# Order of periodic element of general linear group over integers is bounded

This article gives the statement, and possibly proof, of a particular group or type of group (namely, General linear group over integers (?)) satisfying a particular group property (namely, Group in which all elements of finite order have a common bound on order (?)).

## Statement

Suppose $n$ is a natural number. Then, the order of any element of the General linear group (?) $GL(n,\mathbb{Z})$ having finite order is bounded by a function of $n$.

Explicitly, if there exists an element of order $m$, then we can write $n = a_1\varphi(m_1) + a_2\varphi(m_2) + \dots + a_r\varphi(m_r)$ where $\varphi$ denotes the Euler totient function, all $a_i \ge 1$, and the lcm of $m_1,m_2,\dots,m_r$ equals $n$.

This gives a bounding function of the form: the maximum possible order of a $n \times n$ matrix is at most $(1 + n)^{2n}$. In practice, the maximum possible order is much lower.

## Particular cases

Value of $n$ Possible orders of periodic elements in $GL(n,\mathbb{Z})$ Maximum possible order lcm of possible orders
1 1,2 2 2
2 1,2,3,4,6 6 12
3 1,2,3,4,6 6 12
4 1,2,3,4,5,6,8,10,12 12 120
5 1,2,3,4,5,6,8,10,12 12 120
6 1,2,3,4,5,6,7,8,9,10,12,14,15,20,24,30 30 2520

## Proof

### Proof of the Euler totient function partition condition

Given: An element $g \in GL(n,\mathbb{Z})$ of order $m$.

To prove: There exists a partition $n = a_1\varphi(m_1) + a_2\varphi(m_2) + \dots + a_r\varphi(m_r)$ where the lcm of $m_1,m_2,\dots,m_r$ is $m$ and $\varphi$ is the Euler totient function.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 $g$ satisfies the polynomial $t^m - 1$, so the minimal polynomial of $g$ divides $t^m - 1$. $g$ has order $m$
2 The minimal polynomial of $g$ is a product of cyclotomic polynomials $\Phi_{m_1}(t)\Phi_{m_2}(t)\dots \Phi_{m_r}(t)$, where the lcm of $m_1,m_2,\dots,m_r$ is $m$. $t^m - 1 = \prod_{d \mid m} \Phi_d(t)$, $\Phi_d(t)$ are irreducible $g$ has order $m$. Step (1) By Step (1) and the factoring of $t^m - 1$ we conclude that the minimal polynomial must be a product of $\Phi_d(t)$ for a subset of the set of divisors of $m$. Call these divisors $m_1,m_2,\dots,m_r$. They all divide $m$. If the lcm is smaller than $m$, then the order of $g$ is strictly smaller than $m$, a contradiction. Hence the lcm must be exactly $m$.
3 The characteristic polynomial is of the form $\Phi_{m_1}(t)^{a_1}\Phi_{m_2}(t)^{a_2}\dots \Phi_{m_r}(t)^{a_r}$ where $a_1,a_2,\dots,a_r$ are all positive integers, and the lcm of $m_1,m_2,\dots,m_r$ is $m$. minimal polynomial divides characteristic polynomial Step (2) Follows from fact and Step (2).
4 $n = a_1\varphi(m_1) + a_2\varphi(m_2) + \dots + a_r\varphi(m_r)$, where $a_1,a_2,\dots,a_r$ are all positive integers, and the lcm of $m_1,m_2,\dots,m_r$ is $m$. degree of cyclotomic polynomial $\Phi_d$ is $\varphi(d)$, degree of characteristic polynomial of $n \times n$ matrix is $n$ Step (3) Direct from Step (3), compute degree of characteristic polynomial in two different ways.

### Proof of bounding

Given: $n = a_1\varphi(m_1) + a_2\varphi(m_2) + \dots + a_r\varphi(m_r)$, where $a_1,a_2,\dots,a_r$ are all positive integers, and the lcm of $m_1,m_2,\dots,m_r$ is $m$.

To prove: $m \le n^{2n}$.

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Each $\varphi(m_i)$ is bounded by $n$.
2 Each $m_i. $x \le (1 + \varphi(x))^2$ for any natural number $x$. Step (1)
3 $r \le n$
4 $m$, defined as the lcm of $m_1,m_2,\dots,m_r$ is at most $m_1m_2\dots m_r$.
5 $m \le (1 + n)^{2n}$ Steps (3), (4), (5) Step-combination direct.