# Difference between revisions of "Element structure of special linear group of degree two over a finite field"

This article gives specific information, namely, element structure, about a family of groups, namely: special linear group of degree two. This article restricts attention to the case where the underlying ring is a finite field.
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This article describes the element structure of the special linear group of degree two over a finite field of order $q$ and characteristic $p$, where $q$ is a power of $p$. We denote by $r$ the positive integer value $\log_pq$, so $q = p^r$. Some aspects of this discussion, with suitable infinitary analogues of cardinality, carry over to infinite fields and fields of infinite characteristic.

## Summary

Item Value
order $q^3 - q = q(q - 1)(q + 1)$
exponent $p(q^2 - 1)/2 = p(q - 1)(q + 1)/2$ for $p$ odd, $2(q^2 - 1)$ for $p = 2$
number of conjugacy classes Case $q$ odd: $q + 4$
Case $q$ even (and hence, a power of 2): $q + 1$
equals the number of irreducible representations, see also linear representation theory of special linear group of degree two over a finite field
conjugacy class sizes Case $q$ odd: 1 (2 times), $(q^2 - 1)/2$ (4 times), $q(q - 1)$ ($(q - 1)/2$ times), $q(q + 1)$ ($(q - 3)/2$ times)
Case $q$ even: 1 (1 time), $q^2 - 1$ (1 time), $q(q-1)$ ($q/2$ times), $q(q+1)$ ($(q - 2)/2$ times)
number of $p$-regular conjugacy classes (Where $p$ is the characteristic of the field) $q$
equals the number of irreducible representations in that characteristic, see also modular representation theory of special linear group of degree two over a finite field in its defining characteristic
number of orbits under automorphism group Case $q = p \ne 2$ (i.e., prime field for odd prime): $q + 2$ (basically same as the conjugacy classes relative to $GL_2$)
Case $q = p = 2$: 3
Other cases: Complicated
equals number of orbits of irreducible representations under automorphism group, see also linear representation theory of special linear group of degree two over a finite field

## Particular cases

$q$ (field size) $p$ (underlying prime, field characteristic) Case on $q$ group $SL(2,q)$ order of the group (= $q^3 - q$) conjugacy class sizes (ascending order) number of conjugacy classes (= $q + 1$ if $q$ even, $q + 4$ if $q$ odd) element structure page
2 2 even symmetric group:S3 6 1,2,3 3 element structure of symmetric group:S3
3 3 odd special linear group:SL(2,3) 24 1,1,4,4,4,4,6 7 element structure of special linear group:SL(2,3)
4 2 even alternating group:A5 60 1,12,12,15,20 5 element structure of alternating group:A5
5 5 odd special linear group:SL(2,5) 120 1,1,12,12,12,12,20,20,30 9 element structure of special linear group:SL(2,5)
7 7 odd special linear group:SL(2,7) 336 1,1,24,24,24,24,42,42,42,56,56 11 element structure of special linear group:SL(2,7)
8 2 even special linear group:SL(2,8) 504 1,56,56,56,56,63,72,72,72 9 element structure of special linear group:SL(2,8)
9 3 odd special linear group:SL(2,9) 720 1,1,40,40,40,40,72,72,72,72,90,90,90 13 element structure of special linear group:SL(2,9)
The expandable display contains the GAP code to verify this; the theoretical explanation is in the next section. [SHOW MORE]

## Conjugacy class structure

### Number of conjugacy classes

As we know in general, number of conjugacy classes in special linear group of fixed degree over a finite field is PORC function of field size, the degree of this PORC function is one less than the degree of matrices, and we make cases based on the congruence classes modulo the degree of matrices. Thus, we expect that the number of conjugacy classes is a PORC function of the field size of degree 2 - 1 = 1, and we need to make cases based on the congruence class of the field size modulo 2. Moreover, the general theory also tells us that the polynomial function of $q$ depends only on the value of $\operatorname{gcd}(n,q-1)$, which in turn can be determined by the congruence class of $q$ mod $n$ (with $n = 2$ here).

