TITLE: Conjugacy Classes of subgroups in GL(n,p)
QUESTION [6 upvotes]: What is the conjugacy class of subgroups of order $p$ in $GL(n,p)$? (Are all subgroups of order $p$ conjugate in $GL(n,p)$?)

REPLY [2 votes]: (Same as Jack's, different normal form!)
If $A\in\operatorname{GL}(n,p)$ has order $p$, then $A^p-I=0$. It follows that $A$ satisfies the polynomial $X^p-1$, so the minimal polynomial $m_A$ divides $X^p-1=(X-1)^p$. Since every eigenvalue of $A$ is a root of $m_A$, we conclude that the only eigenvalue of $A$ (in any algebraic closure of $\mathbb F_p$), is $1$.
It follows at once that both the Jordan form $J$ of $A$ and a matrix conjugating $A$ to $J$ are in fact both elements of $\operatorname{GL}(n,p)$. If we want to count the elements of $\{A\in\operatorname{GL}(n,p):A^p=I\}$ up to conjugation in $\operatorname{GL}(n,p)$, then, it is enough to count the conjugacy classes of matrices of $\operatorname{GL}(n,p)$ in Jordan canonical form with minimal polynomial dividing $(X-1)^p$. 
This is a pretty simple excercise. The answer is: the number of partitions of $n$ with parts not larger than $p$.
Hmm, this is counting conjugacy classes of matrices and not of subgroups...*