Element structure of special linear group of degree three over a finite field

This article gives specific information, namely, element structure, about a family of groups, namely: special linear group of degree three.
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This article describes the element structure of the special linear group of degree three over a finite field.

We take $q$ as the number of elements in the field and $p$ as the underlying prime number, so $q$ is a power of $p$.

Summary

Item	Value
number of conjugacy classes	Case $q$ is not 1 mod 3 (e.g., $q = 2,3,5,8,9,\dots$): $q^2 + q$ Case $q \equiv 1 \pmod 3$ (e.g., $q = 4,7,13,16,19,\dots$): $q^2 + q + 8$
order	$q^3(q^3 - 1)(q^2 - 1) = q^3(q-1)^2(q+1)(q^2 + q + 1)$
exponent	?

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 3 - 1 = 2, and we need to make cases based on the congruence class of the field size modulo 3. 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 $n = 3$ here).

Value of $\operatorname{gcd}(3,q-1)$	Corresponding congruence classes of $q$ mod 3	Number of conjugacy classes (polynomial of degree 3 - 1 = 2 in $q$)	Additional comments
1	0 or -1 mod 3: 0 mod 3 (e.g. $q = 3,9,27,\dots$) -1 mod 3 (e.g., $q = 2,5,8,\dots$)	$q^2 + q = q(q + 1)$	In this case, we have an isomorphism between linear groups when degree power map is bijective, so $SL(3,q) \cong PGL(3,q) \cong PSL(3,q)$
3	1 mod 3 (e.g., $q = 4,7,13,16,19,\dots$)	$q^2 + q + 8$

General strategy and summary

Further information: Element structure of special linear group over a finite field, conjugacy class of elements with semisimple generalized Jordan block does not split in special linear group over a finite field, splitting criterion for conjugacy classes in special linear group of prime degree over a finite field

Before making the entire table, we recall the general strategy: first, imitate the procedure of element structure of general linear group of degree three over a finite field to determine that $GL(3,q)$-conjugacy classes in $SL(3,q)$. 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 three. Further, the splitting criterion for conjugacy classes in special linear group of prime degree over a finite field, applied to the case of degree three, tells us that:

• When the field size is 1 mod 3, there are three such $GL(3,q)$-conjugacy classes (corresponding to the cube roots of unity) and each splits into three $SL(3,q)$-conjugacy classes. We thus get a total of 9 $SL(3,q)$-conjugacy classes of this type.
• When the field size is not 1 mod 3, there is only one such $GL(3,q)$-conjugacy class and it does not split over $SL(3,q)$.

Summary for field size 1 mod 3

The key feature for fields of size $q \equiv 1 \pmod 3$ is that such fields have three distinct cube roots of unity.

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 $SL_3$ relative to $GL_3$?
Diagonalizable over $\mathbb{F}_q$ with equal diagonal entries, hence a scalar	$\{ a,a,a \}$ where $a^3 = 1$	$(x - a)^3$ for $a$ the eigenvalue	$x - a$	1	3	3	Yes	Yes	No
Diagonalizable over $\mathbb{F}_q$ with one eigenvalue having multiplicity two, other eigenvalue having multiplicity one	$\{ a,a,1/a^2 \}$ where $a \in \mathbb{F}_q^\ast$ but is not a cube root of unity	$(x - a)^2(x - 1/a^2)$	$(x - a)(x - 1/a^2)$	$q^2(q^2 + q + 1)$	$q - 4$	$q^2(q^2 + q + 1)(q - 4)$	Yes	Yes	No
Diagonalizable over $\mathbb{F}_q$ with all distinct diagonal entries	$\{ a,b,c \}$ all distinct, with $abc = 1$	$(x - a)(x - b)(x - c)$	same as characteristic polynomial	$q^3(q + 1)(q^2 + q + 1)$	$(q^2 - 5q + 10)/6$	$q^3(q + 1)(q^2 + q + 1)(q^2 - 5q + 10)/6$	Yes	Yes	No
Diagonalizable over $\mathbb{F}_{q^3}$, not over $\mathbb{F}_q$	Distinct Galois conjugate triple of elements in $\mathbb{F}_{q^3}^\ast$ of norm 1. If one of the elements is $a$, the other two are $a^q$ and $a^{q^2}$. We must have $a^{q^2 + q + 1} = 1$.	irreducible degree three polynomial over $\mathbb{F}_q$	same as characteristic polynomial	$q^3(q-1)^2(q+1)$	$(q - 1)(q + 2)/3 = (q^2 + q - 2)/3$	$q^3(q-1)^3(q+1)(q + 2)/3$	Yes	No	No
One eigenvalue is in $\mathbb{F}_q^\ast$, the other two are in $\mathbb{F}_{q^2} \setminus \mathbb{F}_q$	one element of $\mathbb{F}_q^\ast$, pair of Galois conjugates over $\mathbb{F}_q$ in $\mathbb{F}_{q^2}$, whose norm is reciprocal of that element	product of linear polynomial and irreducible degree two polynomial over $\mathbb{F}_q$	same as characteristic polynomial	$q^3(q - 1)(q^2 + q + 1) = q^6 - q^3$	$q(q - 1)/2 = (q^2 - q)/2$	$q^4(q - 1)^2(q^2 + q + 1)/2$	Yes	No	No
Has Jordan blocks of sizes 2 and 1 with distinct eigenvalues over $\mathbb{F}_q$	$\{ a,a,1/a^2 \}$ with $a\in \mathbb{F}_q^\ast$, $a$ not a cube root of unity	$(x - a)^2(x - 1/a^2)$	same as characteristic polynomial	$q^2(q + 1)(q - 1)(q^2 + q + 1)$	$q - 4$	$q^2(q + 1)(q - 1)(q^2 + q + 1)(q - 4)$	No	No	No
Has Jordan blocks of sizes 2 and 1 with equal eigenvalues over $\mathbb{F}_q$	$\{ a,a,a \}$ with $a^3 = 1$	$(x - a)^3$	$(x - a)^2$	$q(q + 1)(q - 1)^2(q^2 + q + 1) = q^6 - q^4 - q^3 + q$	3	$3q(q + 1)(q - 1)^2(q^2 + q + 1) = 3q^6 - 3q^4 - 3q^2 + 3q$	No	No	No
Has Jordan block of size 3	$\{ a,a,a \}$ with $a^3 = 1$	$(x - a)^3$	same as characteristic polynomial	$(q - 1)(q + 1)(q^2 + q + 1)/3 = (q^4 + q^3 - q - 1)/3$	9	$3(q - 1)(q + 1)(q^2 + q + 1) = 3(q^4 + q^3 - q - 1)$	No	No	Yes
Total (--)	--	--	--	--	$q^2 + q + 8$	$q^3(q^3 - 1)(q^2 - 1)$	--	--	--

