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

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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</math is less than <math>(1 + n)^2. 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.