Prove $|GL(2,p)| = (p^2-1)(p^2-p)$ 
Let $G = GL(2,p)$ and 
  $$P= \{ \begin{bmatrix} 1 & \lambda \\ 0 & \lambda \end{bmatrix} | \lambda \in F \}$$
   where $F$ denotes the field of $p$ elements, $p$ a prime.
Prove that $|GL(2,p)| = (p^2-1)(p^2-p)$ and that $P$ is a Sylow $p$-subgroup of $G$.

I started out by multiplying 
$$\begin{bmatrix} 1 & \lambda \\ 0 & \lambda \end{bmatrix}\begin{bmatrix} 1 & \lambda \\ 0 & \lambda \end{bmatrix} = \begin{bmatrix} 1 & \lambda^2 + \lambda \\ 0 & \lambda^2 \end{bmatrix}$$
But then when I take the determinant of that matrix I just get $\lambda^2$, far from the desired result.
Please can someone point out what I am doing wrong.
I also don't know how to approach the second part of the question, please could someone give me a hint.
 A: Not quite sure about the notation. If $GL(2,P)$ is $GL_2(\mathbb F_p)$, then we can count the amount of all possibilities. It is a $2 \times 2$ matrix, for first column(or row), we have $p^2-1$ different choices, just avoid $0$ for both entries, for second column, we need to choose a vector not in the span of first column, that is:
$$(c,d)^T \neq k(a,b)^T$$
where $k \in \mathbb F_p$.
thus, for each pair we put in first column, we have $p^2 - p$ different choices(we don't need to consider $(0,0)^T$ because $0 \in \mathbb F_p$) for the second column, it implies:
$$|GL_2(\mathbb F_p)| = (p^2-1)(p^2 - p)= p(p+1)(p-1)^2$$
For the next part, you just need to verify.
$$
        \begin{pmatrix}
        1 & a  \\
        0 & a  \\
        \end{pmatrix}^p = 
        \begin{pmatrix}
        1 & \sum_{i=1}^p a^i \\
        0 & a^p  \\
        \end{pmatrix}
$$
Note, $a^p \equiv a \pmod p$, furthermore,
$$\sum_{i=1}^p a^i = \frac{a(1-a^p)}{1-a} \equiv a \pmod p$$
Thus, this element's order is $p$, then it generates a group of order $p$. Now we finish the proof.
