β)χl(α))(χk(α)χl(β) +χk(β)χl(α))





= 4(q−1)

(q−1)² + 1 (q−1)²



 X

α6=β

(χk(α)χl(β) +χk(β)χl(α))(χk(α)χl(β) +χk(β)χl(α))



. Therefore

X

α6=β

(χk(α)χl(β) +χk(β)χl(α))(χk(α)χl(β) +χk(β)χl(α)) = 2(q−1)²−4(q−1) = 2(q−1)(q−3).

Turning back to equation (4.13), we get hχ_k,l, χ_k,li = (q+ 1)

q(q−1)+1

q + 1

2(q−1)²2(q−1)(q−3) = (q+ 1) q(q−1)+1

q +(q−3) (q−1)

= (q+ 1) + (q−1) +q(q−3)

q(q−1) = 2q+q²−3q

q²−q = q²−q q²−q = 1.

This shows that χk,l for 0≤k < l≤q−2 are all irreducible.

Chapter 4 — GL(2, q) and Some of its Subgroups

It is easily seen that χ_k,l =χ_kχ_l=χ_lχ_k=χ_l,k.Since kand l are distinct and they range between 0 andq−2,there are ^{(q−1)(q−2)}₂ irreducible characters χ_k,l.

To see that two characters χ_k,l and χ_k⁰_,l⁰ for 0 ≤ k < l ≤ q−2 and 0 ≤ k⁰ < l⁰ ≤ q −2 with (k, l)6= (k⁰, l⁰) are distinct, refer to Corollary 28.11 of James [40].

This completes the proof for Case I.

Case II: Suppose that χl=χk; that is l=k.By g ∈C ∈ T⁽ⁱ⁾ we mean an element g∈GL(2, q) in the conjugacy classC of type T⁽ⁱ⁾.Then by computations similar toCase I, we have

• Ifg∈C∈ T⁽¹⁾,thenχk,k(g) = (q+ 1)χ²_k(α).

• Ifg∈C∈ T⁽²⁾,thenχ_k,k(g) =χ²_k(α).

• Ifg∈C∈ T⁽³⁾,thenχ_k,k(g) = 2χ_k(α)χ_k(β).

• Ifg∈C∈ T⁽⁴⁾,thenχ_k,k(g) = 0.

Now

hχ_k,k, χ_k,ki = 1

|GL(2, q)|

X

g∈GL(2,q)

χ_k,k(g)χ_k,k(g) = (q−1)(q+ 1)²

q(q−1)²(q+ 1)χ²_k(α)χ²_k(α) + (q−1)(q²−1)

q(q−1)²(q+ 1)χ²_k(α)χ²_k(α) +(q−1)(q−2)q(q+ 1)

2q(q−1)²(q+ 1) (2χ_k(α)χ_k(β))(2χ_k(α)χ_k(β))

= (q+ 1) q(q−1)+1

q +2(q−2)

(q−1) = (q+ 1) + (q−1) + 2q²−4q

q(q−1) = 2q²−2q q²−q) = 2.

Thusχ_k,k is not an irreducible character of GL(2, q).We next setSt:=χ_0,0−1.Therefore

• Ifg∈C∈ T⁽¹⁾,thenSt(g) = (q+ 1)−1 =q.

• Ifg∈C∈ T⁽²⁾,thenSt(g) = 1−1 = 0.

• Ifg∈C∈ T⁽³⁾,thenSt(g) = 2−1 = 1.

• Ifg∈C∈ T⁽⁴⁾,thenSt(g) = 0−1 =−1.

S S

Since

hSt, Sti = 1

|GL(2, q)|

X

g∈GL(2,q)

St(g)St(g) = (q−1)q² q(q−1)²(q+ 1) + (q−1)(q−2)q(q+ 1)

2q(q−1)²(q+ 1) + (q²−q)(q²−q) 2q(q−1)²(q+ 1)

= q

(q−1)(q+ 1) + (q−2)

2(q−1)+ q 2(q+ 1)

= 2q²+q(q+ 1)(q−2) +q²(q−1)

2q(q+ 1)(q−1) = 2q²+q³−q²−2q+q³−q² 2q³−2q

= 2q³−2q 2q³−2q = 1,

we deduce that Stis an irreducible character ofGL(2, q).The character Stdefined above is called an Steinberg character of GL(2, q). In fact for the group GL(n, q), Steinberg [72] defined |P(n)|

irreducible characters corresponding to the partitions ofn. These characters defined by Steinberg come from the action of the group GL(n, q) on some geometric entities. The character St corre- sponds to the partition (1,1)`2.The other Steinberg character of GL(2, q) corresponding to 2`2 is the trivial character ofGL(2, q).Steinberg characters will be studied in more details in Section 5.5.

