# Element structure of general linear group of degree two over a finite field

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This article gives specific information, namely, element structure, about a family of groups, namely: general linear group of degree two.
View element structure of group families | View other specific information about general linear group of degree two
This article gives information on the element structure for a finite field. If you are interested in rings such as $\mathbb{Z}/p^n\mathbb{Z}$ or $\mathbb{F}_p[t]/(t^n)$, or Galois rings, see element structure of general linear group of degree two over a finite discrete valuation ring.

This article gives the element structure of the general linear group of degree two over a finite field. Similar structure works over an infinite field or a field of infinite characteristic, with suitable reinterpretation and modification.

We denote the order (or size) of the field by $q$ and the characteristic of the field by $p$. $q$ is a power of $p$.

## Summary

Item Value
order $(q^2 - q)(q^2 - 1) = q^4 - q^3 - q^2 + q = q(q + 1)(q - 1)^2$
exponent $p(q^2 - 1) = p(q + 1)(q - 1)$
number of conjugacy classes $q^2 - 1 = (q - 1)(q + 1)$

## Particular cases

$q$ (field size) $p$ (underlying prime, field characteristic) general linear group $GL(2,q)$ order of the group (= $(q^2 - 1)(q^2 - q)$) number of conjugacy classes (= $q^2 - 1$) element structure page
2 2 symmetric group:S3 6 3 element structure of symmetric group:S3
3 3 general linear group:GL(2,3) 48 8 element structure of general linear group:GL(2,3)
4 2 direct product of A5 and Z3 180 15
5 5 general linear group:GL(2,5) 480 24 element structure of general linear group:GL(2,5)
7 7 general linear group:GL(2,7) 2016 48 element structure of general linear group:GL(2,7)

## Conjugacy class structure

There is a total of $q(q+1)(q - 1)^2 = q^4 - q^3 - q^2 + q$ elements, and there are $q^2 - 1 = (q - 1)(q +1)$ conjugacy classes of elements.

For background on how this conjugacy class structure can be obtained and also generalized to general linear groups of degree three or more, refer to conjugacy class size formula in general linear group over finite field.

Nature of conjugacy class Eigenvalues Characteristic polynomial Minimal polynomial Size of conjugacy class Number of such conjugacy classes Total number of elements Semisimple? Diagonalizable over $\mathbb{F}_q$?
Diagonalizable over $\mathbb{F}_q$ with equal diagonal entries, hence a scalar $\{a,a \}$ where $a \in \mathbb{F}_q^\ast$ $(x - a)^2$ where $a \in \mathbb{F}_q^\ast$ $x - a$ where $a \in \mathbb{F}_q^\ast$ 1 $q - 1$ $q - 1$ Yes Yes
Diagonalizable over $\mathbb{F}_{q^2}$, not over $\mathbb{F}_q$. Must necessarily have no repeated eigenvalues. Pair of conjugate elements of $\mathbb{F}_{q^2}$ $x^2 - ax + b$, irreducible Same as characteristic polynomial $q(q - 1)$ $\frac{q(q - 1)}{2} = \frac{q^2 - q}{2}$ $q^2(q-1)^2/2 = (q^4 - 2q^3 + q^2)/2$ Yes No
Not diagonal, has Jordan block of size two $a$ (multiplicity two) where $a \in \mathbb{F}_q^\ast$ $(x - a)^2$ where $a \in \mathbb{F}_q^\ast$ Same as characteristic polynomial $q^2 - 1$ $q - 1$ $(q^2 - 1)(q - 1) = q^3 - q^2 - q + 1$ No No
Diagonalizable over $\mathbb{F}_q$ with distinct diagonal entries $\lambda, \mu$ (interchangeable) distinct elements of $\mathbb{F}_q^\ast$ $x^2 - (\lambda + \mu)x + \lambda \mu$ Same as characteristic polynomial $q(q+1)$ $\frac{(q - 1)(q - 2)}{2} = \frac{q^2 - 3q + 2}{2}$ $\frac{q(q+1)(q-1)(q-2)}{2}= \frac{q^4 - 2q^3 - q^2 + 2q}{2}$ Yes Yes
Total NA NA NA NA $q^2 - 1$ $q(q+1)(q-1)^2 = q^4 - q^3 - q^2 + q$

