Finding representatives for the distinct conjugacy classes of matrices of finite order in $GL_2(\mathbb{Z}_p)$. The representatives for the distinct conjugacy classes of matrices of finite order in $GL_2(\mathbb{Q})$ can be found as mentioned in https://www.math3ma.com/blog/rational-canonical-form-example-1
However, if I think of the representatives for the distinct conjugacy classes of matrices of finite order in $GL_2(\mathbb{Z}_p)$, where $p(>2)$ is a prime, then how can I get the conjugacy classes?
What differences should be present when obtaining the conjugacy classes for matrices of finite order in $GL_2(\mathbb{Z}_p)$?
Can someone please help to obtain the results similar to that in the above website, where the matrix satisfies the polynomial $x^n-1$ (same as in the website) for $n \geq 2$ (and $n$ is an odd prime less than $p$, $n|p^2-1$)?
Thanks a lot in advance.
 A: I am assuming that $\Bbb{Z}_p$ are the $p$-adic integers.

For $p\ge 5$ and $A\in GL_2(\Bbb{Z}_p)$ of finite order then the $GL_2(\Bbb{Z}_p)$-conjugacy class is determined by its characteristic polynomial $f\in \Bbb{Z}_p[x]$.

$A$ is diagonalizable over the splitting field,

*

*Either $f$ has a double root and $A$ is diagonal-scalar


*Or $f=(x-\mu)(x-\xi)$ where $\mu,\xi$ are two distinct roots of unity in $\Bbb{Q}_p(\zeta_{p^2-1})$. Thus $f\bmod p \in \Bbb{F}_p[x]$ is separable.
In that case $A\bmod p$ has a cyclic vector $v$, ie. such that $v,Av$ is a basis of $(\Bbb{F}_p)^2$

(here any $v$ which is not an eigenvector)

taking any $u\in (\Bbb{Z}_p)^2,u\equiv v\bmod p$ we get that $u,Au$ is a $\Bbb{Z}_p$-basis of $(\Bbb{Z}_p)^2$.
For another matrix $B\in GL_2(\Bbb{Z}_p)$ with the same charasteristic polynomial, $w,Bw$ is a $\Bbb{Z}_p$-basis of $(\Bbb{Z}_p)^2$, then the matrix $P$ sending $u,Au$ to $w,Bw$ is in $GL_2(\Bbb{Z}_p)$ and satisfies $B=PAP^{-1}$.
If you meant the finite field with $p$-elements then I almost answered too, the only remaining case is when $A\bmod p$ is not diagonalizable, $f=(x-a)^2$, $A-aI$ is nilpotent, from its kernel and an element not in the kernel we get $A = Q \pmatrix{a&b\\0&a}Q^{-1}$ with $b\ne 0$ and $Q\in GL_2(\Bbb{F}_p)$, and $\pmatrix{b&0\\0& 1}^{-1}\pmatrix{a&b\\0&a}\pmatrix{b&0\\0& 1}=\pmatrix{a&1\\0&a}$ whose conjugacy class are every element with characteristic polynomial $(x-a)^2$ and different from $\pmatrix{a&0\\0&a}$.

*

*For the size of the $GL_2(\Bbb{F}_p)$ conjugacy classes, count the number of matrices with a given characteristic polynomial. Given $charpoly(\pmatrix{a&b\\c&d})=x^2-rx+s$ we must have $d=r-a,a(r-a)-s=bc$

*

*if $r^2-4s$ is not a square then $p$ choices for $a$ give $bc\ne 0$ so we have $p-1$ choices for $b$.


*If $r^2-4s$ is a non-zero square. $p-2$ values of $a$ give $bc\ne 0$ thus $p-1$ choices for $b$.
$2$ values of $a$ give $bc=0$ thus $2p-1$ choices for $b,c$.


*if $r^2-4s=0$ then $p-1$ values of $a$ give $bc\ne 0$ thus $p-1$ choices for $b$.
$1$ value of $a$ gives $bc=0$ thus $2p-1$ choices for $b,c$.
Substract one ( the conjucacy class of the scalar matrix $\pmatrix{r/2&0\\0&r/2}$).
