On quasi and weak Steinberg characters of general linear groups

Abstract

Let G be a finite group and r be a prime divisor of the order of G. An irreducible character of G is said to be quasi r-Steinberg if it is non-zero on every r-regular element of G. A quasi r-Steinberg character of degree $\displaystyle |Syl_r(G)|$ is said to be weak r-Steinberg if it vanishes on the r-singular elements of $G.$ In this article, we classify the quasi r-Steinberg cuspidal characters of the general linear group $GL(n,q).$ Then we characterize the quasi r-Steinberg characters of $GL(2,q)$ and $GL(3,q).$ Finally, we obtain a classification of the weak r-Steinberg characters of $GL(n,q).$

1. Introduction

The significance of Steinberg characters to the study of finite groups of Lie type was well established by Curtis, Humphreys and Steinberg in [3, 8, 16] and [17]. Motivated by the intrinsic property of the Steinberg character, Feit introduced the notion of an r-Steinberg character for any finite group G and for any prime divisor r of the order of G (see [5]). Recall that an element of a group is called r-regular if its order is co-prime to $r.$ An irreducible character of G is said to be r-Steinberg if each r-regular element, say $\displaystyle g,$ of G takes the value $\displaystyle \pm{|C_G(g)|_r}$ on it. Here $|C_G(g)|_r$ is the highest power of r dividing the order of the centralizer $C_G(g)$ of g in $G.$

Feit conjectured [5] that if a finite simple group has an r-Steinberg character, then it is isomorphic to a simple group of Lie type in characteristic r. Darafsheh obtained an affirmative answer to this conjecture for the alternating and projective special linear groups (see [4]). Later, Tiep [18] extended the study to the rest of the finite simple groups and gave a positive answer.

In the last decade, several variants of r-Steinberg character have emerged as an important tool for studying the structure of finite groups through their characters (see [13]). One of the variants of r-Steinberg character is an r-vanishing character which is an irreducible character that vanishes on all the r-singular elements of the group. Note here that an element of a group is called r-singular if its order is divisible by r. In particular, if the degree of an r-vanishing character is $|Syl_r(G)|,$ where $\displaystyle |Syl_r(G)|$ is the order of the Sylow r-subgroup of $G,$ then the character is said to be Steinberg-like. In [9], Malle and Zalesski obtained a classification of the Steinberg-like characters of all finite simple groups.

Recently, Paul and Singla [12] introduced the notion of a quasi r-Steinberg character and a weak r-Steinberg character:

Definition 1.1. Let G be a finite group and r be a prime dividing its order. An irreducible character of G is said to be quasi r-Steinberg if it takes non-zero value on every r-regular element of $\displaystyle G.$ Further, a quasi r-Steinberg character which is Steinberg-like is said to be a weak r-Steinberg character.

It follows that the r-Steinberg and the weak r-Steinberg characters are quasi r-Steinberg, but the converse need not be true. The above two variants of r-Steinberg characters were introduced to answer a question posed by Dipendra Prasad that asked whether the existence of a weak r-Steinberg character of a finite group G implies that G is a group of Lie type. It is well known that every finite group of Lie type has a r-Steinberg character for a prime r. Hence if a group does not have a nonlinear quasi r-Steinberg character, then it cannot be a finite group of Lie type of characteristic r. Therefore, one naturally requires the classification of quasi r-Steinberg characters of any finite group G.

In [12], the authors classified all the quasi r-Steinberg characters of symmetric and alternating groups and their double covers. A classification of the quasi r-Steinberg characters for the complex reflection groups has recently been done in [11].

In the present article, we study the quasi and weak r-Steinberg characters of the general linear group $GL(n,q)$ over a finite field F_q, where q is a power of prime p. Since every linear character of G is quasi r-Steinberg for any prime divisor r of $|G|,$ one aims at the classification of the nonlinear quasi r-Steinberg characters of $G.$ The paper is divided into six sections. In the second section, we introduce the notation which we follow throughout the article. In the third section, we classify the quasi r-Steinberg cuspidal characters of $GL(n,q)$ and the following is the first main result of the paper:

Theorem 1.2. Let $\displaystyle n \geq 2$ be an integer and $\displaystyle r$ be a prime divisor of the order of $\displaystyle GL(n,q).$ Then a cuspidal character of $\displaystyle GL(n,q)$ is quasi r-Steinberg if and only if one of the following holds:

1. 1. n = 2 and $\displaystyle q = r^{\beta}+1,$ for some $\displaystyle \beta \in \mathbb{N}.$

2. 2. n = 3 and $q = 3,$ when r = 2.

3. 3. n = 3 and $q = 2,$ when r = 3.

In the fourth and the penultimate section, we characterize the quasi r-Steinberg characters of $GL(2,q)$ and $GL(3,q),$ respectively.

Let χ_l be a character of $\displaystyle F_q^{*}$ indexed by $\displaystyle l \in \{0, 1, \ldots, (q-2)\}$. Further, let $\chi_k^{(1)}$ denote a linear and χ_t denote a cuspidal character of $GL(2,q)$ indexed by some $\displaystyle k \in \{0, 1, \ldots, (q-2)\}$ and $\displaystyle t \in \{1, 2, \ldots, (q^2-1)\}.$ Assume that $L_{(2,1)}$ denotes the standard Levi complement of the standard parabolic subgroup (say $P_{(2,1)}$) of $GL(3,q).$ We denote the characters of $GL(3,q)$ obtained by parabolic induction of the irreducible characters $\chi_k^{(1)} \bigotimes \chi_l$ and $\chi_t \bigotimes \chi_l$ of $L_{(2,1)}$ by $\displaystyle \theta_{k,l}^{(4)}$ and $\displaystyle \theta_{t,l}^{(7)}$, respectively. The following is the second main result of the paper, which gives a classification of the weak r-Steinberg characters of $GL(n,q)$.

Theorem 1.3. Let $\displaystyle r$ be a prime divisor of $|GL(n,q)|$ different from $p.$ Then an irreducible character χ of $GL(n,q)$ is weak r-Steinberg if and only if one of the following holds:

1. 1. χ is a parabolically induced character of $GL(2,q)$ where $(q+1)$ is an r-power, for some odd prime r.

2. 2. $\displaystyle \chi$ is a character of type $ \displaystyle \theta_{k,l}^{(4)}$ of $GL(3,q),$ where

   • • If q is odd, then $(1+q+q^2)$ is an r-power, $3 \nmid (q-1)$ and $(l-k)$ is not invertible in $\displaystyle \mathbb{Z}_{q-1}.$

   • • If q is even, then $(1+q+q^2)$ is an r-power.

3. 3. $\displaystyle \chi$ is the character $ \theta_{t,l}^{(7)}$ of $GL(3,2)$ and r = 7.

4. 4. χ is a cuspidal character of $GL(3,2)$ and r = 3.

A proof of Theorem 1.3 is included in § 6.