We know from Proposition 2.3.3 that a product of a linear character by an irreducible charac- ter is an irreducible character. Thus by tensoring the q−1 linear charactersλk with the Steinberg characterSt, we getq−1 irreducible characters of degree q. In the following we give the values of these q−1 irreducible characters on the classesGL(2, q)

• Ifg∈C∈ T⁽¹⁾,thenλ_kSt(g) =qχ²_k(α).

• Ifg∈C∈ T⁽²⁾,thenλkSt(g) = 0.

• Ifg∈C∈ T⁽³⁾,thenλ_kSt(g) =χ_k(α)χ_k(β).

• Ifg∈C∈ T⁽⁴⁾,thenλ_kSt(g) =−χ_k(r^q+1).

The q−1 irreducible characters λkSt are all distinct, because for the primitive (q−1)th root of unity ε,we have λkSt

ε 0 0 1

!!

=χk(ε),which gives distinct values for 0≤k≤q−2.

Note thatλkSt(g) =χk,k−λk, ∀ 0≤k≤q−2.We may denoteλkSt by ψk.

So far we have found

• q−1 linear characters.

• q−1 irreducible characters of degreeq.

Chapter 4 — GL(2, q) and Some of its Subgroups

• ^{(q−1)(q−2)}₂ irreducible characters of degree q+ 1.

Thus so far we have (q−1) + (q−1) + ^{(q−1)(q−2)}₂ = ^(q−1)(q+2)₂ irreducible characters. Since there areq²−1 irreducible characters ofGL(2, q),we need to find ^q2₂^−q additional irreducible characters.

Moreover, if we add up the squares of the degrees of the characters we have found so far, we get (q−1) + (q−1)q²+ (q−1)(q−2)

2 (q+ 1)² = 2q−2 + 2q³−2q²+q⁴−q³−3q²+q+ 2 2

= q⁴+q³−5q²+ 3q

2 .

Now

|GL(2, q)| −q⁴+q³−5q²+ 3q

2 =q(q−1)²(q+ 1)−q⁴+q³−5q²+ 3q

2 = q²−q

2 (q−1)². It will be shown that each of the remaining ^q2₂^−q characters will have the degree q−1.

We aim to find the remaining ^q2₂^−q irreducible characters ofGL(2, q) by using the characters of the groupF^∗_q2.The groupF^∗_q2 =hσiis embedded into GL(2, q) byσ7→k_σ,wherek_σ = σ 0

0 σ^q

! .Now the group K = hk_σi =

( σ^s 0 0 σ^qs

!

|1≤s≤q²−1 )

is an isomorphic copy of F^∗_q² in GL(2, q) of index q² −q. If s is a multiple of q+ 1, that is s = (q + 1)j for some 1 ≤ j ≤ q −1, then σ^qs = σ^q(q+1)j = σ^q2jσ^qj = σ^jσ^qj = σ^(q+1)j = σ^s. Thus when s = (q+ 1)j, the elements kσs = kσ^s = σ^(q+1)j 0

0 σ^(q+1)j

!

are the scalar matrices in GL(2, q). On the other hand, if s is not a multiple ofq+ 1,thenσ^qs 6=σ^s.Note that there are (q²−1)−(q−1) =q²−q such integerssand consequently,q²−q non-scalar kσ inK.

Therefore, K meets GL(2, q) only on classes of typeT⁽¹⁾ and T⁽⁴⁾.

Since K is cyclic, its irreducible characters are known (Theorem 2.2.4). If θ_k ∈ Irr(K), then we let φ_k =θ_k↑^GL(2,q)_K and χ_k=θ_k↓^K

F^∗q.Now using Proposition 2.5.5 we obtain

• Ifg∈C∈ T⁽¹⁾,thenφ_k(g) = ^|CGL(2,q)_|C ^(g)|

K(g)| θ_k(α) = (q²−q)χ_k(α).

• Ifg∈C∈ T⁽²⁾,thenφ_k(g) = 0.

• Ifg∈C∈ T⁽³⁾,thenφ_k(g) = 0.

• Ifg∈C∈ T⁽⁴⁾,thenφ_k(g) =|C_GL(2,q)(g)|

θk(r)

|C_K(r)|+_|C^θk(rq)

K(r^q)|

=θ_k(r) +θ_k(r^q).

S S

In the next step, we find the remaining irreducible characters ofGL(2, q).For 1≤k≤q²−1 such thatq+ 1-k,let π_k:=χ0,−kψ_k−φ_k−χ_0,k and note thatπ_kis a combination of known characters of GL(2, q).In Table 4.5 we list the values of π_k on classes ofGL(2, q).