## Order information

### Information of elements of an exact order

Type of element Case of order $d$ dividing $q - 1$ Case of order $d$ dividing $q^2 - 1$, not $q - 1$ Case of order $pd$, $d$ divides $q -1$.
Central, diagonal over $\mathbb{F}_q$ with equal diagonal entries $\varphi(d)$ conjugacy classes, $\varphi(d)$ elements. 0, 0 0, 0
Diagonalizable over $\mathbb{F}_{q^2}$, not $\mathbb{F}_q$ 0 conjugacy classes, 0 elements $\varphi(d)/2$ conjugacy classes, $q(q-1)\varphi(d)/2$ elements 0 conjugacy classes, 0 elements
Diagonalizable over $\mathbb{F}_q$, distinct diagonal entries $\varphi(d)[d - (\varphi(d) + 1)/2]$ conjugacy classes, $q(q+1)\varphi(d)[d - (\varphi(d) + 1)/2]$ elements 0 conjugacy classes, 0 elements 0 conjugacy classes, 0 elements
Jordan block of size two 0 conjugacy classes, 0 elements 0 conjugacy classes, 0 elements $\varphi(d)$ conjugacy classes, $(q^2 - 1)\varphi(d)$ elements
Total $\varphi(d)[d + (1 - \varphi(d))/2]$ conjugacy classes, $\varphi(d)[1 + q(q+1)d - q(q+1)(\varphi(d) + 1)/2]$ elements $\varphi(d)/2$ conjugacy classes, $q(q-1)\varphi(d)/2$ elements $\varphi(d)$ conjugacy classes, $(q^2 - 1)\varphi(d)$ elements

## Central elements

The center is a subgroup of order $q - 1$, and its elements are diagonal matrices of the form:

$\{ \begin{pmatrix} a & 0 \\ 0 & a \\\end{pmatrix} : a \in \mathbb{F}_q^\ast \}$

The subgroup is isomorphic to the multiplicative group of the field of $q$ elements, and since multiplicative group of finite field is cyclic, it is a cyclic group of order $q - 1$.

### Real and rational conjugacy

We have the following:

Item Value
Number of equivalence classes under real conjugacy in the whole group for elements in the center $(q + 1)/2$ if $q$ is odd, $q/2$ if $q$ is even.
Number of real elements in the center 2 if $q$ is odd, 1 if $q$ is even
Number of equivalence classes under rational conjugacy in the whole group for elements in the center $\sigma_0(q - 1)$ where $\sigma_0$ is the divisor count function
Number of rational elements in the center 2 if $q$ is odd, 1 if $q$ is even

### Order information

For a given $d$ dividing $q - 1$:

Item Value
Number of elements of order precisely $d$ $\varphi(d)$ (the Euler totient function)
Number of elements of order dividing $d$ $d$

## Elements diagonalizable over $\mathbb{F}_{q^2}$ but not $\mathbb{F}_q$

These elements have pairs of distinct eigenvalues over $\mathbb{F}_{q^2}$ that are conjugate over $\mathbb{F}_q$. The unique non-identity automorphism of $\mathbb{F}_{q^2}$ over $\mathbb{F}_q$ is the map $x \mapsto x^q$, so these two elements are $q^{th}$ powers of each other, i.e., if one of them is $\alpha$, the other one is $\alpha^q$.