Value of $\operatorname{gcd}(2,q-1)$ Corresponding congruence classes of $q$ mod 2 Number of conjugacy classes (polynomial of degree 2 - 1 = 1 in $q$) Additional comments
1 0 mod 2 (e.g., $q = 2,4,8,\dots$) $q + 1$ In this case, we have an isomorphism between linear groups when degree power map is bijective, so $SL(2,q) \cong PGL(2,q) \cong PSL(2,q)$
2 1 mod 2 (e.g., $q = 3,5,7,\dots$) $q + 4$

### General strategy and summary

Before making the entire table, we recall the general strategy: first, imitate the procedure of element structure of general linear group of degree two over a finite field to determine that $GL(2,q)$-conjugacy classes in $SL(2,q)$. Then, use the splitting criterion for conjugacy classes in the special linear group to determine which of these conjugacy classes split, and how much.

In this case, the fact that conjugacy class of elements with semisimple generalized Jordan block does not split in special linear group over a finite field tells us that the only conjugacy class that might split is the conjugacy class with a Jordan block of size two. Further, the splitting criterion for conjugacy classes in special linear group of prime degree over a finite field, applied to the case of degree two, tells us that:

• When the field size is odd, there are two such $GL(2,q)$-conjugacy classes (repeated eigenvalue 1, and repeated eigenvalue -1) and each splits into two $SL(2,q)$-conjugacy classes. We thus get a total of 4 $SL(2,q)$-conjugacy classes of this type.
• When the field size is even, there is only one such $GL(2,q)$-conjugacy class and it does not split over $SL(2,q)$.

### Summary for odd field size

The key feature for fields of odd size is that there exist two distinct square roots of unity in such fields: 1 and -1.

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 $\mathbb{F}_q$? Splits in $SL_2$ relative to $GL_2$?
Diagonalizable over $\mathbb{F}_q$ with equal diagonal entries, hence a scalar $\{ 1,1 \}$ or $\{ -1,-1\}$ $(x - a)^2$ where $a \in \{ -1,1 \}$ $x - a$ where $a \in \{ -1,1\}$ 1 2 2 Yes Yes No
Not diagonal, has Jordan block of size two $1$ (multiplicity 2) or $-1$ (multiplicity 2) $(x - a)^2$ where $a \in \{ -1,1 \}$ Same as characteristic polynomial $(q^2 - 1)/2$ 4 $2(q^2 - 1)$ No No Yes (two conjugacy classes over $GL_2$, each splits into two over $SL_2$)
Diagonalizable over $\mathbb{F}_{q^2}$, not over $\mathbb{F}_q$. Must necessarily have no repeated eigenvalues. Pair of conjugate elements of $\mathbb{F}_{q^2}$ of norm 1 $x^2 - ax + 1$, irreducible Same as characteristic polynomial $q(q - 1)$ $(q - 1)/2$ $q(q-1)^2/2 = (q^3 - 2q^2 + q)/2$ Yes No No
Diagonalizable over $\mathbb{F}_q$ with distinct (and hence mutually inverse) diagonal entries $\lambda, 1/\lambda$ where $\lambda \in \mathbb{F}_q \setminus \{ 0,1,-1 \}$ $x^2 - (\lambda + 1/\lambda)x + 1$ Same as characteristic polynomial $q(q+1)$ $(q - 3)/2$ $q(q+1)(q-3)/2 = (q^3 - 2q^2 - 3q)/2$ Yes Yes No
Total NA NA NA NA $q + 4$ $q(q+1)(q-1) = q^3 - q$ $q(q^2 - 2q - 1) + 2$ elements
$q$ conjugacy classes
$q(q^2 - 3q - 2)/2 + 2$ elements
$(q + 1)/2$ conjugacy classes
$2(q^2 - 1)$elements
4 conjugacy classes

