There is an $m$ such that $M^{m}-I_{n}$ is not invertible for all $M \in GL({n,q})$ We have a general linear group over a finite field. I need to show that for every $M$ in my group I can find an integer $m$ such that $$M^{m}-I_{n}$$ is not invertible.
I know this happens because of finite field, since after finding the suitable $m$ all entries of $M^{m}$ either become 0 or 1 ( of course the diagonal should be 1) and hence $M^{m}-I_{n}=0$ which will not be invertible, but this is not proof and not sure even its make sense. please help.

We can actually do better than this: Since the group $G := GL(n, q)$ is finite, for every $M \in G$ we have $M^{|G|} = I_n$, and so $M^{|G|} - I_n$ is actually the zero matrix. Hence, we may take $m$ to be
$$|G| = (q^n - 1) (q^n - q) \cdots (q^n - q^{n - 1}) .$$ Probably one can improve on this in general. For $n = q = 2$, for example, one can take $m$ to be $3 < 6 = |GL(2, 2)|$, and it is perhaps a more interesting question to ask what this minimum is as a function of $n, q$.
