# Elements of finite order of $\mathrm{PGL}(n,\mathbb{Q})$

For some research work, I need to know the classification of elements of finite order of $\mathrm{PGL}(n,\mathbb{Q})$, up to conjugation.

Since I essentially need $n\le 4$, I think that I can show it by hand, using cyclotomic extensions and Galois theory, but is there some work in the literature on this?

EDIT: Looking at the possible orders is essentially trivial in $\mathrm{GL}(n,\mathbb{Q})$, by just looking at the cyclotomic polynomials. The conjugacy classes require a little more work but are easy exercises, at least in low dimension. For $\mathrm{PGL}(n,\mathbb{Q})$, the case of orders prime to $n$ follows essentially from the case of $\mathrm{GL}(n,\mathbb{Q})$, the orders are more interesting.

• there seems to be a lot known about finite subgroups of $GL_n(\mathbb{Q})$, cf. e.g. references in ams.org/journals/proc/1997-125-12/S0002-9939-97-04283-4/… E.g. each of them is conjugate to one in $GL_n(\mathbb{Z})$. – Dima Pasechnik Jun 1 '14 at 10:32
• Isn't this just a question about rational canonical form? The question is certainly easier for elements of finite order in ${\rm GL}(n,\mathbb{Q}),$ but even there, calculating the maximum possible order of an element of finite order in ${\rm GL}(n,\mathbb{Q})$ is quite subtle. – Geoff Robinson Jun 1 '14 at 11:01
• @GeoffRobinson: what's so subtle in $GL_n(Q)$? you need to list $L_n$, the set of all cyclotomic polynomials of degree $\le n$, then from $L_n$ you list $L'_n$, the set of all subsets of $L_n$ whose sum of degrees is $\le n$. For each $\{\Phi_{n_1},\dots,\Phi_{n_k}\}$ in $L'_n$, you get an element of order lcm$(n_1,\dots,n_k\}$ in $GL_n(Q)$. Of course I don't claim it's algorithmically efficient when $n$ is very large. – YCor Jun 1 '14 at 11:26
• @Yves Cornulier: I just meant that that lcm is not so easy to explicitly evaluate, although theoretically, as you say, it is a formality. – Geoff Robinson Jun 1 '14 at 11:30
• @Jérémy: my guess is that a natural approach would be to start up to conjugation in $PGL_n(C)$, and then to understand when two elements in $PGL_n(Q)$ are conjugate over $C$ are conjugate over $Q$: since this problem is trivial in $GL_n$, one can expect it to we well encoded (in Galois cohomology?) in $PGL_n$. – YCor Jun 1 '14 at 11:37

I figured I'd write up the elementary observations here, since no one else has: If $g^k = \mathrm{Id}$ in $PGL_n$, then $g^k = a \mathrm{Id}$ in $GL_n$ for some nonzero $a$. So the minimal polynomial of $g$ divides $x^k-a$. Let $p_1 p_2 \cdots p_r$ be the factorization of $x^k-a$ over $\mathbb{Q}$. We can use rational canonical form to write down a $\deg p_i \times \deg p_i$ matrix $g_i$ (over $\mathbb{Q}$) with characteristic polynomial $p_i$. Then, for any nonnnegative integers $a_i$ such that $\sum a_i \deg p_i = n$, we can take the block diagonal matrix whose entries are $a_i$ copies of $g_i$. Conversely, if $g^k = a$, we can break $g$ into blocks according to the irreducible factors of $x^k-a$.

So, what remains is to analyze the degrees of the $p_i$. Let $K$ be the splitting field of $x^k-a$ over $\mathbb{Q}$ and let $G$ be the Galois group. Write $\zeta$ for a primitive $k$-th root of $1$ and $\alpha$ for a chosen $k$-th root of $a$ inside $K$. Then every element of $G$ is of the form $\zeta^i \alpha \mapsto \zeta^{ui+v}$ for $u \in (\mathbb{Z}/k)^{\times}$ and $v \in \mathbb{Z}/k$. So $G$ is a subgroup of $(\mathbb{Z}/k)^{\times} \ltimes (\mathbb{Z}/k)$. Also, $G$ surjects onto $\mathrm{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$, so, for every $u \in \mathbb{Z}/k^{\times}$, the group $G$ contains an element of the form $i \mapsto ui+v$. The problem is to describe the orbits of such a group on $\mathbb{Z}/k$.

For example, when $k=4$, then $G$ is a subgroup of $\{ \pm 1 \} \ltimes (\mathbb{Z}/4)$. If it isn't the whole group (in which case $x^4-a$ is irreducible), then it is either the group generated by $i \mapsto -i$ (in which case $x^4-a$ factors as $(\mbox{linear}) (\mbox{linear})(\mbox{quadratic})$, like $x^4-1$), or the group generated by $i \mapsto -i+1$ (in which case $x^4-a$ factors as $(\mbox{quadratic}) (\mbox{quadratic})$, like $x^4+4=(x^2+2x+2)(x^2-2x+2)$), or else $\{ \pm 1 \} \times 2 \mathbb{Z}/4$, in which case (in which case $x^4-a$ factors as $(\mbox{quadratic}) (\mbox{quadratic})$, like $x^4-4$).