2. Notation and preliminaries

For a finite group $G,$ we denote the set of its irreducible characters by $Irr(G).$ We say that the conjugacy class $[g]$ of an element g of $GL(n,q)$ is primary if its characteristic polynomial has a unique irreducible factor. Further, g is said to be irreducible if its characteristic polynomial over F_q is irreducible of degree n. For any $\displaystyle n \geq 1,$ assume that $\displaystyle F_{q^n}^{*} = \langle \epsilon_n \rangle.$ Then we denote the irreducible element $diag(\epsilon_n, \epsilon_n^q, \ldots, \epsilon_n^{q^{n-1}})$ of $GL(n,q)$ by $\displaystyle E_n$. Let $\displaystyle \hat \ : \ F_{q^n}^{*} \rightarrow \mathbb{C}^{*}$ be the homomorphism defined by the rule $\displaystyle \hat (\epsilon_{n}^{s}) = \hat{\epsilon_{n}^{s}} = e^{s \frac{2 \pi i}{q^n-1}},$ where $\displaystyle 0 \leq s \leq (q^n-2).$

The character tables of $\displaystyle GL(2,q)$ and $\displaystyle GL(3,q)$ can be found in [15] and are also included in Appendix of this paper. We follow [1] for the notation of the conjugacy classes and the characters of these groups. For any partition $\displaystyle \lambda = (\lambda_1, \lambda_2, \ldots, \lambda_m)$ of $n,$ let P_λ denote the standard parabolic subgroup of $GL(n,q).$ The unipotent radical and the Levi complement of P_λ are denoted by U_λ and $L_{\lambda},$ respectively. Assume that for any $\displaystyle 1 \leq i \leq m,$ $\displaystyle \chi_{\lambda_i} \in Irr(GL(\lambda_i, q)).$ An irreducible character of $GL(n,q)$ obtained by parabolic induction of the character of $\displaystyle P_{\lambda}$ lifted from $\displaystyle \bigotimes_{i=1}^{m} \chi_{\lambda_i} \in Irr(L_{\lambda})$ is called a parabolically induced character and is denoted by $\displaystyle \bigodot_{i=1}^{m} \chi_{\lambda_i}.$

We first list the types of the conjugacy classes of $GL(2,q)$ and $GL(3,q)$. We denote by $C^{(1)}$ and $C^{(2)}$ the types of conjugacy classes of the elements of $GL(2,q)$ whose characteristic polynomial is $\displaystyle (x-\alpha)^2,$ for some $\displaystyle \alpha \in F_{q}^{*},$ parameterized by the partitions $(1,1) $ and (2) of $2,$ respectively. We let $\displaystyle C^{(3)}$ denote the type of the conjugacy classes of elements having distinct eigen values in $F_q^{*}$. Further, note that for some $\displaystyle y \in F_{q^2} \setminus F_q,$ the matrix $$v = \begin{pmatrix} 0 & 1 \\ -y^{q+1} & y+y^{q} \end{pmatrix}$$ is an element of $GL(2,q).$ We denote by $C^{(4)}$ the types of conjugacy classes constituted by such irreducible elements of $GL(2,q)$.

Further, let $T^{(1)},$ $T^{(2)}$ and $T^{(3)}$ denote the types of conjugacy classes of the elements of $GL(3,q)$ with characteristic polynomial $\displaystyle (x-\alpha)^3,$ for some $\displaystyle \alpha \in F_{q}^{*};$ parameterized by the partitions $(1,1,1), (2,1) $ and (3) of $3,$ respectively. For any $\displaystyle \alpha \neq \beta \neq \gamma \neq \alpha \in F_q^{*},$ we denote by $T^{(4)}$ and $T^{(5)}$ the types of conjugacy classes of the elements whose characteristic polynomial is $\displaystyle (x-\alpha)^2 (x - \beta),$ for the partitions $(1,1)$ and (2) of $2,$ respectively. Further, $T^{(6)}$ denotes the conjugacy class of elements of the form $\displaystyle diag(\alpha, \beta, \gamma)$ and $\displaystyle T^{(7)}$ denotes the conjugacy class of the elements with characteristic polynomial $\displaystyle (x^2+ax+b)(x-\alpha),$ where $\displaystyle (x^2+ax+b)$ is an irreducible polynomial over $F_q.$ Finally, $\displaystyle T^{(8)}$ corresponds to the conjugacy class of irreducible elements.

Now we determine the notation for the characters of $GL(2,q)$ and $GL(3,q)$. Let $\displaystyle 0 \leq k, l \leq (q-2)$ be some integers. Then the linear character of $GL(2,q)$ indexed by k is denoted as $\chi_k^{(1)}.$ The character of degree q obtained by tensoring the linear character $\chi_k^{(1)}$ with the Steinberg character $St^{(1,1)}$ is denoted by $\displaystyle \chi_{k}^{(2)}.$ Further, for any $\displaystyle k \neq l,$ the irreducible character parabolically induced from $\chi_k \bigotimes \chi_l \in Irr(L_{(1,1)})$ is denoted by $\displaystyle \chi_{k,l}.$ Let T be the set of integers $\displaystyle 1 \leq t \leq (q^2-1)$ such that $(q+1) \nmid t$ and tq is excluded whenever t is included. Clearly, $\displaystyle |T| = \frac{q^2-q}{2}$. It is known that the number of cuspidal characters of $GL(2,q)$ is same as the size of $T.$ A cuspidal character indexed by some $\displaystyle t \in T$ is denoted by $\displaystyle \chi_{t}.$

We denote the linear character of $GL(3,q)$ indexed by k as $\theta_k^{(1)}.$ The types of characters obtained by tensoring the Steinberg characters $St^{(2,1)}$ and $St^{(1,1,1)}$ with the linear character $\theta_k^{(1)}$ are denoted by $\displaystyle \theta_k^{(2)}$ and $\displaystyle \theta_k^{(3)},$ respectively. For any integer $\displaystyle m \in \{0, 1, \ldots, (q-2)\},$ let $\displaystyle \theta_{k,l}^{(4)}, \theta_{k,l}^{(5)}$ and $ \theta_{k,l,m}^{(6)}$ denote the characters $\displaystyle \chi_{k}^{(1)} \bigodot \chi_{l}, \ \chi_{k}^{(2)} \bigodot \chi_{l}$ and $ \chi_{k,m} \bigodot \chi_l (k \neq l \neq m \neq k)$, respectively. Further, for any $t \in T$ we let $\theta_{t,l}^{(7)}$ denote the character $\chi_{t} \bigodot \chi_l.$ Finally, we consider the set V of integers $v \in \{0, 1, \ldots, q^3-1\}$ such that $ (q^2+q+1) \nmid v$ and vq, vq ² are excluded whenever v is being chosen. Then the cuspidal characters of $GL(3,q)$, indexed by $v \in V,$ are denoted by $\theta_{v}^{(8)}$ and are $|V| = \frac{q^3-q}{3}$ many.

We recall here that a cuspidal character of $GL(n,q)$ vanishes on all the non-primary conjugacy classes. For a proof of this fact, one can refer to [7, pp. 64, 67 (Theorem 3.2 and Proposition 4.1)].

3. Quasi Steinberg cuspidal characters of $GL(n,q)$

In this section, we classify the quasi r-Steinberg cuspidal characters of $\displaystyle GL(n,q).$ Observe that if $\displaystyle r=p,$ then for any $n \geq 2$ and $q \geq 3,$ the matrix $\displaystyle A = diag(\epsilon_1, 1, \ldots, 1)$ is an r-regular element of $\displaystyle GL(n,q).$ Since conjugacy class of A is not primary, a cuspidal character would vanish on it. Therefore, no cuspidal character is quasi p-Steinberg. Now assume that r is a prime different from $p.$

We first obtain the quasi r-Steinberg cuspidal characters of $GL(2,q).$ When $q=2,$ $\displaystyle |GL(2,2)| = 6$ and so r can be either 2 or 3. As $r \neq p, r=3.$ Since it follows from the character table that the cuspidal character of $GL(2,2)$ does not vanish on any conjugacy class, it implies that it is quasi 3-Steinberg. Thus we now characterize the quasi r-Steinberg cuspidal characters of $GL(2,q)$ when $\displaystyle q \geq 3:$

Proposition 3.1. Let r be a prime dividing the order of the group $GL(2,q),$ where $\displaystyle q \geq 3$ is a prime power. Then a cuspidal character of $GL(2,q)$ is quasi r-Steinberg if and only if $\displaystyle q=(r^{\beta}+1),$ for some $\displaystyle \beta \in \mathbb{N}.$

Proof. Let χ_t be a cuspidal character of $GL(2,q)$. Note that if $\displaystyle r \ \nmid \ (q-1),$ then $A = diag(\epsilon_1, 1)$ is an r-regular element. Since $\chi_t(A) = 0$, χ_t is not quasi r-Steinberg. Therefore, assume that $\displaystyle (q-1) = r^{\beta} \gamma,$ where $\beta \geq 1$ and $(r, \gamma) = 1.$ Now the following cases arise:

Case I: $\displaystyle r \ | \ (q+1).$ As r also divides $(q-1)$, r = 2. If $\displaystyle \gamma \gt 1,$ then $\displaystyle A_1 = diag(\epsilon_1^{\frac{q-1}{\gamma}}, 1)$ is 2-regular and its conjugacy class is not primary. Since $\displaystyle \chi_t(A_1) = 0, \chi_t$ is not quasi 2-Steinberg.

