How many conjugacy classes does the group $GL(3,\mathbb{Z}_p)$ have, if $p$ is a prime?
Until now I have a (very broad) lower boud, using that the number of conjugacy classes in a group is at least as much as the number of elements in the center. Now, the center of a group is the set of elements that commute with every other element of the group, and in the case of $GL(n,\mathbb{Z}_p)$ the scalar matrices are the elements with this property. Therefore the number of conjugacy classes of $GL(3,\mathbb{Z}_p)$ is at least $p−1$.
Also, I noticed that the number of conjugacy classes is the number of cosets of the centralizer of an element of the group, which is the same as the index of the centralizer. So I was thinking I could pick an element in $GL(3,\mathbb{Z}_p)$ and find the centralizer and its index. Am I on the right path?
If two matrices are similar, they have the same characteristic polynomials $\lambda^3+a_1\lambda^2+a_2\lambda+a_3$. The number of such polynomials is $p^2(p-1)$. So the number of conjugacy classes is at least $p^3-p^2$.
But in fact the number is bigger. Every matrix over any field is similar to unique block-diagonal matrix in the natural (sometimes called regular) normal form (unique up to the order of blocks). The blocks correspond to divisors of the characteristic polynomial. So for each polynomial of degree $3$ as above, we need to count the number of possible normal forms. Then add these numbers up, and we get the number of all conjugacy classes.
The calculation has been done in classical group theory books (say, in Dickson), and, more recently, in this paper. The total number can also be found here:
It is equal to $p^3-p$.