### Summary for $p = 2$, $q$ a power of 2

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 $\mathbb{F}_q$? Splits in $SL_2$ relative to $GL_2$?
Diagonalizable over $\mathbb{F}_q$ with equal diagonal entries, hence a scalar. $1, 1$ $x^2 + 1$ $x + 1$ 1 1 1 Yes Yes No
Diagonalizable over $\mathbb{F}_{q^2}$, not over $\mathbb{F}_q$. Must necessarily have no repeated eigenvalues. Pair of conjugate elements of $\mathbb{F}_{q^2}$ of norm 1 $x^2 - ax + 1$, irreducible $x^2 - ax + 1$, irreducible $q(q - 1)$ $q/2$ $q^2(q - 1)/2 = (q^3 - q^2)/2$ Yes No No
Not diagonal, has Jordan block of size two $1$ (multiplicity 2) $x^2 + 1$ $x^2 + 1$ $q^2 - 1$ 1 $q^2 - 1$ No No No
Diagonalizable over $\mathbb{F}_q$ with distinct (and hence mutually inverse) diagonal entries $\lambda, 1/\lambda$ where $\lambda \in \mathbb{F}_q \setminus \{ 0,1 \}$ $x^2 - (\lambda + 1/\lambda)x + 1$ $x^2 - (\lambda + 1/\lambda)x + 1$ $q(q + 1)$ $(q - 2)/2$ $q(q+1)(q-2)/2 = (q^3 - q^2 - 2q)/2$ Yes Yes No
Total NA NA NA NA $q + 1$ $q(q + 1)(q - 1) = q^3 - q$ $(q + 1)(q - 1)^2 = q^3 - q^2 - q + 1$

## Automorphism class structure

We have that special linear group of degree two has a class-inverting automorphism. In particular, any such group is a group in which every element is automorphic to its inverse. Also, special linear group of degree two is ambivalent iff -1 is a square, where an ambivalent group is a group in which every element is conjugate to its inverse.

We discuss below the automorphism class structure, i.e., the orbit structure under the action of the automorphism group.

### Summary for odd characteristic $p$, field size $q$

We let $q = p^r$. Note that if $q = p$, then $r = 1$.

Nature of automorphism class Number of automorphism classes of this type Number of conjugacy classes within each automorphism class Size of each conjugacy class Size of each automorphism class Total number of elements across all such automorphism classes
Diagonalizable over $\mathbb{F}_q$ with equal diagonal entries, hence a scalar 2 1 1 1 2
Not diagonal, has Jordan block of size two 2 2 $(q^2 - 1)/2$ $q^2 - 1$ $2(q^2 - 1)$
Diagonalizable over $\mathbb{F}_{q^2}$, not over $\mathbb{F}_q$. Must necessarily have no repeated eigenvalues. (complicated, depends on $r$) (complicated, depends on $r$) $q(q - 1)$ $q(q-1)^2/2 = (q^3 - 2q^2 + q)/2$
Diagonalizable over $\mathbb{F}_q$ with distinct (and hence mutually inverse) diagonal entries (complicated, depends on $r$) (complicated, depends on $r$) $q(q + 1)$ $q(q+1)(q-3)/2 = (q^3 - 2q^2 - 3q)/2$

## Central elements

The center is a subgroup of order either 1 or 2, depending on whether $q$ is even or odd. For odd $q$, the center is given by:

$\left\{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \\\end{pmatrix}, \begin{pmatrix} -1 & 0 \\ 0 & -1 \\\end{pmatrix} \right\}$

For even $q$ (i.e., the characteristic is 2 and $q$ is a power of 2), the center is the trivial group, comprising only the identity element.

## Jordan block of size two

The conjugacy classes of this type are the only ones that split in the special linear group relative to the general linear group (in the odd characteristic case), i.e., where there are elements of $SL_2$ that are conjugate in $GL_2$ but not in $SL_2$.

Over $GL_2$, there are two conjugacy classes in odd characteristic (which collapse to one class in even characteristic):

• The conjugacy class of $\begin{pmatrix} 1 & 1 \\ 0 & 1 \\\end{pmatrix}$. In particular, this conjugacy class includes all matrices of the form $\begin{pmatrix} 1 & \lambda \\ 0 & 1 \\\end{pmatrix}$ and also all matrices of the form $\begin{pmatrix} 1 & 0 \\ \lambda & 1 \\\end{pmatrix}$ where $\lambda \in \mathbb{F}_q^\ast$.
• The conjugacy class of $\begin{pmatrix} -1 & 1 \\ 0 & -1 \\\end{pmatrix}$. In particular, this conjugacy class includes all matrices of the form $\begin{pmatrix} -1 & \lambda \\ 0 & -1 \\\end{pmatrix}$ and also all matrices of the form $\begin{pmatrix} -1 & 0 \\ \lambda & -1 \\\end{pmatrix}$ where $\lambda \in \mathbb{F}_q^\ast$.