On the other hand, if $\displaystyle (q-1)=2^{\beta},$ then $\displaystyle (q+1) = 2^{\beta}+2.$ If $\displaystyle \beta=1,$ then $\displaystyle q=3.$ Note that the only 2-regular elements of $GL(2,3)$ are the identity matrix, say $I_2,$ and the matrix $$A_2= \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$. Since no cuspidal character vanishes on both of these elements, it follows that all the cuspidal characters of $GL(2,3)$ are quasi 2-Steinberg. But if $\displaystyle \beta \gt 1,$ then $F_{q}^{*}$ is a 2-group and in this case the only 2-regular element of $GL(2,q)$ is A ₂. The conjugacy class of A ₂ is of type $C^{(2)}$ and no cuspidal character vanishes on $C^{(2)}.$

Now we examine whether χ_t vanishes on some irreducible 2-regular element of $GL(2,q)$ or not. Since $\displaystyle (q^2-1) = 2(q-1)(2^{\beta-1}+1),$ the 2-regular elements of $\displaystyle F_{q^2}^{*}$ are of the form $\displaystyle \epsilon_{2}^{2(q-1)l}$ for some divisor l of $\displaystyle (2^{\beta-1}+1) = \frac{(q+1)}{2}.$ Thus an irreducible 2-regular element, say $A_3,$ of $GL(2,q)$ is of form $\displaystyle diag \Big(\epsilon_{2}^{2(q-1)l}, \epsilon_{2}^{2q(q-1)l} \Big)$. Note that $\displaystyle \chi_t(A_3) = \Big( \hat{\epsilon}_{2}^{2t(q-1)l} + \hat{\epsilon}_{2}^{2qt(q-1)l} \Big) = 0$ if and only if $\displaystyle 2t(q-1)l = \frac{(q^2-1)}{2} - 2t(q-1)l.$ It further implies that $\displaystyle 4tl = 2^{\beta-1}+1,$ which is not possible. Therefore, every cuspidal character is quasi 2-Steinberg.

Case II: $\displaystyle r \ \nmid \ (q+1).$ First assume that $\displaystyle (q-1)$ has a prime divisor, say $\displaystyle r^{'},$ which is different from $r.$ Then $\displaystyle A_4 = diag(\epsilon_1^{\frac{q-1}{r^{'}}}, 1)$ is an r-regular element whose conjugacy class is not primary. It follows from the previous argument that no cuspidal character is quasi r-Steinberg in this case.

Now assume that $\displaystyle (q-1) = r^{\beta},$ for some $\displaystyle \beta \in \mathbb{N}.$ Since $\displaystyle r \ \nmid \ (q+1), r$ is some odd prime. As $\displaystyle (q-1) = r^{\beta},$ no element in $C^{(1)}$ and $C^{(3)}$ is r-regular. Also, A ₂ is the only r-regular element whose conjugacy class is of type $C^{(2)}.$ One can check that no cuspidal character vanishes on it. Further, since $\displaystyle (q^2-1)=r^{\beta}(r^{\beta}+2),$ the only irreducible r-regular element of $GL(2,q)$ is of form $A_5 = diag \Big(\epsilon_{2}^{r^{\beta}l}, \epsilon_{2}^{qr^{\beta}l} \Big),$ where l is some divisor of $(r^{\beta}+2)$. Note that $\displaystyle \chi_t(A_5) = 0$ if and only if $\displaystyle \frac{(r^{\beta}+2)}{tl} = 4,$ which is not possible since r is odd. Therefore, even in this case every cuspidal character of is quasi r-Steinberg.

Note that it follows from Theorem 49.8 in [2] that the degree of a cuspidal character of $GL(n,q)$ is $\displaystyle \prod_{i=1}^{n-1} (q^i-1)$. Thus the degree of a cuspidal character of $GL(2,q)$ is $(q-1)$. Now the above discussion leads to the following:

Remark 3.2. If a cuspidal character of $GL(2,q)$ is quasi r-Steinberg, then its degree is an r-power.

In the following, we classify the quasi r-Steinberg cuspidal characters of $GL(n,q):$

Proof of Theorem 1.2

Letϕ be a cuspidal character of $GL(n,q)$. Assume that $\displaystyle q \geq 3.$ If $\displaystyle r \ \nmid \ (q-1),$ then $A = diag(\epsilon_1, 1, \ldots, 1)$ is an r-regular element of $GL(n,q)$. Since ϕ vanishes on A, it is not quasi r-Steinberg. On the other hand, when $\displaystyle r \ \mid \ (q-1),$ the following cases arise:

Case I: $\displaystyle r \ \nmid \ (q+1).$ If $\displaystyle n \geq 3,$ then $\displaystyle B_1 = diag(\epsilon_2^{q-1}, \epsilon_2^{q(q-1)}, 1, \ldots, 1)$ is an r-regular element of $GL(n,q).$ Since the conjugacy class $[B_1]$ is not primary, it follows that ϕ is not quasi r-Steinberg.

Case II: $\displaystyle r \ | \ (q+1).$ Since $\displaystyle r \ \mid \ (q-1),$ r = 2. Assume that $\displaystyle (q-1) = 2^{\beta} \gamma,$ where $\displaystyle \beta \geq 1$ and $(\gamma,2)=1.$ If $\displaystyle \gamma \gt 1,$ then $\displaystyle B_2 = diag \Big(\epsilon_1^{\frac{q-1}{\gamma}}, 1, \ldots, 1 \Big)$ is 2-regular and its conjugacy class is not primary. Therefore, the previous argument implies that ϕ is not quasi 2-Steinberg.

But if $\displaystyle \gamma = 1,$ then $\displaystyle (q-1) = 2^{\beta}.$ If $\displaystyle \beta=1,$ then $\displaystyle q=3.$ Now for any $\displaystyle n \gt 3,$ $\displaystyle (3^n-3^{n-3}) = 3^{n-3}(13)(2)$ is a divisor of $\displaystyle |GL(n,3)|.$ Now as $$\displaystyle B_3 = \begin{pmatrix} 0 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 2 & 1 \end{pmatrix}$$ is of order $13,$ the element $\displaystyle B_4 = diag(B_3, 1, \ldots, 1) \in GL(n,3)$ is 2-regular. Therefore, for any $n \gt 3,$ no cuspidal character of $GL(n,3)$ is quasi 2-Steinberg. When $n=3,$ the only 2-regular elements of $GL(3,3)$ are the elements of order 13. Now note that the degree of any cuspidal character of $GL(3,3)$ is $(3-1)(3^2-1)=16$ and it follows from the character table of $GL(3,3)$ that no character of degree 16 vanishes on any element of order 13. Therefore, every cuspidal character of $GL(3,3)$ is quasi 2-Steinberg.

Further, if $\displaystyle \beta \gt 1,$ then $\displaystyle (q+1) = 2(2^{\beta-1}+1).$ Observe that for any $\displaystyle n \geq 3,$ the conjugacy class of the 2-regular element $\displaystyle diag(\epsilon_2^{2(q-1)}, \epsilon_2^{2q(q-1)}, 1, \ldots, 1)$ is not primary. Therefore, even in this case no cuspidal character is quasi r-Steinberg.

Now consider the group $GL(n,2),$ where $n \geq 3.$ Note that if $r \neq 3,$ then $(q+1) = 3.$ Therefore, the argument in Case I provides that no cuspidal character of $GL(n,2)$ is quasi r-Steinberg. If $r=3,$ then for any $\displaystyle n \geq 4, B_4 = diag(E_3, 1, \ldots, 1)$ is a 3-regular element of $GL(n,2)$. As $\phi(B_4) = 0, \phi$ is not quasi 3-Steinberg. Further, it follows from the character table that every cuspidal character $GL(3,2)$ is quasi 3-Steinberg. Now the result follows from Proposition 3.1.

