# 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