Number of conjugacy classes in symplectic group of fixed degree over a finite field is PORC function of field size

Statement

Suppose $n$ is an even natural number, i.e., $n = 2m$ for some natural number $m$ . Then, there exists a PORC function $f$ of degree $m = n/2$ such that, for any prime power $q$ , the number of conjugacy classes in the symplectic group $Sp(n,q)$ (i.e., the symplectic group of degree $n$ over the finite field of size $q$ ) is $f(q)$ .

A PORC function is a polynomial on residue classes -- it looks like different polynomial functions on different congruence classes modulo a particular number. In this case, we only need to consider congruence classes modulo $n$ to define the PORC function. In fact, for a field size of $q$ , the polynomial depends only on the value $\operatorname{gcd}(\operatorname{lcm}(2,m),q - 1)$ .

Particular cases

$n$ (degree of symplectic group)	$m = n/2$ (degree of PORC function, also number used in Chevalley notation)	$\operatorname{lcm}(2,m)$	Possibilities for $\operatorname{gcd}(\operatorname{lcm}(2,m),q - 1)$	Corresponding congruence classes mod $\operatorname{lcm}(2,m)$ for $q$	Corresponding polynomials in PORC function of $q$ giving number of conjugacy classes in $Sp(n,q)$	More information
2	1	2	1 2	0 1	$q + 1$ $q + 4$	See element structure of special linear group of degree two over a finite field (note that $Sp(2,q) = SL(2,q)$ ).
4	2	2	1 2	0 1	$q^2 + 2q + 3$ $q^2 + 5q + 10$	See element structure of symplectic group of degree four over a finite field