The conjugacy class is parameterized by the pair $\{ \alpha, \alpha^q \}$. Here is more information:

Item Value Explanation
Trace of conjugacy class $a = \alpha + \alpha^q$, is an element of $\mathbb{F}_q$ Trace is sum of eigenvalues
Norm or determinant of conjugacy class $b = \alpha \cdot \alpha^q = \alpha^{q + 1}$, is an element of $\mathbb{F}_q^\ast$ Product of eigenvalues
Minimal polynomial of conjugacy class $(x - \alpha)(x - \alpha^q)$ which equals $x^2 - ax + b$, $a,b$ the trace and norm as above. Distinct eigenvalues
Characteristic polynomial of conjugacy class Same as minimal polynomial Distinct eigenvalues
Centralizer of any element of conjugacy class It is isomorphic to the multiplicative group $\mathbb{F}_{q^2}^\ast$ which is a cyclic group of order $q^2 - 1$ Any choice of element gives a particular interpretaiton of a two-dimensional vector space over $\mathbb{F}_q$ as a one-dimensional vector space of $\mathbb{F}_{q^2}$. The elements that commute with it are therefore precisely the linear maps on a one-dimensional vector space, which is the group $GL(1,q^2) \cong \mathbb{F}_q^\ast$
Size of conjugacy class $\frac{(q - 1)^2q(q + 1)}{q^2 - 1} = q(q - 1)$ It equals the index of the centralizer

Here is information on the collection of all such conjugacy classes:

Item Value Explanation
number of conjugacy classes of this type $\frac{q^2 - q}{2} = \frac{q(q - 1)}{2}$ First explanation: The conjugacy classes are classified by pairs $\{ \alpha,\alpha^q \}$ of elements in $\mathbb{F}_{q^2}$ but not in $\mathbb{F}_q$. The number of elements is $q^2 - q$, so the number of pairs is $(q^2 - q)/2$.
Second explanation: The conjugacy classes are characterized by irreducible monic quadratic polynomials. The total number of monic quadratics is $q^2$. Of these, $q$ have repeated roots, and $q(q - 1)/2$ have distinct roots over the base field. This leaves $q^2 - q - q(q-1)/2 = q(q - 1)/2$ monic quadratics that are irreducible.
total number of elements of this type $\frac{q^2(q-1)^2}{2} = \frac{q^4 - 2q^3 + q^2}{2}$ Multiply the size of each conjugacy class with the number of conjugacy classes of this type
number of centralizers, or equivalently, number of different ways of realizing $\mathbb{F}_{q^2}^\ast$ as a centralizer in this fashion $q(q - 1)/2$ (check)

### Order information

A number $d$ occurs as the order of an element in one of these conjugacy classes if $d \mid q^2 - 1$ but $d$ does not divide $q - 1$. More information below:

Count Number of conjugacy classes Number of elements (obtained by multiplying number of conjugacy classes by size of each conjugacy class, which is $q(q - 1)$
Elements of order precisely $d$ (if $d \mid q^2 - 1$ but $d$ does not divide $q - 1$) $\varphi(d)/2$, $\varphi(d)$ being the Euler totient function. $q(q-1)\varphi(d)/2$.
Elements of order dividing $d$  ?

## Elements diagonalizable over $\mathbb{F}_q$ with distinct diagonal entries

Note that we are including here only invertible elements, hence both eigenvalues must be in $\mathbb{F}_q^\ast$.

Each such conjugacy class is specified by an unordered pair of distinct elements of $\mathbb{F}_q^\ast$, say $\lambda, \mu$.