Table 4.5: Values ofπk on classes ofGL(2, q)

T⁽¹⁾ T⁽²⁾ T⁽³⁾ T⁽⁴⁾

χ0,−k (q+ 1)χ−k(α) χ−k(α) χ−k(α) +χ−k(β) 0 ψ_k qχ²_k(α) 0 χ_k(α)χ_k(β) −χ_k(r^q+1) χ0,−kψ_k (q²+q)χ−k(α) 0 χ_k(α) +χ_k(β) 0

χ_0,k (q+ 1)χ_k(α) χ_k(α) χ_k(α) +χ_k(β) 0

φk q²−qχk(α) 0 0 θk(r) +θk(r^q)

π_k (q−1)χ_k(α) −χ_k(α) 0 −(θ_k(r) +θ_k(r^q))

There are (q²−1)−(q−1) =q²−q characters π_k of GL(2, q).We have hπ_k, π_ki = 1

|GL(2, q)|

X

g∈GL(2,q)

π_k(g)π_k(g) = (q−1)²(q−1) q(q−1)²(q+ 1) + (q−1)(q−1)(q+ 1)

q(q−1)²(q+ 1) + q(q−1) 2q(q−1)²(q+ 1)

X

r∈F^∗_q2\F^∗q

(θ_k(r) +θ_k(r^q))(θ_k(r) +θ_k(r^q))

= (q−1) q(q+ 1)+1

q + 1

2(q−1)(q+ 1) X

r∈F^∗_q2\F^∗q

(θ_k(r) +θ_k(r^q))(θ_k(r) +θ_k(r^q)) (4.14)

and note that replacingr withr^q in an element of a class of typeT⁽⁴⁾ produces the same conjugacy class. To evaluate the sum in the last term of the right hand side of equation (4.14), we use the following two groups. Let K be the subgroup of GL(2, q) defined earlier, which is isomorphic to F^∗_q2 and suppose thatr ∈F^∗_q2.Also, let

H=

( r 0 0 r^q

!

|r∈F^∗q

)

=

( r 0 0 r

!

|r∈F^∗q

) .

Then K and H are both abelian groups of ordersq²−1 and q−1 respectively withH < K.Now for any r∈F^∗_q2 and for fixed k,the functionγ^(k) :K−→GL(2,C) given by

γ^(k)

r 0 0 r^q

!!

= θk(r) 0 0 θ_k(r^q)

!

is a representation of K.Then γ^(k)=γ₁^(k)⊕γ₂^(k),where γ₁^(k), γ₂^(k):K −→GL(1,C) given by

Chapter 4 — GL(2, q) and Some of its Subgroups

γ₁^(k)

r 0 0 r^q

!!

= θ_k(r),

γ₂^(k)

r 0 0 r^q

!!

= θ_k(r^q).

We have γ₁^(k), γ₂^(k) ∈Irr(K).Also χ_γ(k)

r 0 0 r^q

!!

=θ_k(r) +θ_k(r^q).

We deduce thatχ_γ(k)is a sum of two non-equivalent irreducible characters. Also note thatχ_γ(k)↓^K_H = 2θ_k↓^K_H.

Now

2 = D

χ_γ(k), χ_γ(k)

E

= 1

|K|

X

g∈K

χ_γ(k)(g)χ_γ(k)(g)

= 1

q²−1 X

r∈F^∗_q2

(θ_k(r) +θ_k(r^q))(θ_k(r) +θ_k(r^q)).

Thus

X

r∈F^∗_q2

(θk(r) +θk(r^q))(θk(r) +θk(r^q)) = 2(q²−1).

Similarly

4 =

2θ_k↓^K_H,2θ_k↓^K_H

H =D

χ_γ(k)↓^K_H, χ_γ(k)↓^K_HE

H = 1

|H|

X

g∈H

χ_γ(k)↓^K_H(g)χ_γ(k)↓^K_H(g)

= 1

q−1 X

r∈F^∗q

(θ_k(r) +θ_k(r^q))(θ_k(r) +θ_k(r^q)).

Thus

X

r∈F^∗q

(θ_k(r) +θ_k(r^q))(θ_k(r) +θ_k(r^q)) = 4(q−1).

Hence

X

r∈F^∗_q2\F^∗q

(θ_k(r) +θ_k(r^q))(θ_k(r) +θ_k(r^q)) = 2(q²−1)−4(q−1) = 2(q−1)².

Now returning back to equation (4.14), we get hπ_k, π_ki = (q−1)

q(q+ 1)+1

q + 2(q−1)²

2(q−1)(q+ 1) = (q−1)²+ (q−1)(q+ 1) +q(q−1)² q(q−1)(q+ 1)

= (q−1)[q−1 +q+ 1 +q²−q]

q(q−1)(q+ 1) = q(q+ 1) q(q+ 1) = 1.