4. Quasi Steinberg characters of $GL(2,q)$

In this section, we obtain a characterization of the nonlinear quasi r-Steinberg characters of $\displaystyle GL(2,q).$ In this direction, we start with the parabolically induced characters of $GL(2,q)$. We first make the following remark towards this end:

Remark 4.1. Let n be an odd prime and $\displaystyle r$ be a prime divisor of $|GL(n,q)|$ different from p. Assume that $\displaystyle r \ | \ (q^i-1),$ for some $\displaystyle 1 \lt i \lt n.$ If r is also a divisor of $(q^n-1),$ then since $\displaystyle gcd(q^i-1, q^n-1) = (q^{gcd(i,n)}-1) = (q-1),$ $\displaystyle r \ | \ (q-1).$ It further implies that $\displaystyle r \ \nmid \ \frac{(q^i-1)}{q-1}.$ Indeed if r divides $\displaystyle \frac{(q^i-1)}{q-1},$ then since $\displaystyle gcd \Big(\frac{q^i-1}{q-1}, q^n-1 \Big) = 1,$ $\displaystyle r \ | \ 1,$ which is not possible.

In the following, we establish a necessary condition for a parabolically induced character to be quasi r-Steinberg:

Lemma 4.2. Let n be a prime and $\displaystyle r (\neq p)$ be a prime divisor of $|GL(n,q)|.$ If a parabolically induced character ψ of $GL(n,q)$ is quasi r-Steinberg, then r is the only prime divisor of $\displaystyle \frac{(q^n-1)}{(q-1)}$.

Proof. Assume that n is an odd prime. If r is $2,$ then since $\displaystyle \frac{(q^n-1)}{(q-1)} = \sum_{j=0}^{n-1} q^j$ is odd, $\displaystyle r \nmid \frac{(q^n-1)}{(q-1)}.$ Thus $\displaystyle B = diag(\epsilon_n^{q-1}, \epsilon_n^{q(q-1)}, \ldots, \epsilon_n^{q^{n-1}(q-1)})$ is an r-regular element of $GL(n,q).$ Note that the order $\displaystyle \sum_{j=0}^{n-1} q^j$ of $\epsilon_n^{q-1}$ does not divide $\displaystyle (q^i-1),$ for any $\displaystyle 1 \leq i \lt n.$ Therefore, $\epsilon_n^{q-1} \in F_{q^{n}}^{*} \setminus F_{q^{i}}^{*} \ \forall \ 1 \leq i \lt n$ and so B is an irreducible element of $GL(n,q)$. Since the conjugacy class of an irreducible element of $GL(n,q)$ intersects $\displaystyle P_{\lambda}$ trivially, $\displaystyle \psi(B) = 0.$ Therefore, ψ is not quasi r-Steinberg.

Now consider the case when r is an odd prime. Note that if $\displaystyle r \ | \ (q+1),$ then $\displaystyle r \ \nmid \sum_{j=0}^{n-1} q^j$. Thus the previous argument implies that ψ is not quasi r-Steinberg. Now let $\displaystyle r \ \nmid (q+1).$ Suppose $\displaystyle \sum_{j=0}^{n-1} q^j = r^{\delta}k,$ where $(r,k)=1$ and $\delta \geq 0$ is an integer. If $k=1,$ then $\displaystyle \frac{(q^n-1)}{(q-1)}$ is an r-power and our claim is established. On the other hand, if $k \gt 1,$ then the order of $\displaystyle \epsilon = \epsilon_n^{(q-1)r^{\delta}}$ is k and so $\displaystyle B^{'} = diag(\epsilon, \epsilon^q, \ldots, \epsilon^{q^{n-1}})$ is an r-regular element of $GL(n,q)$. Further, note that $B^{'}$ is irreducible if and only if $\displaystyle o(\epsilon) = k \ \nmid \ (q^i-1),$ for any $1 \leq i \leq (n-1).$ Since $\displaystyle \Big(q^i-1, \frac{(q^n-1)}{(q-1)} \Big) = 1,$ it follows that $\displaystyle \frac{(q^i-1)}{k}$ is not an integer. Thus $B^{'}$ is an irreducible r-regular element and ψ vanishes on it. It implies that no parabolically induced character of $GL(n,q)$ is quasi r-Steinberg in this case.

Finally, we consider the case of $\displaystyle n=2.$ If $\displaystyle r \ \nmid \ (q+1),$ then the element $\displaystyle diag(\epsilon_2^{(q-1)}, \epsilon_2^{q(q-1)})$ is an irreducible r-regular element of $GL(2,q)$ and the previous argument again implies that no parabolically induced character is quasi r-Steinberg. Further, assume that $\displaystyle (q+1) = r^{\beta} \gamma,$ where $\displaystyle \beta \geq 1$ and $\displaystyle (r, \gamma) = 1.$ If $\displaystyle \gamma \gt 1,$ then in the following we prove that there always exists an r-regular element on which ψ vanishes:

• • $r \nmid (q-1)$: If $\gamma \ | \ (q-1),$ then $\displaystyle \gamma = 2.$ Let $\displaystyle (q-1) = 2^{\alpha} \kappa,$ where $\displaystyle (2, \kappa) = 1.$ Then $\displaystyle diag(\epsilon_2^{\frac{(q^2-1)}{2^{\alpha+1}}}, \epsilon_2^{\frac{q(q^2-1)}{2^{\alpha+1}}})$ is an irreducible r-regular element. If $\displaystyle \gamma \nmid \ (q-1),$ then $\displaystyle diag(\epsilon_2^{\frac{(q^2-1)}{\gamma}}, \epsilon_2^{\frac{q(q^2-1)}{\gamma}})$ is an irreducible r-regular element.

• • $r \mid (q-1)$: If $\displaystyle \gamma \ | \ (q-1),$ then $\displaystyle \gamma = 2,$ which is not possible as r = 2. Therefore, $\displaystyle \gamma \nmid \ (q-1);$ and hence $\displaystyle s_3$ is an irreducible r-regular element.

Therefore, ψ is not quasi r-Steinberg. Further, for $\gamma = 1,$ $\displaystyle (q+1) = r^{\beta}$. Now the result follows.

We now classify the quasi r-Steinberg characters of $GL(2,q)$:

Theorem 4.3. Let $q \geq 3$ be a prime power, r be a prime dividing $\displaystyle |GL(2,q)|$ and $\displaystyle \chi \in \Big\{\chi_k^{(2)}, \chi_{k,l} \ | \ 0 \leq k \lt l \leq (q-2) \Big\}.$

1. 1. If $\displaystyle r=p,$ then χ is quasi p-Steinberg if and only if $\displaystyle \chi$ is of type $\chi_k^{(2)}.$

2. 2. If $\displaystyle r \neq p,$ then χ is quasi r-Steinberg if and only if $\displaystyle \chi$ is of type $\chi_{k,l}$ and $\displaystyle q = (r^{\delta}-1).$

Proof. First assume that $\displaystyle r = p.$ Note that any character of degree q vanishes only on the elements of the form $$ t_{\alpha} = \begin{pmatrix} \alpha & 1 \\ 0 & \alpha \end{pmatrix} \in C^{(2)},$$ for some $\displaystyle \alpha \in F_{q}^{*}.$ Since $\displaystyle o(t_{\alpha}) = lcm(o(\alpha), p),$ no element of $C^{(2)}$ is p-regular. Therefore, for any $\displaystyle 0 \leq k \leq (q-2),$ $\displaystyle \chi_k^{(2)}$ is a quasi p-Steinberg character. Also, a parabolically induced character vanishes on the irreducible r-regular element E ₂ and hence is not quasi p-Steinberg.

