I was asked to show in an exercise that for a prime $ p $, in the group $ GL(n,F_p) $ of invertible matrices of dimension $ 1 \leq n \leq p $ mod $ p $, that no matrix can have order $ p^2 $, meaning that no invertible matrix over $F_p$ has $ A^{p^2}=I $ and for all $ k\leq p^2 $ one has $ A^k\neq I $.
I thought about the equation $ A^{p^2} - I = 0 $ and how the minimal polynomial must divide it. The polynomial in the equation decomposes as $ \lambda ^ {p^2} -1 = (\lambda -1)(1+\lambda +\dots+\lambda^{p^2-1}) $ but I cannot proceed from here. Can anyone please help? I thank all helpers.
Since $\lambda^{p^2}-1=(\lambda-1)^{p^2}$, all eigenvalues of $A$ are $1$. If $N$ is the nilpotent part of $A$, then $A=I+N$. We then have that $A^p=I+N^p=I$ since $N^p=0$ given that $n\leq p$, so $A$ has order at most $p$.