### Characteristic 2 case

In this case, both conjugacy classes collapse into a single conjugacy class. We have:

Item Value Explanation
Trace of conjugacy class 0
Norm or determinant of conjugacy class 1
Minimal polynomial of conjugacy class $(x - 1)^2$ Repeated eigenvalue
Characteristic polynomial of conjugacy class $(x - 1)^2 = x^2 + 1$
Centralizer of any element in conjugacy class in $SL_2$ Isomorphic to additive group of $\mathbb{F}_q$ For the element $\begin{pmatrix}1 & 1 \\ 0 & 1 \\\end{pmatrix}$, the centralizer is all matrices of the form $\begin{pmatrix} 1 & t \\ 0 & 1 \\\end{pmatrix}$ where $t \in \mathbb{F}_q$.
Centralizer of any element in conjugacy class in $GL_2$ Isomorphic to direct product of additive and multiplicative groups of $\mathbb{F}_q$. Explicitly, it is given by $\{ \begin{pmatrix} a & b \\ 0 & a \\\end{pmatrix}, a \in \mathbb{F}_q^\ast, b \in \mathbb{F}_q \}$
Size of conjugacy class $q(q - 1)(q + 1)/q = (q - 1)(q + 1) = q^2 - 1$ 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 1
total number of elements of this type $q^2 - 1$

### Odd characteristic case

In this case, each conjugacy class splits further into two, giving a total of four conjugacy classes. We have:

Item Value Explanation
Trace of conjugacy class 2 or -2, depending on whether the eigenvalues are 1 or -1
Norm or determinant of conjugacy class 1
Minimal polynomial of conjugacy class $x - 1$ or $x + 1$, depending on whether the eigenvalues are 1 or -1 Repeated eigenvalue
Characteristic polynomial of conjugacy class $(x - 1)^2 = x^2 - 2x + 1$ or $(x + 1)^2 = x^2 + 2x + 1$
Centralizer of any element in conjugacy class in $SL_2$ Isomorphic to direct product of additive group of $\mathbb{F}_q$ and cyclic group of order two For the element $\begin{pmatrix}1 & 1 \\ 0 & 1 \\\end{pmatrix}$, the centralizer is all matrices of the form $\begin{pmatrix} 1 & t \\ 0 & 1 \\\end{pmatrix}$ and all matrices of the form $\begin{pmatrix} -1 & t \\ 0 & -1 \\\end{pmatrix}$ where $t \in \mathbb{F}_q$
Centralizer of any element in conjugacy class in $GL_2$ Isomorphic to direct product of additive and multiplicative groups of $\mathbb{F}_q$. Explicitly, it is given by $\{ \begin{pmatrix} a & b \\ 0 & a \\\end{pmatrix}, a \in \mathbb{F}_q^\ast, b \in \mathbb{F}_q \}$
Size of conjugacy class $q(q - 1)(q + 1)/(2q) = (q - 1)(q + 1)/2 = (q^2 - 1)/2$ size of conjugacy class equals index of centralizer

Given two matrices:

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

the following rule works to determine whether they are conjugate in the special linear group: the elements are conjugate if and only if $a/b$ is a square in $\mathbb{F}_q^\ast$.

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

Item Value Explanation
number of conjugacy classes of this type 4 each of the two conjugacy classes relative to $GL_2$ splits into two in $SL_2$.
total number of elements of this type $2(q^2 - 1)$

## 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 $\{ \alpha, \alpha^q \}$.

The element must have determinant 1, so this forces $(\alpha)(\alpha^q) = 1$, so $\alpha^{q+1} = 1$ and $\alpha^q = \alpha^{-1}$.