Further, let $\displaystyle r \neq p.$ In this case, t ₁ is an r-regular element of $GL(2,q).$ Since $\displaystyle \chi_k^{(2)}$ vanishes on t ₁, it is not quasi r-Steinberg. Now we study when a parabolically induced character of $GL(2,q)$ is quasi r-Steinberg. In this direction, a necessary condition provided by Lemma 4.2 is that $\displaystyle (q+1) = r^{\beta},$ for some $\beta \in \mathbb{N}.$

Let $\chi_{k,l}$ be a parabolically induced character, where $0 \leq k \lt l \leq (q-2)$. It follows from the character table that it takes 0 value only on the elements whose conjugacy class is of type $\displaystyle C^{(3)}$ or $\displaystyle C^{(4)}.$ Observe that if r is some odd prime, then $r \ \nmid \ (q-1)$. Indeed, if $r \ | \ (q-1),$ then $r=2,$ which is not the case. Now since $\displaystyle (q^2-1)=(q-1)r^{\beta},$ it follows that no irreducible element of $GL(2,q)$ is r-regular. Therefore, the elements whose conjugacy class is of type $\displaystyle C^{(4)}$ are not r-regular.

Further, let $\displaystyle V = diag(\alpha, \beta)$ has conjugacy class of type $C^{(3)}.$ Then $\chi_{k,l}(V) = \hat{\alpha}^k \hat{\beta}^l + \hat{\alpha}^l \hat{\beta}^k = 0$ if and only if $F_{q}^{*}$ contains the primitive second root of unity. Since r being odd implies q is even, this is also not the case. Therefore, $\displaystyle \chi_{k,l}$ is quasi r-Steinberg.

Now if $\displaystyle (q+1) = 2^{\delta},$ then $\displaystyle (q-1) = 2(2^{\delta-1}-1).$ Note that $\displaystyle \delta = 2$ implies q = 3. Since there is no 2-regular element in $C^{(3)}$ and $C^{(4)}$, the parabolically induced characters of $GL(2,3)$ are quasi 2-Steinberg. Further, if $\displaystyle \delta \geq 3,$ then there is no 2-regular element in $C^{(4)}.$ On the other hand, assume that the conjugacy class of $\displaystyle W = diag(\epsilon_1^{s_1}, \epsilon_1^{s_2}),$ for some $\displaystyle 0 \leq s_1 \lt s_2 \leq (q-2),$ is of type $C^{(3)}.$ Then $\displaystyle \chi_{k,l}(W) = \hat{\epsilon_1}^{s_1k+s_2l} + \hat{\epsilon_1}^{s_1l+s_2k} = 0$ if and only if $\displaystyle (s_2-s_1)(l-k) = \frac{(q-1)}{2}.$ Observe that a pair of such distinct elements s ₁ and s ₂ exists if $\displaystyle (l-k)$ is invertible in $\displaystyle \mathbb{Z}_{q-1}$. Further, note that for any $\displaystyle 1 \leq i \leq 2, \ o(\epsilon_i^{s_i}) = \frac{(q-1)}{s_i}$ and so $\displaystyle o(W) = lcm \Big( \frac{(q-1)}{s_1}, \frac{(q-1)}{s_2} \Big).$ Since s ₁ and s ₂ have different parity, 2 divides $o(W).$ Therefore, W is not a 2-regular element. Thus an element of $C^{(3)}$ on which $\chi_{k,l}$ vanishes, is not 2-regular. It implies that if $\displaystyle (l-k)$ is invertible in $\displaystyle \mathbb{Z}_{q-1},$ then $\chi_{k,l}$ is quasi 2-Steinberg. On the other hand, if $\displaystyle (l-k)$ is not invertible in $\displaystyle \mathbb{Z}_{q-1},$ then there does not exist any element in $\displaystyle C^{(3)}$ on which $\displaystyle \chi_{k,l}$ vanishes. Now the result follows.

5. Quasi Steinberg characters of $GL(3,q)$

In this section, we classify the quasi r-Steinberg characters of $GL(3,q).$ Since the discussion on the cuspidal characters has been done in § 3, we now continue the study for the remaining nonlinear characters of $GL(3,q)$:

Theorem 5.1. Let us denote by S the set constituted by all the irreducible characters of $GL(3,q)$ except the linear and cuspidal ones. If $r = p,$ then the only characters in S that are quasi p-Steinberg are of type $\displaystyle \theta_k^{(3)}$. If $r \neq p,$ then we have the following:

1. 1. If q is even, then the characters of type $\displaystyle \theta_{k,l}^{(4)}$ are quasi r-Steinberg if and only if $(1+q+q^2)$ is an r-power.

2. 2. If q is odd, then the characters of type $\displaystyle \theta_{k,l}^{(4)}$ are quasi r-Steinberg if and only if $(l-k)$ is not invertible in $\displaystyle \mathbb{Z}_{q-1}, (1+q+q^2)$ is an r-power and $3 \nmid (q-1).$

3. 3. The characters of type $\displaystyle \theta^{(7)}_{t,l}$ are quasi r-Steinberg if and only if r = 7 and q = 2.

Proof. Case I: r = p. Note that $\displaystyle B = diag(E_2, 1)$ is an r-regular element and its conjugacy class is of type $T^{(7)}.$ It follows from the character table of $GL(3,q)$ that any character of type $\displaystyle \theta_k^{(2)}$ vanishes on $B.$ Therefore, it is not quasi p-Steinberg. Now note that a character of type $\theta_k^{(3)}$ takes 0 value only on the conjugacy classes of type $\displaystyle T^{(2)}, T^{(3)}$ and $T^{(5)}.$ Since there are no p-regular elements in these classes, $\theta_k^{(3)}$ is quasi p-Steinberg. Further, any parabolically induced character is not quasi p-Steinberg as it vanishes on the irreducible r-regular element $\displaystyle E_n.$

Case II: r ≠ p. Consider the element $$ C = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$$. Since its order is $p,$ it is r-regular and its conjugacy class is of type $T^{(3)}.$ Since the characters $\displaystyle \theta_k^{(2)}, \theta_k^{(3)}$ and $\displaystyle \theta_k^{(5)}$ vanish on $T^{(3)},$ they are not quasi r-Steinberg. Now we study the parabolically induced characters $\theta^{(4)}_{k,l},$ $\theta^{(6)}_{k,l,m}$ and $\theta^{(7)}_{t,l}.$

Observe that if $\displaystyle r \ | \ (q+1),$ then $\displaystyle r \ \nmid (q^2+q+1).$ Thus it follows from Lemma 4.2 that no parabolically induced character is quasi r-Steinberg.

Remark 5.2. For an odd prime $n,$ if a parabolically induced character of $GL(n,q)$ is quasi r-Steinberg, then $\displaystyle r \ \nmid \ (q+1).$

Therefore, in the following, we assume that r does not divide $\displaystyle (q+1):$

The characters of type $\theta^{(6)}_{k,l,m}$

Note that the order of the element $\displaystyle D_1 = diag(\epsilon_2^{(q-1)}, \epsilon_2^{q(q-1)}, 1)$ is $(q+1)$. Since $r \ \nmid \ (q+1), D_1$ is an r-regular element. It follows from the character table that any character of type $\theta^{(6)}_{k,l,m}$ vanishes on D ₁ and hence is not quasi r-Steinberg.

The following lemma gives a necessary condition for a parabolically induced character of $GL(3,q)$ to be quasi r-Steinberg:

Lemma 5.3. Let $\displaystyle q \geq 3.$ If a parabolically induced character of $GL(3,q)$ is quasi r-Steinberg, then $r \ \nmid \ (q-1).$

Proof. On contrary, assume that $\displaystyle q = (1+rm),$ for some $m \in \mathbb{N}$. Then $\displaystyle (1+q+q^2) = (3 + 3rm+r^2m^2).$ Clearly, if r is different from $3,$ then $r \nmid (1+q+q^2).$ Further, if $\displaystyle r=3,$ then $(1+q+q^2) = 3(1+3(m+m^2)),$ which is not a 3-power. In either case, we get a contradiction to Lemma 4.2.