Item Value Explanation
Trace of conjugacy class $a = \lambda + \mu$, is an element of $\mathbb{F}_q$ Trace is sum of eigenvalues
Norm or determinant of conjugacy class $b = \lambda\mu$, is an element of $\mathbb{F}_q^\ast$ Product of eigenvalues
Minimal polynomial of conjugacy class $(x - \lambda)(x - \mu)$ which equals $x^2 - ax + b$, $a,b$ the trace and norm as above. Distinct eigenvalues
Characteristic polynomial of conjugacy class Same as minimal polynomial Distinct eigenvalues
Centralizer of any element of conjugacy class It is isomorphic to $\mathbb{F}_q^\ast \times \mathbb{F}_q^\ast$, which is a group of order $(q - 1)^2$ This is easiest to see for the representative that is diagonal: its centralizer is the subgroup of diagonal matrices.
Size of conjugacy class $(q - 1)^2q(q + 1)/(q - 1)^2 = q(q + 1)$ It equals the index of the centralizer

Here is combined information for all conjugacy classes:

Item Value Explanation
number of conjugacy classes of this type $\binom{q - 1}{2} = (q - 1)(q - 2)/2 = (q^2 - 3q + 2)/2$ We need to pick for the eigenvalues two distinct elements $\lambda, \mu$ from the set $\mathbb{F}_q^\ast$ which has size $q - 1$.
total number of elements of this type $q(q+1)(q-1)(q-2)/2= (q^4 - 2q^3 - q^2 + 2q)/2$ Multiply number of conjugacy classes and size of each conjugacy class.
number of centralizers $q(q + 1)/2$ (check)

### Order information

A natural number $d$ occurs as the order of an element in one of these conjugacy classes if and only if $d$ divides $q - 1$ but is greater than $1$. For such a value of $d$, we have:

Count Number of conjugacy classes Number of elements (obtained by multiplying number of conjugacy classes with size of each conjugacy class)
order precisely $d$ $\varphi(d)[d - \frac{\varphi(d) + 1}{2}]$ where $\varphi$ is the Euler totient function $q(q + 1)\varphi(d)[d - \frac{\varphi(d) + 1}{2}]$
order dividing $d$ $\binom{d}{2} = d(d - 1)/2$ $q(q+1)d(d-1)/2$

## Elements with Jordan block of size two

These are elements conjugate to elements of the form:

$\{ \begin{pmatrix} a & 1 \\ 0 & a \\\end{pmatrix}: a \in \mathbb{F}_q^\ast \}$

Each conjugacy class is parameterized by the value of $a$.

Item Value Explanation
Trace of conjugacy class $2a$
Norm or determinant of conjugacy class $a^2$
Minimal polynomial of conjugacy class $(x - a)^2$ Repeated eigenvalue
Characteristic polynomial of conjugacy class $(x - a)^2 = x^2 - 2ax + a^2$
Centralizer of any element in conjugacy class Isomorphic to direct product of additive group of $\mathbb{F}_q$ and multiplicative group $\mathbb{F}_q^\ast$ For the element $\begin{pmatrix}a & 1 \\ 0 & a \\\end{pmatrix}$, the centralizer is all matrices of the form $\begin{pmatrix} s & t \\ 0 & s \\\end{pmatrix}$ where $s \in \mathbb{F}_q^\ast$ and $t \in \mathbb{F}_q$.
Size of conjugacy class $q(q - 1)^2(q + 1)/(q(q - 1)) = (q - 1)(q + 1) = q^2 - 1$ size of conjugacy class equals index of centralizer

Here is the information on the collection of all conjugacy classes:

Item Value Explanation
number of conjugacy classes of this type $q - 1$ The parameter varies over $\mathbb{F}_q^\ast$
total number of elements of this type $(q^2 - 1)(q - 1) = q^3 - q^2 - q + 1$ multiply number of conjugacy classes with size of each conjugacy class
number of centralizers $q + 1$ First, note that all centralizers can be attained from just one conjugacy class. Next, note that $q - 1$ conjugates of any element lie in its centralizer, hence have the same centralizer. The total number of centralizers is thus $(q^2 - 1)/(q-1)$.

### Order information

The possible orders are $dp$, where $d$ divides $q - 1$ and $p$ is the underlying prime (the characteristic) whose power is $q$. For each order $dp$, the number of conjugacy classes is $\varphi(d)$, the Euler totient function.