Item Value Explanation
Trace of conjugacy class $a = \alpha + \alpha^q = \alpha + \alpha^{-1}$, is an element of $\mathbb{F}_q$ Trace is sum of eigenvalues
Norm or determinant of conjugacy class 1
Minimal polynomial of conjugacy class $x^2 - ax + 1$
Characteristic polynomial of conjugacy class Same as minimal polynomial
Centralizer of any element of conjugacy class It is the intersection of the multiplicative group $\mathbb{F}_{q^2}^\ast$ which is a cyclic group of order $q^2 - 1$ (acting as linear transformations by viewing $\mathbb{F}_{q^2}$ as a two-dimensional vector space over $\mathbb{F}_q$), and the special linear group. This becomes a cyclic subgroup of order $q + 1$ in $\mathbb{F}_{q^2}^\ast$.
Size of conjugacy class $q(q+1)(q-1)/(q+1) = q(q-1)$ 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 $(q - 1)/2$ if $q$ odd, $q/2$ if $q$ even First explanation: The conjugacy classes are classified by pairs $\{ \alpha, \alpha^q \}$ where $\alpha^{q + 1} = 1$ and $\alpha \notin \mathbb{F}_q$. There are $q + 1$ elements of order dividing $q + 1$ in $\mathbb{F}_q^\ast$ (because it is cyclic). Of these, depending on whether $q$ is even or odd, 1 or 2 elements live inside $\mathbb{F}_q^\ast$. The remaining elements occur in pairs $\{\alpha,\alpha^q \}$, so we get $(q + 1 - (1 \operatorname{or} 2))/2$ which simplifies based on the cases.
Second explanation: There are $q$ monic polynomials of degree 2 with constant term 1. We need to figure out how many of them are irreducible, and we can do this by subtracting the number that already factor. When $q$ is odd, we get $2 + (q - 3)/2 = (q + 1)/2$ polynomials that factor over $\mathbb{F}_q$, so the difference is $q - (q +1)/2 = (q - 1)/2$. When $q$ is even, we get $1 + (q - 2)/2 = q/2$ polynomials that factor, so the difference is $q - q/2 = q/2$.
(Addendum to second explanation): The $2 + (q - 3)/2$ and $1 + (q - 2)/2$ arise as (number of elements that square to 1) + (number of distinct pairs of element and its inverse)
total number of elements of this type $q(q-1)^2/2 = (q^3 - 2q^2 + q)/2$ if $q$ odd, $q^2(q - 1)/2 = (q^3 - q^2)/2$ if $q$ even Multiply the size of each conjugacy class with the number of conjugacy classes of this type

## Elements diagonalizable over $\mathbb{F}_q$ with distinct and hence mutually inverse entries

Each such conjugacy class is specified by an unordered pair of distinct elements of $\mathbb{F}_q^\ast$, say $\lambda, \lambda^{-1}$. Note that the eigenvalues must be inverses because their product needs to be 1 for the element to be in the special linear group.

Item Value Explanation
Trace of conjugacy class $a = \lambda + \lambda^{-1}$, is an element of $\mathbb{F}_q$ Trace is sum of eigenvalues
Norm or determinant of conjugacy class 1
Minimal polynomial of conjugacy class $(x - \lambda)(x - \lambda^{-1})$ which equals $x^2 - ax + 1$, $a$ the trace 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 $\mathbb{F}_q^\ast$, which is a group of order $(q - 1)$ This is easiest to see for the representative that is diagonal: its centralizer is the subgroup of diagonal matrices of determinant 1, i.e., matrices of the form $\begin{pmatrix} s & 0 \\ 0 & s^{-1} \\\end{pmatrix}$.
Size of conjugacy class $(q - 1)q(q + 1)/(q - 1) = q(q + 1)$ 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 $(q - 3)/2$ for odd $q$, $(q - 2)/2$ for $q$ a power of 2 We consider $\mathbb{F}_q^\ast$ (order $q - 1$) and remove the elements that are equal to their own inverse (1 such element in characteristic 2, 2 such elements in odd characteristic). Then, we divide by 2 because we are interested in unordered pairs of elements and their inverses.
total number of elements of this type $q(q+1)(q-3)/2= (q^3 - 2q^2 - 3q)/2$ if $q$ odd, $q(q+1)(q-2)/2 = (q^3 - q^2 - 2q)/2$ if $q$ even Multiply number of conjugacy classes and size of each conjugacy class.