If $P$ is an invertible linear operator on a finite dimensional vector space over a finite field, then there is some $n>0$ such that $P^n=I$

Question

From Berkeley problems in Mathematics:

problem 7.4.21: let $P$ be a linear operator on a finite dimensional vector space over a finite field. show that if $P$ is invertible, then $P^n$ = $I$ for some positive integer $n$.

I think there is a mistake in the above statement: it does not hold true if P is a scalar multiple of the identity operator.

But is the statement correct otherwise? i.e. is the following modified statement correct?

let $P$ be a linear operator on a finite dimensional vector space over a finite field. assume $P$ is not a scalar multiple of the identity operator. show that if $P$ is invertible, then $P^n$ = $I$ for some positive integer $n$.

Answer 1

The statement is true even if $P$ is a scalar operator. Note that the field of scalars in question is finite.

Consider the group $G$ of invertible linear operators on the vector space in question. Since the space is finite-dimensional & the scalar field is finite, the vector space itself is finite, and the space of linear operators is finite. Hence $G$, being a subset of the space of all operators, is finite too. Now the result follows from general group theory: for any $P \in G$ we have $P^{|G|} = I$, the identity operator.

Answer 2

I'll make a try:

Take $P,P^2,P^3,...$. Then these can't be all different because the have entries from a finite field. Then $P^t=P^s\Rightarrow P^n=I$ since $P$ is invertible