In other words, if $\displaystyle (1+q+q^2)$ is an r-power, then $\displaystyle r \ \nmid \ (q-1).$ Also if $\displaystyle r \ \nmid \ (q-1),$ then since we have already observed that $\displaystyle r \ \nmid \ (q+1),$ it implies that $\displaystyle r \ \nmid (q^2-1)$ as well. With this note, we now continue our investigation:

The characters of type $\theta^{(4)}_{k,l}$

It follows from the character table that the classes on which a character of type $\displaystyle \theta_{k,l}^{(4)}$ can vanish are of type $T^{(5)}$ or $T^{(6)}.$ Therefore, if the value of $\displaystyle \theta_{k,l}^{(4)}$ on every r-regular element from these conjugacy classes is non-zero, then it follows from Lemma 4.2 that it is quasi r-Steinberg if and only if $\displaystyle (1+q+q^2)$ is an r-power.

Let $\displaystyle A_{\alpha, \beta}$ and $\displaystyle B_{\alpha, \beta, \gamma}$ be some elements of $GL(3,q)$ whose conjugacy class is of type $T^{(5)}$ and $T^{(6)},$ respectively. Note that $\displaystyle \theta_{k,l}^{(4)}(A_{\alpha, \beta}) = (\hat{\alpha^{k+l}} \hat{\beta^l} + \hat{\beta^{2l}} \hat{\alpha^k})$ and $\displaystyle \theta_{k,l}^{(4)}(B_{\alpha, \beta, \gamma}) = \hat{\alpha^{k}} \hat{\beta^l} \hat{\gamma^l} + \hat{\alpha^{l}} \hat{\beta^k} \hat{\gamma^l} + \hat{\alpha^{l}} \hat{\beta^l} \hat{\gamma^k}.$ Therefore, if $\displaystyle \theta_{k,l}^{(4)}$ vanishes on some conjugacy class of type $T^{(5)}$ or $T^{(6)},$ then $F_q^{*}$ contains a primitive second or third root of unity, respectively.

Now the following corollary to Lemma 4.2 classifies the quasi r-Steinberg characters of type $\displaystyle \theta_{k,l}^{(4)}$ of $GL(3,q)$ for even q:

Corollary 5.4. Let q be even. Then $\displaystyle \theta_{k,l}^{(4)}$ is quasi r-Steinberg character of $GL(n,q)$ if and only if $\displaystyle (1+q+q^2)$ is a power of $r.$

Proof. Since q is even, $F_q^{*}$ does not contain the primitive second root of unity and hence $\displaystyle \theta_{k,l}^{(4)}$ does not vanish on $T^{(5)}$. Now we check whether $F_{q}^{*}$ contains a primitive third root of unity. If q is of the form $\displaystyle (3m+2),$ then $\displaystyle 3 \ \nmid \ (q-1).$ As $F_{q}^{*}$ does not have a primitive third root of unity, it follows that $\displaystyle \theta_{k,l}^{(4)}$ does not vanish on $T^{(6)}$. Thus it follows from Lemma 4.2 that $\displaystyle \theta_{k,l}^{(4)}$ is quasi r-Steinberg if and only if $\displaystyle (1+q+q^2)$ is a power of $r.$ On the other hand, if $\displaystyle q = (3m+1),$ then $\displaystyle (1+q+q^2) = 3(1+3m^2+3m)$ is not a prime power. So $\displaystyle \theta_{k,l}^{(4)}$ is not a quasi r-Steinberg character of $GL(3,q)$ for such q.

Example. Since $\displaystyle q=8$ is of the form $(3m+2)$ and $\displaystyle (1+q+q^2) = 73$ is a prime, any character of type $\displaystyle \theta_{k,l}^{(4)}$ is a quasi 73-Steinberg character of $GL(3,8).$

Let us now consider the case when q is odd. It follows from Lemma 5.3 that if $\displaystyle \theta_{k,l}^{(4)}$ is quasi r-Steinberg, then $\displaystyle r \ \nmid \ (q-1).$ Therefore, the elements in the conjugacy classes of type $\displaystyle T^{(4)}, T^{(5)}, T^{(6)}$ and $T^{(7)}$ are r-regular.

If $\displaystyle \alpha = \epsilon_1^{s_1}$ and $\displaystyle \beta = \epsilon_1^{s_2}$ are some elements of $F_{q^*}$ with $\displaystyle 1 \leq s_1 \lt s_2 \leq (q-1),$ then $\displaystyle \theta_{k,l}^{(4)}(A_{\alpha, \beta}) = \hat{\epsilon_1}^{s_1(k+l)+s_2l} + \hat{\epsilon_1}^{(2s_1l+s_2k)}.$ Therefore, $\displaystyle \theta_{k,l}^{(4)}(A_{\alpha, \beta}) = 0$ if and only if $\displaystyle (s_1-s_2)(k-l) = \frac{(q-1)}{2}.$ It further implies that if k and l are such that $(k-l)$ is not invertible in $\mathbb{Z}_{q-1},$ then $\displaystyle \theta_{k,l}^{(4)}$ does not vanish on any element whose conjugacy class is of type $T^{(5)}$. On the other hand, if $\displaystyle (k-l, q-1) = 1,$ then for the choice of $\displaystyle s_1 = \Big( \frac{q-1}{2(k-l)} + s_2 \Big), \theta_{k,l}^{(4)}(A_{\alpha, \beta}) = 0.$

Here note that if $\displaystyle \theta_{k,l}^{(4)}$ is quasi r-Steinberg, then $\displaystyle 3 \ \nmid \ (q-1).$ Indeed if $\displaystyle 3 \ \mid \ (q-1),$ then it follows from a similar discussion as in Corollary 5.4 that $\displaystyle (1+q+q^2)$ is not an r-power. Thus we obtain the following necessary conditions for $\displaystyle \theta_{k,l}^{(4)}$ to be quasi r-Steinberg:

Corollary 5.5. Let q be odd. If a character of type $\displaystyle \theta_{k,l}^{(4)}$ is quasi r-Steinberg character of $GL(3,q),$ then r is the only prime divisor of $\displaystyle (1+q+q^2),$ $(l-k)$ is not invertible in $\mathbb{Z}_{q-1}$ and $\displaystyle 3 \ \nmid \ (q-1).$

Now we obtain the sufficient conditions for a character of type $\displaystyle \theta_{k,l}^{(4)}$ to be quasi r-Steinberg. Clearly, if $\displaystyle 3 \nmid (q-1),$ then $\displaystyle \theta_{k,l}^{(4)}(B_{\alpha, \beta, \gamma}) \neq 0 \ \forall \ \alpha, \beta, \gamma \in F_{q}^{*}.$ Therefore, the sufficient conditions for a character $\displaystyle \theta_{k,l}^{(4)}$ to be quasi r-Steinberg are that r is the only prime divisor of $\displaystyle (1+q+q^2),$ $(l-k)$ is not invertible in $\mathbb{Z}_{q-1}$ and $\displaystyle 3 \ \nmid \ (q-1).$ It now establishes the parts (i) and (ii) of the Theorem.

We now make a remark about the existence of infinitely many primes q such that many characters of type $\displaystyle \theta^{(4)}_{k,l}$ are quasi r-Steinberg characters of $GL(3,q),$ for some $r.$ For instance, in addition to the aforementioned sufficient conditions, let us assume that q is such that $\displaystyle 4 \nmid (q-1).$ Since q is odd, it implies that 2 is not invertible in $\mathbb{Z}_{q-1}.$ Therefore, as a consequence of the above discussion, we have that whenever $\displaystyle k = (l+2), \ \theta_{k,l}^{(4)}(A_{\alpha, \beta}) \neq 0 \ \forall \ \alpha, \beta \in F_{q}^{*}.$ This now leads to the following corollary which establishes our remark:

Corollary 5.6. As per Bunyakovsky conjecture there exist infinitely many q for which $\displaystyle (1+q+q^2)$ is a prime. Then any such q is of the form $\displaystyle q \not\equiv 1 \mod 12$ and if $\displaystyle (1+q+q^2) = r$(a prime), then the character $\displaystyle \theta^{(4)}_{k,l},$ where $\displaystyle k =(l+2),$ is a quasi r-Steinberg character of $GL(3,q).$

