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 $a_{i}\geq 1$, 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: $m\leq n^{2n}$.

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