A question regarding the group $G = GL_2(\mathbb Z/ p\mathbb Z)$. For any prime $p$ , consider the group $G =\mathrm{ GL}_2(\mathbb Z/ p\mathbb Z)$. Then which of the following are true?
1) $G$ has an element of order $p$.
2) $G$ has exactly one element of order $p$.
3) $G$ has no p-Sylow subgroups.
4) Every element of order $p$ is conjugate to a matrix $$A = \left[\begin{matrix} 1 & a  \\ 0 & 1  \end{matrix}\right]$$ where $a\in  (\mathbb Z/ p\mathbb Z)^*$.
My try:  2 & 3 are false obviously. And 1 is true. But I am confused  about 4. 4 is true for $p=2$,  but will the statement be true for every prime?

The following is a description of the group $G =\mathrm{ GL}_2(\mathbb Z/ p\mathbb Z)$ that answers all your questions:
1) The order $n$ of $G$ is $n=p(p+1)(p-1)^2$. Indeed , for the first row we have the choice between $p^2-1$ possibilities. For the second row we have the choice of $p^2$ vectors minus the $p$ vectors who are linearly dependent on the first vector.
2)The matrices of the form $\left[\begin{matrix} 1 & a  \\ 0 & 1  \end{matrix}\right]$ define a Sylow-p subgroup $P$ of order $p$ since the maximal exponent of $p$ occuring in n is $1$.
3) The normalizer $N$ of $P$ contains, apart from $P$ itself, the group $D$ of diagonal matrices which has order $(p-1)^2$. Since $P \cap D = {I_2}$ the order of $N$ is at least $p(p-1)^2$.
4) By the orbit/stabilizer theorem the orbit of P by the inner automorphisms of $G$ has at  most $p+1$ elements (all the Sylow-p subgroups). Since every element of order $p$ lies in a Sylow-p subgroup it must be conjugate to a matrix of the form defined in 2).
PS: As an exercise try to use the same kind of argument to prove the same facts for $G =\mathrm{ GL}_k(\mathbb Z/ p\mathbb Z)$ for an arbitrary positive integer $k$.