Summary for field size not 1 mod 3

In this case, the only cube root of 1 is 1. In particular, the field does not have primitive cube roots of unity.

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 $SL_3$ relative to $GL_3$?
Diagonalizable over $\mathbb{F}_q$ with equal diagonal entries, hence a scalar	$\{ 1,1,1 \}$	$(x - 1)^3$	$x - 1$	1	1	1	Yes	Yes	No
Diagonalizable over $\mathbb{F}_q$ with one eigenvalue having multiplicity two, other eigenvalue having multiplicity one	$\{ a,a,1/a^2 \}$ where $a \in \mathbb{F}_q^\ast$ is not 1	$(x - a)^2(x - 1/a^2)$	$(x - a)(x - 1/a^2)$	$q^2(q^2 + q + 1)$	$q - 2$	$q^2(q^2 + q + 1)(q - 2)$	Yes	Yes	No
Diagonalizable over $\mathbb{F}_q$ with all distinct diagonal entries	$\{ a,b,c \}$ all distinct, with $abc = 1$	$(x - a)(x - b)(x - c)$	same as characteristic polynomial	$q^3(q + 1)(q^2 + q + 1)$	$(q - 2)(q - 3)/6$	$q^3(q + 1)(q^2 + q + 1)(q - 2)(q - 3)/6$	Yes	Yes	No
Diagonalizable over $\mathbb{F}_{q^3}$, not over $\mathbb{F}_q$	Distinct Galois conjugate triple of elements in $\mathbb{F}_{q^3}^\ast$ of norm 1. If one of the elements is $a$, the other two are $a^q$ and $a^{q^2}$. We must have $a^{q^2 + q + 1} = 1$.	irreducible degree three polynomial over $\mathbb{F}_q$	same as characteristic polynomial	$q^3(q-1)^2(q+1)$	$q(q + 1)/3 = (q^2 + q)/3$	$q^4(q-1)^2(q+1)^2/3$	Yes	No	No
One eigenvalue is in $\mathbb{F}_q^\ast$, the other two are in $\mathbb{F}_{q^2} \setminus \mathbb{F}_q$	one element of $\mathbb{F}_q^\ast$, pair of Galois conjugates over $\mathbb{F}_q$ in $\mathbb{F}_{q^2}$, whose norm is reciprocal of that element	product of linear polynomial and irreducible degree two polynomial over $\mathbb{F}_q$	same as characteristic polynomial	$q^3(q - 1)(q^2 + q + 1) = q^6 - q^3$	$q(q - 1)/2 = (q^2 - q)/2$	$q^4(q - 1)^2(q^2 + q + 1)/2$	Yes	No	No
Has Jordan blocks of sizes 2 and 1 with distinct eigenvalues over $\mathbb{F}_q$	$\{ a,a,1/a^2 \}$ with $a\in \mathbb{F}_q^\ast$, $a$ not 1	$(x - a)^2(x - 1/a^2)$	same as characteristic polynomial	$q^2(q + 1)(q - 1)(q^2 + q + 1)$	$q - 2$	$q^2(q + 1)(q - 1)(q^2 + q + 1)(q - 2)$	No	No	No
Has Jordan blocks of sizes 2 and 1 with equal eigenvalues over $\mathbb{F}_q$	$\{ 1,1,1 \}$	$(x - 1)^3$	$(x - 1)^2$	$q(q + 1)(q - 1)^2(q^2 + q + 1) = q^6 - q^4 - q^2 + q$	1	$q(q + 1)(q - 1)^2(q^2 + q + 1) = q^6 - q^4 - q^2 + q$	No	No	No
Has Jordan block of size 3	$\{ 1,1,1 \}$	$(x - 1)^3$	same as characteristic polynomial	$(q - 1)(q + 1)(q^2 + q + 1) = (q^4 + q^3 - q - 1)$	1	$(q - 1)(q + 1)(q^2 + q + 1) = q^4 + q^3 - q - 1$	No	No	Yes
Total (--)	--	--	--	--	$q^2 + q$	$q^3(q^3 - 1)(q^2 - 1)$	--	--	--