How to compute the number of Sylow p-subgroups of $GL_{n}(F_{p})$ ？ Let $F_{p}$ be the field of p-elements and $GL_{n}(F_{p})$ the group of invertible
n by n matrices.
(1) Compute the order of $GL_{n}(F_{p})$ .
(2)Find a Sylow p-subgroup of $GL_{n}(F_{p})$.
(3) Compute the number of Sylow p-subgroups.
I got stuck in how to compute the number. From (2) the number of sylow  p-subgroup cannot be 1  since there are at least two groups：the upper triangular matrixs with diagonal entries 1 and the lower forms. The key to solve it as I think is using the second and third Sylow theorem.One possible value is p+1. But I don't know what to do next.
 A: You know that a Sylow $p$-subgroup is of the form
$$T = \left\{\begin{bmatrix}
1 & a_{12} & a_{13} & \dots & a_{1,n-1} & a_{1n}\\
0 & 1 & a_{23}  & \dots &a_{2,n-1} & a_{2n}\\
&&&\ddots\\
0 & 0 & 0  & \dots &1 & a_{n-1n}\\
0 & 0 & 0 & \dots & 0 & 1\\
\end{bmatrix}
: a_{ij} \in F_{p} \right\}
$$
Find its normaliser to be
$$N = \left\{\begin{bmatrix}
a_{11} & a_{12} & a_{13} & \dots & a_{1,n-1} & a_{1n}\\
0 & a_{22} & a_{23}  & \dots &a_{2,n-1} & a_{2n}\\
&&&\ddots\\
0 & 0 & 0  & \dots &a_{n-1,n-1} & a_{n-1,n}\\
0 & 0 & 0 & \dots & 0 & a_{nn}\\
\end{bmatrix}
: a_{ij} \in F_{p}, a_{ii} \ne 0 \right\}.
$$
The index of the normalizer in $\operatorname{GL}_{n}(F_{p})$ is the number of the $p$-Sylow subgroups.

To determine the normaliser of $T$, let $e_{i}$ be the standard basis, let $V_{i} = \langle e_{1}, e_{2}, \dots, e_{i} \rangle$. Note that $T V_{i} \subseteq V_{i}$, and that the $V_{i}$ are the only subspaces $U$ such that $T U \subseteq U$. If $g \in N_{G}(T)$, then for $t \in T$ we have $g^{-1} t g = s  \in T$, so that $g s V_{i} = g V_{i} = t (g V_{i})$. So $U = g V_{i}$ is  a subspace such that $T U \subseteq U$, from which it follows that $g V_{i} = V_{i}$ for each $i$, which means $g \in N$.
Thanks to i707107 for noticing a misprint.