Proof. Bunyakovsky conjecture predicts that there exist infinitely many q for which $\displaystyle (1+q+q^2)$ is a prime. We first note that such a q cannot be of the form $\displaystyle (12m+1).$ Indeed, then $\displaystyle (1+q+q^2) = 3(1+12m(1+4m))$, which cannot be a prime. It implies that for any q for which $(1+q+q^2)$ is a prime, say $r,$ $\displaystyle q \not\equiv 1 \mod(12).$ Now it follows from Corollary 5.5 and the discussion following it, that for any such $q,$ $\displaystyle \theta^{(4)}_{l+2,l}$ is a quasi r-Steinberg character of $GL(3,q).$

The characters of type $\theta^{(7)}_{t,l}$

It follows from character table that $\displaystyle \theta^{(7)}_{t,l}$ vanishes on all the conjugacy classes of type $\displaystyle T^{(6)}$ and $\displaystyle T^{(8)}.$ Now we study the values of $\theta^{(7)}_{t,l}$ on the conjugacy classes of type $T^{(7)}$. Note that for $\displaystyle E_{i,\alpha} = diag(F_i, \alpha),$ where $\displaystyle F_i = diag(\epsilon_2^i, \epsilon_2^{qi})$ is an irreducible in $\displaystyle GL(2,q), [E_{i,\alpha}]$ is of type $T^{(7)}$. Here $\displaystyle \theta^{(7)}_{t,l}(E_{i,\alpha}) = \hat{\alpha}^t (\hat{\epsilon_2}^{li} + \hat{\epsilon_2}^{lqi}).$ If $\theta^{(7)}_{t,l}$ vanishes on $E_{i,\alpha},$ then $F_{q^2}^{*}$ contains the primitive second root of unity. Note that if $\theta^{(7)}_{t,l}$ is quasi r-Steinberg, then $\displaystyle r \ \nmid (q-1)$ (see Lemma 5.3). Now the further discussion is as follows:

• • q ≥ 4: Let $\displaystyle \alpha, \beta, \gamma$ be distinct elements in $F_q^{*}.$ Now since $\theta^{(7)}_{k,l}$ vanishes on the r-regular element $\displaystyle B_2 = diag(\alpha, \beta, \gamma),$ it is not quasi r-Steinberg.

• • q = 2: Note that $\displaystyle (1+q+q^2) = 7.$ Since $(q-1) = 1,$ there is no element in $T^{(6)};$ and as $F_{2^2}^{*}$ does not contain the primitive second root of unity, $\displaystyle \theta^{(7)}_{t,l}$ does not vanish on $T^{(7)}$ either. Since there does not exist any irreducible 7-regular element in $GL(3,2),$ it follows that $\displaystyle \theta^{(7)}_{t,l}$ is a quasi 7-Steinberg character of $GL(3,2).$ Further, if $\displaystyle r$ is some prime different from $7,$ then it follows from Lemma 4.2 that $\displaystyle \theta^{(7)}_{t,l}$ is not quasi r-Steinberg.

• • q = 3: In this case, $(1+q+q^2) = 13$. Note that the element $\displaystyle E_{2,1} = diag(F_1, 1),$ where $\displaystyle F_1 = diag(\epsilon_2^2, \epsilon_2^{6}),$ is a 13-regular element whose class is of type $T^{(7)}.$ Since $\displaystyle \theta^{(7)}_{t,l}$ vanishes on $\displaystyle E_{2,1}$, it is not quasi 13-Steinberg. Now one can conclude from Lemma 4.2 that $\displaystyle \theta^{(7)}_{t,l}$ is not a quasi r-Steinberg character of $GL(3,3),$ for any r.

6. Weak Steinberg characters of general linear groups

In this section, we classify the weak r-Steinberg characters of $GL(n,q).$ Towards this end, we first make the following remark:

Remark 6.1. If $\displaystyle \chi$ is a weak r-Steinberg character of $\displaystyle GL(n,q),$ then $\displaystyle r \ \nmid \ (q-1).$ Indeed if $\displaystyle r \ \mid \ (q-1),$ then the central elements $\displaystyle \alpha I_n,$ where $\displaystyle 1 \neq \alpha \in F_q^{*},$ are r-singular. It implies that $\displaystyle \chi(\alpha I_n) = 0,$ which is a contradiction.

Recall that by definition, a weak r-Steinberg character is quasi r-Steinberg. In the following, we first determine which quasi r-Steinberg characters of $GL(2,q)$ and $GL(3,q)$ are weak r-Steinberg:

Lemma 6.2. Let $\displaystyle q \geq 3$ be a prime power. Then a nonlinear quasi r-Steinberg character χ of $GL(2,q)$ is weak r-Steinberg if and only if one of the following holds:

1. 1. χ is of type $\chi_k^{(2)}$ and $r=p.$

2. 2. χ is a parabolically induced character and $(q{+}1)$ is an r-power, for some odd prime r.

Proof. Note that $|Syl_p(GL(2,q))| = q.$ If $r=p,$ then the characters of type $\chi_k^{(2)}$ are weak p-Steinberg as their degree is q and they vanish on all p-singular elements of $GL(2,q)$. Now we consider the case when r is some prime other than $p.$ In this direction, first recall that a cuspidal character of $GL(2,q)$ is quasi r-Steinberg if and only if $\displaystyle (q-1) = r^{\beta}.$ Since $\displaystyle |GL(2,q)|= q(q-1)^2(q+1), |Syl_r(GL(2,q))| \geq (q-1)^2.$ Since the degree of a cuspidal character is $(q-1)$, it is not weak r-Steinberg.

Further, note that it follows from Theorem 4.3 that a parabolically induced character $\displaystyle \chi_{k,l}$ is quasi r-Steinberg if and only if its degree, $\displaystyle (q+1),$ is an r-power. If $r=2,$ then $ (q-1)$ is also divisible by 2 and hence $\displaystyle \chi_{k,l}(1) \neq |Syl_r(GL(2,q))|.$ On the other hand, if $\displaystyle r \neq 2,$ then q is even and $\displaystyle \chi_{k,l}(1) = |Syl_r(GL(2,q))|.$ Since there does not exist any r-singular element in $C^{(1)}, C^{(2)}$ and $C^{(3)},$ $\displaystyle \chi_{k,l}$ is r-vanishing too. Hence the result follows.

Lemma 6.3. A cuspidal character of $GL(3,q)$ is weak r-Steinberg if and only if q = 2 and r = 3. Otherwise, an irreducible nonlinear character of $GL(3,q)$ is weak r-Steinberg if and only if it is quasi r-Steinberg.

Proof. Since the degree of a cuspidal character is $\displaystyle (q-1)(q^2-1),$ it follows from Theorem 1.2 that a quasi r-Steinberg cuspidal character of $GL(3,q)$ has a prime power degree if and only if $(q,r) \in \{(3,2), (2,3)\}.$ If $q=3,$ then since $2 \ | \ (3-1),$ it follows from Remark 6.1 that no cuspidal character of $GL(3,3)$ is weak 2-Steinberg. Further, if $q=2,$ then the degree of a cuspidal character of $GL(3,2)$ is 4 which is $|Syl_3(GL(3,2))|.$ Now since any cuspidal character of $GL(3,2)$ is 3-vanishing, it is weak 3-Steinberg. Hence the first part of result is established.