At this point, it isn't clear what to do next, and the question is also a bit unfocused. Here are some (in my opinion) natural questions:

• Is is true that $x^k-a$ always has a factor of degree $\geq \phi(k)$? UPDATE: No. $x^8-16 = (x^2-2)(x^2+2)(x^2-2x+2)(x^2+2x+2)$, and $\phi(8) = 4$.

• For fixed $k$, what is the smallest $n$ for which $PGL_n(\mathbb{Q})$ has an element of order $k$? It is NOT $\phi(k)$: We can build elements of order $15$ in $GL_6$ as the direct sum of elements of orders $3$ and $5$ in $GL_2$ and $GL_4$, even though $\phi(15) = 8$.

• I don't have an example of a case where $PGL_n(\mathbb{Q})$ has an element of order $k$ but $GL_n(\mathbb{Q})$ doesn't. I imagine such a thing exists, but it would be good to have an example.

• For your Q2, I think it's sum over $p$ dividing $k$ of $\phi(p^r)$, where $r$ is the highest power of $p$ dividing $k$, except if $p=2$ and $r=1$ in which case you can ignore that term. This is because you have to include each prime power, and multiplying them multiplies their $\phi$, but adding is always less than multiplying when you're multiplying things at least $2$. – Will Sawin Aug 31 '14 at 20:12

Let $\lambda^k=a$. Consider how the Galois group of the splitting field of the minimal polynomial of $\lambda$ acts on the roots. As David Speyer points out, this is a subgroup of $(\mathbb Z/k)^\times \ltimes (\mathbb Z/k)$. If this subgroup has a nontrivial intersection with $(\mathbb Z/k)$, then there is a $l$ dividing $k$ sucht hat for any root $\lambda$ of the minimal polynomial, and $l$th root of unity $\mu$, $\mu \lambda$ is also a root.

So the degree of the minimal polynomial of $\lambda$ is $l$ times the degree of the minimal polynomial of $\lambda^l$. Now we may reduce to the case where the Galois group has trivial intersection with $\mathbb Z/k$.

Observation 1: In this case the Galois group is abelian, so the size of the orbit is the size of the Galois group, so the size of the orbit is at least $\phi$ of the order of the element. So no new orders of elements of $PGL_n(\mathbb Q)$ appear that aren't orders of elements of $GL_n(\mathbb Q)$.

Observation 2: If $k$ is odd, then there is some element of the form $x \to 2x+b$ in the Galois group. This has the unique fixed point $x=-b$. Because the Galois group is abelian, every other element fixes this point. So the $k$th roots of $a$ are some rational number times the $k$th roots of unity.

Full classification: Let $m \in (\mathbb Z/k)^\times$ be an element such that $m-1 \in 2 (\mathbb Z/k)^\times$. Then there is an element $x \to mx+b$ in the Galois group. If $b$ is even, then this element has two fixed points, the solutions to $(m-1)x+b=0$, so every other element either fixes or reverses those points. What this means is that there is a quadratic extension of $\mathbb Q$ fixed by the index $2$ subgroup of the Galois group that fixes those two points, and that those two points are of the form $q \sqrt{D}$ where $q$ is rational and $\sqrt{D}$ generates that quadratic extension. One can easily classify all sets of roots of this type.

I see my argument in the case $b$ odd doesn't actually work so I don't have a full classification.

• You can avoid the case $b$ odd: Replacing $x^n-a$ by $x^{2n}-a^2$, we may assume that our equation has a real root. Choose that real root to call $\alpha$, then complex conjugation acts by $\zeta^x \alpha \mapsto \zeta{-x} \alpha$, and that puts you in the case $b$ even. At this point, I think I know everything that can happen when factoring $x^n-a$ (including fun things like $x^{10} - 5^5$) but I'm having trouble organizing it into a succinct answer. If you can, please go ahead. – David E Speyer Sep 1 '14 at 20:45
• PS: There is not always an orbit of size $\geq \phi(n)$. Check out $x^8-16$. – David E Speyer Sep 1 '14 at 20:46

It seems to me that the question is equivalent to classifying up to conjugacy in ${\rm GL}(n,\mathbb{Q}),$ those matrices $M$ which satisfy $M^{d} = \frac{u}{v}I,$ where $d$ is a positive integer, $u$ is a positive integer, $v$ is anon-zero integer, ${\rm gcd}(u,v) = 1$ and neither $u$ nor $v$ is divisible by the $d$-th power of any prime, and furthermore, no lower power of $M$ is scalar. Then it becomes a question of determining what the possible rational canonical forms for $M$ can be. Maschke's theorem still goes through in this situation, suitably interpreted, and the minimum polynomial of $M$ is multiplicity free. The possible factors for the characteristic polynomial of such an $M$ are easy to determine.

• Yes, this is the natural start. By looking at the determinant you get $\det(M)^d=(u/v)^m$. If $m$ and $d$ are prime, you can then reduce to the case of elements of finite order of $\mathrm{GL}(n,\mathbb{Q})$. The other cases are more interesting. – Jérémy Blanc Jun 1 '14 at 16:15

Look for The Minkowski sequence

• Thanks, but it seems that the text that you mention only gives orders of finite subgroups of $\mathrm{GL}(n,\mathbb{Q})$, which is essentially trivial for cyclic groups. – Jérémy Blanc Jun 1 '14 at 16:20