A general result for finite order matrices with integer coefficients?

Question

The following question was given as an exercise in many orals for french preparatory classes:

Let $A \in \mathcal{M}_2(\mathbb{Z})$ such that there exists $p \in \mathbb{N}^*$ with $A^p = I_2$.

Show that $A^{12} = I_2$

I showed this result by looking at the eigenvalues of $A$ and using the fact that its trace is an integer we can narrow down exactly what eigenvalues $A$ can have, and then we can conclude.

However, I want to generalize the exercise into this result:

Let $n \in \mathbb{N}^*$ and $A \in \mathcal{M_n}(\mathbb{Z})$, can we show that there exists a certain sequence $b_n$ such that if $A$ verifies the existence of $p$ such that $A^p = I_n$, then necessarily $A^{b_n} = I_n$ as well? Can we have a formula for $b_n$?

I have the following sketch of a proof, but am not satisfied with it as it does not really give any formula for $b_n$:

Suppose $A^p = I_n$ for some integer $p$, therefore the polynomial $P = X^p - 1$ annhilates $A$, meaning that $A$ is diagonalizable in $\mathbb{C}$ and its eigenvalues are all $p$-th roots of unity.

Let $\chi_A$ be $A$'s characteristic polynomial, we have $\displaystyle \chi_A = \prod_{\omega \in Sp(A)}\left(X-w \right)$, and $|\omega| = 1$

Additionally, each coefficient of $\chi_A$ is an integer, and from Vieta's formulas those coefficients are the sum of complex numbers of modulus $1$, therefore using triangle inequality each coefficient is bounded by some number (independent of $p$).

Therefore, we have only finitely many coefficients possible, and therefore only finitely many roots (eigenvalues) possible. Hence, there are only finitely many $p$'s such that $A^p = I_n$, and so taking $b_n$ to be the $\mathrm{lcm}$ of the possible $p$'s, we do get $A^{b_n} = I_n$

Already, is this proof correct? And is there a more constructive one that could actually build the sequence $b_n$?

So far, I've (possibly wrongly) calculated $b_2 = 12$ (from the exercise), $b_3 = 12$, and $b_4 = 120$

I've thought of using cyclotomic polynomials, but I'm really not comfortable enough with those to actually get anything done.

Answer 1

If $A^p=I_n,$ then each eigenvalue of the matrix $A$ must be a $p$-th root of unity. This means, each eigenvalue of the matrix $A$ must be a root of a cyclotomic polynomial.

The minimal polynomial of an algebraic number divides each polynomial with rational coefficients that has this number as its root. This can be shown by means of polynomial long division.

This means: If $A^p=I_n$ and $\lambda$ is a $p$-th root of unity with associated cyclotomic polynomial $q$ ($q$ is the minimal polynomial of $\lambda$), then $q$ divides the characteristic polynomial of $A,$ because the characteristic polynomial of $A$ has rational coefficients (it even has integer coeffcients.) Consequently, $\deg(q)\leq n.$

On the other hand, we cannot narrow down the set of possible eigenvalues even more. For each cyclotomic polynomial $q$ with a degree of at most $n,$ we can find a matrix $A\in\mathcal{M}_n(\mathbb{Z})$ such that there is a $p$ with $A^p = I_n.$ Just use the companion matrix of $q$ as the upper left block of $A$, the identity matrix as lower right block of $A$ and set $p$ to the index of the given cyclotomic polynomial.

Conclusion: Given the information "There is a number $p$ such that $A^p=I_n$", we can say that $A$ has eigenvalues which are the $p$-th roots of unity, and $\varphi(p)\leq n$ ($\varphi$ is Euler's totient function, which is the degree of the $p$-th cyclotomic polynomial). In order to guarantee $A^{b_n}=I_n$ without knowing $p,$ we must set $b_n=\mathrm{lcm}\left\{p\;|\;\varphi(p)\leq n\right\}.$