# Finite order elements of $\mathrm{GL}_d(\mathbb{Z})$ that are conjugate to powers of themselves

$$\DeclareMathOperator\GL{GL}$$Let $$A\in \GL_d(\mathbb{Z})$$ have finite order $$n.$$ Suppose that $$k\in \mathbb{Z}$$ is relatively prime to $$n.$$ Is $$A^k$$ conjugate to $$A$$ in $$\GL_d(\mathbb{Z})$$?

For $$d\leq 4$$ the answer is yes. Indeed the papers "On the finite subgroups of $$\GL(3,\mathbb{Z})$$" by K. Tahara, 1971 and "Conjugacy Classes of Torsion in $$\GL_n(\mathbb{Z})$$", by Q. Yang, 2015 list all torsion elements up to conjugacy for the cases $$d=2,3$$ and $$d=4$$ respectively. There are not many cases that need to be checked, so I checked each case and the answer turns out to be "yes" in all of these cases. But I have no general argument for why things work out when $$d\leq 4$$, only computations.

• Just as a motivation, if I'm correct, this is always true in $\mathrm{GL}_d(\mathbf{Q})$, due to the fact that the set of roots of every cyclotomic polynomial is stable under inversion. (And this is clearly false in $\mathrm{GL}_d(\mathbf{C})$ for every $d\ge 1$, just use a scalar matrix of finite order $\ge 3$.)
– YCor
Feb 4 at 10:24
• @YCor : What you say is true, but I think you need a little more than stable under inversion: the roots of irreducible cyclotomic polynomials are all Galois conjugate to each other (Galois groups being over $\mathbb{Q}$). Feb 4 at 10:29
• @GeoffRobinson Oh, yes, I focussed on $A$ being conjugate to $A^{-1}$ after Alex B.'s answer. So, I should have said that the set of roots of $\Phi_n$ is stable under $x\mapsto x^m$ for $m$ coprime to $n$.
– YCor
Feb 4 at 10:37

The question is equivalent to the following question: let $$C_n$$ be a cyclic group of order $$n$$, let $$g$$ be a generator, and let $$\rho\colon C_n\to {\rm GL}_d(\mathbb{Z})$$ be an integral representation of $$C_n$$, defined by $$g\mapsto A$$. Let $$\sigma\in {\rm Aut}(C_n)$$ be defined by $$g\mapsto g^k$$. Then is the integral representation $$\rho\circ\sigma$$ of $$C_n$$ isomorphic to $$\rho$$? Let me now explain a general construction of representations for which the answer is "no".
Set $$d=\phi(n)$$, the Euler phi function. For example if $$n$$ is prime, then we will have $$d=n-1$$. Fix a primitive $$n$$-th root of unity $$\zeta_n$$, and let $$I$$ be an ideal in $$\mathbb{Z}[\zeta_n]$$, the ring of integers of the $$n$$-th cyclotomic field $$\mathbb{Q}(\zeta_n)$$. Let $$C_n$$ act on $$I$$ by letting $$g$$ act by multiplication by $$\zeta_n$$. Since $$I$$ is an ideal, multiplication by $$\zeta_n$$ preserves it as a set, so this defines a representation $$\rho_I\colon C_n\to {\rm GL}_d(\mathbb{Z})$$. Moreover, it is not hard to see that if $$I$$ and $$J$$ are two such ideals, then $$\rho_I$$ is isomorphic to $$\rho_J$$ if and only if $$I$$ and $$J$$ represent the same class in the ideal class group of $$\mathbb{Q}(\zeta_n)$$. Finally, given $$k$$ coprime to $$n$$, there exists an element $$\sigma$$ of the Galois group of $$\mathbb{Q}(\zeta_n)$$ such that $$\sigma\zeta_n = \zeta_n^k$$. This $$\sigma$$ also defines an automorphism of $$C_n$$ via $$g\mapsto g^{k}$$. It is easy to see that $$\rho_I\circ \sigma=\rho_{\sigma^{-1}I}$$. Therefore, to produce an example of the type you desire, you can take a cyclotomic field in which the Galois group acts non-trivially on the class group and let $$I$$ be a representative of any class in the class group that is not preserved by the Galois group.
The smallest example that you can produce this way has $$n=23$$, $$d=22$$. Indeed, the cyclotomic field $$\mathbb{Q}(\zeta_{23})$$ is the first cyclotomic field with non-trivial class group, which has order $$3$$, and complex conjugation acts by $$-1$$ on it. So take any non-principal ideal $$I$$ in $$\mathbb{Z}[\zeta_{23}]$$, write down a $$\mathbb{Z}$$-basis for it, and write out the matrix $$A$$ of multiplication by $$\zeta_{23}$$ on this basis. Then $$A$$ will not be conjugate to $$A^{-1}=A^{22}$$ in $${\rm GL}_{22}(\mathbb{Z})$$.