Observe that the degree of a character of type $\theta_k^{(2)}$ is not a prime power and that any character of type $\theta_k^{(3)}$ is weak r-Steinberg if and only if $r=p.$ Recall that if a parabolically induced character of $GL(3,q)$ is quasi r-Steinberg, then $(1+q+q^2)$ is an r-power. It follows that $r \neq p.$

Assume that the integers $k, l$ and t are such that the characters $\displaystyle \theta_{k,l}^{(4)}$ and $\displaystyle \theta_{t,l}^{(7)}$ are quasi r-Steinberg (c.f. Theorem 5.1). Since they satisfy the degree condition, we check whether they are r-vanishing or not. It follows from Lemma 5.3 that $\displaystyle r \ \nmid \ (q-1)$ and hence no element in the classes of type $\displaystyle \{T^{(i)} \ | \ 1 \leq i \leq 6 \}$ is r-singular. Further, if $\displaystyle r \ | \ (q+1),$ then no parabolically induced character of $GL(3,q)$ is quasi r-Steinberg. Therefore, $\displaystyle r \ \nmid \ (q^2-1).$ It implies that no element in any class of type $\displaystyle T^{(7)}$ is r-singular. Moreover, as r is a divisor of $GL(3,q),$ it follows that $r \ | \ (q^3-1).$ Hence $GL(3,q)$ has irreducible r-singular elements. Since $\displaystyle \theta_{k,l}^{(4)}$ and $\displaystyle \theta_{t,l}^{(7)}$ vanish on them, they are r-vanishing. Therefore, the quasi r-Steinberg parabolically induced characters of $GL(3,q)$ are weak r-Steinberg.

Now we characterize the weak r-Steinberg characters of $GL(n,q).$ Recall that an irreducible character χ of $\displaystyle GL(n,q)$ is of the form $\displaystyle (g_1^{n_1} g_2^{n_2} \ldots g_k^{n_k}),$ where for any $\displaystyle 1 \leq i \leq k,$ the degree of the simplex g_i is $\displaystyle d_i$ and $\displaystyle \sum_{i=1}^{k} d_i n_i = n$ (see [6]). Further, it follows from [4] that if the degree of χ is some prime power, then

(6.1)\begin{equation} \chi(1) = \frac{\psi_n(q)}{\psi_{n_1}(q^{d_1}) \psi_{n_2}(q^{d_2}) \ldots \psi_{n_k}(q^{d_k})}, \end{equation}

where $\displaystyle \psi_{x}(q^{y}) = \prod_{i=1}^{x} (q^{iy}-1).$ Now we state the Zsigmondy’s theorem (see Theorem 3 in [14]), which is crucial to the upcoming discussion.

Let a and l be integers greater than 1. Then there exists a prime divisor s of $\displaystyle (a^l-1)$ such that $\displaystyle s \ \nmid \ (a^k-1) \ \forall \ 0 \lt k \lt l$, except exactly in the following cases:

• • $\displaystyle l = 2, a = (2^t - 1),$ where $\displaystyle t \geq 2.$

• • $\displaystyle l = 6, a = 2.$

We are now in a position to classify the weak r-Steinberg characters of $GL(n,q):$

Proof of Theorem 1.2:

Let χ be a weak r-Steinberg character of $GL(n,q).$ Since $\displaystyle r \neq p, \ \chi(1) = |(q^n-1)(q^{n-1}-1) \ldots (q-1)|_r.$ The following cases arise:

Case I: $\displaystyle r \nmid (q^n-1).$ By Zsigmondy’s theorem there exists a prime divisor s of $\displaystyle (q^n-1)$ such that $\displaystyle s \ \nmid \ (q^m-1) \ \forall \ m \lt n;$ except for $\displaystyle (n, q) = (2, 2^t - 1)$ and $(6, 2).$ Clearly, s is different from $r.$ Since $\displaystyle \chi(1)$ is an r-power and $\displaystyle s \mid (q^n-1),$ it follows from (6.1) that there exists some $\displaystyle 1 \leq j \leq k$ such that $\displaystyle s \ | \ \psi_{n_j}(q^{d_j}).$ Without loss of generality, assume that j = 1. Since s is a prime, s divides $\displaystyle (q^{id_1}-1)$ for some $\displaystyle 1 \leq i \leq n_1.$ As $\displaystyle s \ \nmid \ (q^m-1) \ \forall \ m \lt n,$ it implies that $id_1 \gt (n-1).$ Thus $n_1d_1=n.$ Also note that $d_1 = 1$ implies that $\chi(1) = 1$. So $d_1 \geq 2.$

If $\displaystyle n=2,$ then $\displaystyle r \ | \ (q^2-1).$ But since $\displaystyle r \nmid (q^n-1),$ it follows that $\displaystyle n \geq 3.$ First assume that n = 3. If $\displaystyle r \nmid (q^3-1),$ then it follows from Lemma 4.2 that χ is not a parabolically induced character. Also, Lemma 6.3 implies that the cuspidal characters of $GL(3,2)$ are weak r-Steinberg if and only if r = 3. Further, note that it follows from Equation (6.2) that the only choice of $\displaystyle (n,q)$ for which $\chi(1)$ is a prime power is $(4,2)$. The only character in the character table of $GL(4,2)$ whose degree is a prime power is the character of degree 7. But this character is not quasi 7-Steinberg.

If $\displaystyle (n, q) = (2, 2^t - 1),$ it follows from Lemma 6.2 that χ is not weak r-Steinberg. Now assume that $\displaystyle (n, q) = (6, 2).$ Since $GL(6,2)$ is a quasi-simple group, it follows from Theorem 1.2 in [10] that $GL(6,2)$ does not have any character whose degree is a prime power.

Case II: $\displaystyle r \mid (q^n-1).$ Since $\displaystyle r \nmid (q-1)$, we have that $\displaystyle r \nmid (q^{n-1}-1).$ In this case, the previous argument for a = q and $l=(n-1)$ yields that k = 2 and g ₁ is a simplex of degree d ₁ with $\displaystyle n_1d_1=(n-1),$ where q and n are such that $\displaystyle (q, n-1) \notin \{(2^t - 1, 2), (2,6) \}.$ Clearly, the degree of g ₂ is 1 and hence the degree of χ is

(6.2)\begin{equation} \frac{(q^n-1)(q^{n-1}-1)\ldots(q-1)}{(q^{d_1}-1)(q^{2d_1}-1)\ldots(q^{n_1d_1}-1)(q-1)} = (q^n-1)(q^{n-2}-1)\ldots(q^{n-d_1}-1). \end{equation}

If $\displaystyle n=2,$ then the degree of $\chi,$ $\chi(1) = (q+1).$ It follows from Lemma 6.2 that χ is weak r-Steinberg if and only if $(q+1)=r^{\beta},$ for an odd prime $r.$ Further, if $\displaystyle n=3,$ then we have the following:

1. 1. If $d_1 = 1,$ then $\chi(1) = (q^2+q+1)$;

2. 2. If $d_1 = 2,$ then $\displaystyle \chi(1) = (q^3-1).$

Now Lemma 6.3 gives the conditions under which the characters of these degrees are weak r-Steinberg. Finally, assume that $n \geq 4.$ If $d_1 \geq 2,$ then both $(q^n-1)$ and $(q^{n-2}-1)$ occur as divisors of $\chi(1)$ and hence are r-powers. This contradicts the fact that $\displaystyle r \ \nmid \ (q-1).$ Therefore, $\displaystyle d_1 = 1$ and hence $\displaystyle \chi = (g_1^{n-1}g_2)$ has degree $\displaystyle \chi(1) = \frac{q^n-1}{q-1}.$ Since

\begin{align*} \chi(1) = |(q^n-1)(q^{n-1}-1)\cdots(q-1)|_r, \end{align*}

it further implies that $\displaystyle r \ \nmid \ (q^i-1),$ for any $\displaystyle 1 \leq i \leq (n-1).$

Note that if $\displaystyle n \geq 4$ is an even integer, then $\displaystyle diag(E_2, E_2, \ldots, E_2) \notin P_{n-1,1}$. Further, if $\displaystyle n \geq 5$ is an odd integer, then $\displaystyle diag(E_3, E_2, \ldots, E_2)$ intersects $\displaystyle P_{n-1,1}$ trivially. Since in either case χ vanishes on some r-regular element, it is not weak r-Steinberg.

Now if $\displaystyle (n-1, q) = (2, 2^t - 1),$ then n = 3. The weak Steinberg characters of $GL(3,q)$ have been classified in Lemma 6.3. If $\displaystyle (n-1, q) = (6, 2),$ then since r is a divisor of $\displaystyle (2^7-1),$ $\displaystyle r = 127.$ Now note that it follows from the character table that $GL(7,2)$ does not have a character of degree 127. Now the result follows.