S S

Thus π_k ∈ Irr(GL(2, q)). From Table 4.5, we can see that π_kq = π_k. This restricts the number of π_k’s to ^q2₂^−q. We claim that these characters are all distinct. Assume that q + 1 - k, l and k6∼=l, lq mod(q²−1).We show thatπ_k6=π_l.Recall that the character χ_γ(k) of K and H is given on elementg= r 0

0 r^q

!

byχ_γ(k)(g) =θ_k(r) +θ_k(r^q).Now

• ifg∈H,that is r ∈F^∗q (thusr^q=r),thenχ_γ(k)(g) = 2θ_k(r)

• ifg∈K\H,that is r∈F^∗_q2\F^∗_q (thusr^q6=r),thenχ_γ(k)(g) =θk(r) +θk(r^q).

Sincek6∼=l, lq mod(q²−1),we haveχ_γ(k) andχ_γ(l) are distinct. Thus eitherθk(r)6=θl(r) for some r∈F^∗q orθ_k(r) +θ_k(r^q)6=θ_l(r) +θ_l(r^q) for some r ∈F^∗_q2 \F^∗q.Thusπ_k 6=π_l.

Finally, to be consistent with the notation given in the character table ofGL(2, q),shown in Table 4.2, we useαb^k, α∈F^∗_q to denoteχk(α).The same applies for the elementsr ∈F^∗_q2,wherebr^kmeans θ_k(r) for θ_k ∈ Irr(F^∗_q2). Also, the irreducible characters λ_k, λ_kSt, χ_k,l and π_k will be renamed toχ⁽¹⁾_k , χ⁽²⁾_k , χ⁽³⁾_k,l and χ⁽⁴⁾_k respectively. Hence, Table 4.2 is the character table of GL(2, q).This

completes the proof of the Theorem.

Summary and Discussion

It is well known from elementary theory of ordinary representations that the number of irreducible characters is the same as the number of the conjugacy classes of the finite group G. In general, there is no way of associating a conjugacy class to each irreducible character. However, we do have a very natural correspondence between the conjugacy classes and the irreducible characters of the groupGL(2, q).The groups F^∗q,F^∗_q2; and their character groups Ch(F^∗q) and Ch(F^∗_q2); are used respectively to parameterize the conjugacy classes and the irreducible characters of GL(2, q) as follows. To give a representative of a class in the first two types of classes, we use only one element α ∈F^∗q and the same for the first two types of characters, we use only one character χ∈Ch(F^∗q).

Note that the union of conjugacy classes of the first type will give the center of the groupGL(2, q), while the union of characters of the first type form a group isomorphic to the center of the group GL(2, q). To represent a class of type T⁽³⁾, we have used two distinct elements α, β ∈ F^∗_q where the conjugacy class is unaltered if we interchangeα withβ in the class. We have used two distinct charactersχ_k, χ_l∈Ch(F^∗q) to parameterize a character of the third type and we have seen that the product of χk, χl∈Ch(F^∗_q) is commutative. Finally, to obtain a class of typeT⁽⁴⁾,we made use of the elements r ∈ F^∗_q2 which are not in F^∗q and whenever we choose such r, we exclude r^q because r and r^q give the same conjugacy class. Also to produce characters of the fourth type, we used charactersθ∈Ch(F^∗_q²) which do not decompose into characters inCh(F^∗q) and whenever we choose such kto index a character of the fourth type, we excludekq from the indexing set becausek and

Chapter 4 — GL(2, q) and Some of its Subgroups

kq give the same character. So we can say that there is a complete duality between the conjugacy classes and the irreducible characters of the groupGL(2, q).Letεandθbe generators of the groups F^∗q and F^∗_q2 respectively. Table 4.6 shows the association of a class to an irreducible character of GL(2, q).

Table 4.6: Duality between irreducible characters and conjugacy classes of GL(2, q) Irreducible Character Corresponding Conjugacy Class

χ⁽¹⁾_k T_k⁽¹⁾ = ε^k 0 0 ε^k

!

χ⁽²⁾_k T_k⁽²⁾ = ε^k 1 0 ε^k

!

χ⁽³⁾_k,l T_k,l⁽³⁾= ε^k 0 0 ε^l

!

χ⁽⁴⁾_k T_k⁽⁴⁾ = 0 1

−θ^(k+1)q θ^k+θ^kq

!

As a final remark, the above duality between the conjugacy classes and irreducible characters of the groupGL(2, q) will be satisfied in general for all groupsGL(n, q).A similar table of duality for the group GL(3, q) will be given in Table 5.13.