Finite matrix power over $\Bbb F_q$

Question

What is largest $s\in\Bbb N$ such that a matrix $M\in\Bbb F_q^{n\times n}=\Bbb F_{p^r}^{n\times n}$ could satisfy $$M^i\neq I,\quad\forall i\in\Bbb Z_+:0<i<s$$ $$M^0=M^s=I?$$

Answer 1

The answer is $q^{n}-1,$ which we prove by induction on $n$ ( though we allow $q$ to vary), the result being clear when $n=1$ (for any $q$). Suppose then that $n >1$ and that the minimum polynomial of $M$ in $F_{q}[x]$ is $\prod_{i=1}^{r} f_{i}(x)^{m_{i}}$, where the $f_{i}(x)$ are distinct irreducibles, of respective degrees $d_{i}$. Then we have $n = \sum_{i=1}^{r}m_{i}d_{i}$ and $C_{{\rm GL}(n,q)}(M)$ is isomorphic to a subgroup of $\prod_{i=1}^{r}{\rm GL}(m_{i},q^{d_{i}}).$ If $r >1$ or $d_{i} >1$ for any $i,$ then by induction, the order of $M$ is at most $\prod_{i=1}^{r} (q^{d_{i}m_{i}}-1) < q^{n}-1.$ Hence we may suppose that $M$ has minimum polynomial $(x-\lambda)^{n}$ for some $\lambda \in F.$ Then the order of $M$ is at most $n(q-1) <(q^{n}-1)$ ( for note that the semisimple part of $M$ has order at most $q-1$ and the unipotent part of $M$ certainly has order at most $n$ ( the latter being equality only when $n$ is a power of $p$)).