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What is the largest (finite) order of an element of $GL_{10}(\mathbb{Q})$?

The following is my guess: Let $A \in GL_{10}(\mathbb Q)$ is of finite order $n$ (say). Then $A^n=I$, or every eigenvalue of $A$ is an $n$-th root of $1$. Thus an eigenvalue $\lambda$ of $A$ has degree $\leq\phi(n)$ over $\mathbb Q$. Equality holds only when $\lambda$ is primitive $n$-th root of $1$. OTOH $\lambda$ also satisfies the characteristic polynomial of $A$, which is of degree $10$. Thus if the characteristic polynomial is irreducible (e.g., $A$ is companion matrix of $X^{10}+X^9+\cdots+X+1$) then degree of $\lambda=10$, then we must have $\phi(n)\leq10$. The maximum $n$ such that $\phi(n)=10$ is $15$.
If I am doing some mistake please correct me. Thanks.

Ben Grossmann
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user371231
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  • It’s not too bad a first guess, but $\phi(15)=8$, though… also, if the characteristic polynomial is irreducible, then $\lambda$ has degree $10$… why does this imply that $\phi(n) \leq 10$? Unless you restrict yourself to the case where there is some $\lambda$ such that $A$ and $\lambda$ have the same order… which isn’t always true. There is also this reliance on the irreducibility of the characteristic polynomial that doesn’t work too well either. – Aphelli Jul 15 '21 at 09:34
  • $x^8 + x^7 - x^5 - x^4 - x^3 + x + 1$ is a factor of $x^{30}-1$ and no lower $x^n-1$. So if you find a linear transformation with that as minimal polynomial... Somehow I imagine that's easier in $GL_8$. $x^{10} - x^9 + x^8 - x^7 + x^6 - x^5 + x^4 - x^3 + x^2 - x + 1$ is probably the best degree 10 polynomial to aim for, being a factor of $x^{22}-1$. – Arthur Jul 15 '21 at 10:04
  • Hi @Arthur Just asking: How do you get those polynomials? Are there any systematic way of getting those? – user371231 Jul 15 '21 at 10:15
  • I'm searching for "cyclotomic polynomials". Those are the polynomials that appear when you factor $x^n-1$ into irreducible factors. WolframAlpha gives them out directly. – Arthur Jul 15 '21 at 10:18
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    @Arthur: it’s not the highest $n$, actually. Not by far. – Aphelli Jul 15 '21 at 11:59
  • Interestingly, repeating this analysis for $GL_n(\Bbb Q)$ for $n = 1,2,\dots$ yields the sequence $$ 2,6,6,12,12,30,30,60,60,120,\dots $$ which does not appear in the OEIS. – Ben Grossmann Jul 15 '21 at 17:11
  • @Mindlack For some reason I didn't think about multiplying together different cyclotomic polynomials. – Arthur Jul 15 '21 at 17:39

1 Answers1

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Briefly: the answer is $120$. One such matrix is $$ A = \pmatrix{0&1\\& \ddots & \ddots \\ \\ \\ \\ \\ \\ \\ &&&&&&&\ddots & \ddots \\ &&&&&&&&0&1 \\-1&-2 & -3 & -3 & -4 & -4 & -4 & -3 & -3 & -2}. $$

Here is an approach to the problem.

The matrix $A$ will satisfy $A^n = I$ if and only if the minimal polynomial of $A$ divides $x^n - 1$. Notably, $x^n - 1$ is a product of cyclotomic polynomials (which are irreducible over $\Bbb Q$), so the minimal polynomial of $A$ must be a product of these. Because $A$ has size $10$, its minimal polynomial has degree at most $10$.

Let $\Phi_n$ denote the $n$th cyclotomic polynomial. The degree of $\Phi_n$ is $\phi(n)$ (where $\phi$ denotes the Euler totient function). For a product $p = \Phi_{n_1} \cdots \Phi_{n_k}$ with $n_1,\dots,n_k$ distinct, the smallest $n$ for which $x^n - 1$ is divisible by $p$ will be $n = \mathrm{lcm}(n_1,\dots,n_k)$.

Putting all that together, we can frame the problem as follows:

Maximize $\mathrm{lcm}(n_1,\dots,n_k)$ over all sets of distinct positive integers $n_1,\dots,n_k$ satisfying the constraint that $\phi(n_1) + \cdots + \phi(n_k) \leq 10$.

This maximal gcd will be equal to the largest possible finite order of an element of $GL_{10}(\Bbb Q)$.

Notably, there are lower bounds on $\phi$ that make it possible to obtain a solution by brute force. In particular, it is known that $\phi(n) \geq \sqrt{n/2}$. Thus, each $n_k$ must satisfy $$ \sqrt{n_k/2} \leq \phi(n_k) \leq 10 \implies n_k \leq 200. $$

After a systematic search I have found that the highest possible order is $120$. This maximum is attained with a matrix whose minimal polynomial is equal to $$ \Phi_3\Phi_5\Phi_8 = (x^2 + x + 1)(x^4 + x^3 + x^2 + x + 1)(x^4 + 1). $$ For instance, we can take $A$ to be the associated companion matrix.

Here's the Python code I used to compute this answer:

from math import gcd

def gcds(args):
    if len(args) == 2:
        return gcd(args)
    else:
        return gcds(gcd(args[:2]),args[2:])


def lcm(args):
    if len(args) == 1:
        return args[0]
    if len(args) == 2:
        return args[0]args[1]//gcd(args)
    else:
        return lcm(lcm(args[:2]),*args[2:])


def Ephi(n):
    s = 0
    for k in range(n):
        if gcd(k,n) == 1:
            s += 1
    return s


def get_list(deg, start = 1, M = 200):
    if deg == 0:
        return [[]]
    arr = []
    for k in range(start,M+1):
        phi = Ephi(k)
        if phi <= deg:
            new_deg = deg - phi
            new_M = 2new_deg*2
            arr += [[k] + L for L in get_list(new_deg, start = k+1,  M = new_M)] # start=k allows repeats
    return arr


if name == 'main':


n = 10 # set value of n (in GL(n,Q)) here

M = 2*n**2
arr = get_list(n, M=M)
print('    number of non-derogatory elements with finite order (up to iso): ' + str(len(arr)))
m = 0
for L in arr: 
    x = lcm(*L)
    if x &gt; m:
        m = x
        mL = L
print('    max order: %d'%m) #prints maximal order
s = ''
for k in mL:
    s += 'Phi_%d '%k
# minimal polynomial of element with maximal order as product of 
# cyclotomic polynomials
print('minimal polynomial of maximal element: ' + s)

There are a few other candidates for the minimal polynomial. All together, the valid possibilies are $$ \Phi_3\Phi_5\Phi_8,\\ \Phi_3\Phi_8\Phi_{10},\\ \Phi_5\Phi_6\Phi_8,\\ \Phi_6\Phi_8\Phi_{10}. $$

Relevant paper: Finite Groups of Matrices Whose Entries Are Integers

Ben Grossmann
  • 225,327