For any prime $p,$ consider the group $G = \text {GL}_2 \left (\Bbb Z / p \Bbb Z \right ).$ Then show that every element of $G$ of order $p$ is conjugate to a matrix $\begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix},$ where $a \in \left (\Bbb Z/p \Bbb Z \right )^*.$
My attempt $:$ Here $\left \lvert G \right \rvert = (p^2 - 1) (p^2 - p) = p (p - 1)^2 (p + 1).$ So all the $p$-Sylow subgroups of $G$ are of order $p$ and hence they are all cyclic, generated by any of it's non-identity element. I have found that $\text {ord} (A) = p,$ where $A = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}.$ Let $H = \left \langle A \right \rangle$ be the $p$-Sylow subgroup generated by $A.$ Now let $B$ be any other element in $G$ of order $p.$ Then it generates some $p$-Sylow subgroup, say $K.$ By Sylow's second theorem it follows that $H$ and $K$ are conjugates of each other i.e. there exists $P \in G$ such that $P^{-1} A' P = B,$ for some $A' \in H.$ So there exists $0 \lt r \lt p$ such that $A' = A^r.$ But this implies that $B = \left (P^{-1} A P \right )^r,$ for some $0 \lt r \lt p.$ If $r = 1,$ then we are through. What will happen if $1 \lt r \lt p\ $? Any help in this regard will be appreciated.
Thanks for your time.