If $T$ be an invertible linear operator on a finite-dimensional vector space over a finite field , then $T^n$ is the identity operator?

Question

If $T$ be an invertible linear operator on a finite-dimensional vector space over a finite field , then is it true that $T^n = I$ ( the identity operator) for some positive integer $n$ ?

Answer 1

We note that any field has an algebraic closure. With that in mind, any operator $T^n$ over a finite field $k$ can be written using Jordan normal form with entries in the closure $\bar k$.

Thus, it suffices to verify that there is a particular $n$ such that $J^n = I$ for each Jordan block $J$. Suppose that $\bar k$ has characteristic $p$. Let $J$ be the Jordan block $$ J = \pmatrix{\lambda&1\\&\lambda&1\\&&\ddots&\\&&&\lambda} $$ with $\lambda \in \bar k \setminus \{0\}$ we note that $$ J^n = \pmatrix{\lambda^n& n\lambda^{n-1} & n(n-1)\lambda^{n-2} &\cdots\\ 0&\lambda^n & n\lambda^{n-1} & \cdots\\ &&\ddots } $$ Verify that $J^p$ is diagonal. It follows that $(J^p)^{(p-1)} = I$.

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So, in summary: if $T$ is an invertible operator on a finite-dimensional vector space over a finite field of characteristic $p$, then $T^{p(p-1)} = I$.

Answer 2

Yes. The group of automorphisms of a finite group is finite.

A corrected version of Omnomnomnom's argument is as follows. An element of $T \in \text{GL}_n(\mathbb{F}_q)$ has a decomposition into Jordan blocks over $\overline{\mathbb{F}_q}$ which can be written $J = \lambda + S$ where $\lambda \in \overline{\mathbb{F}_q}$ and $S$ is the off-diagonal part of the Jordan block. We compute that

$$J^{q^i} = \lambda^{q^i} + S^{q^i}$$

and hence that $J^{q^i}$ is diagonal provided that $q^i \ge n$. Let $i$ be the smallest positive integer with this property, and let $j$ be the smallest positive integer such that $\lambda$ lies in $\mathbb{F}_{q^j}$. Since $\lambda$ is a root of the characteristic polynomial of $T$, which has degree $n$, we are guaranteed that $j \le n$. Then we find that

$$J^{q^i (q^j - 1)} = 1.$$

If the characteristic polynomial of $T$ factors into irreducible factors of degrees $n = j_1 + j_2 + \dots + j_k$, then we conclude that $T$ has order dividing

$$q^i \text{ lcm}(q^{j_1} - 1, q^{j_2} - 1, \dots q^{j_k} - 1).$$

The maximum possible value of this function as the $j_i$ vary can be thought of as a $q$-analogue of Landau's function, which is the largest possible order of an element of $S_n$. The term in front $q^i$ is at most $qn$ and can be dropped if $T$ is diagonalizable.

You can get a quicker and easier bound on the order by computing the order of $\text{GL}_n(\mathbb{F}_q)$, which is a standard and entertaining exercise. Since there are $q^{n^2}$ not-necessarily-invertible $n \times n$ matrices over $\mathbb{F}_q$ that gets you an upper bound pretty quickly (but not a divisibility condition).