# Order of invertible matrices

I recently came across an interesting problem in Artin which says:

If $A \in GL_2(\mathbb{Z})$ is of finite order then it has order $1,2,3,4,6$. I was looking for a generalization of this problem. For example, if they were invertible finite order matrices with rational entries then how big they have to be such that it has order $1,2,3,4,6$? How about if it has complex entries?

## 2 Answers

There is a paper of Burgisser which discusses finite order elements in matrix groups. The results he proves imply

There appear in the groups $GL(n,\mathbb{Q})$, $GL(n,\mathbb{Z})$, $SL(n,\mathbb{Q})$, $SL(n,\mathbb{Z})$, $Sp(n,\mathbb{Q})$ and $Sp(n,\mathbb{Z})$ exactly the same finite orders of elements.

Furthermore, he describes exactly those orders which are possible :

Let $m = \prod_{i=1}^h p_i^{a_i}$, where the primes $p_i$ satisfy $p_i < p_{i+1}$ and where $a_i \geq 1$. There is a matrix $A\in Sp_n(\mathbb{Z})$ of order $m$ if and only if $$(a)\quad \sum_{i=2}^h \varphi(p_i^{a_i}) \leq n \text{ if } m\equiv 2\pmod{4}$$ $$(b)\quad\sum_{i=1}^h \varphi(p_i^{a_i}) \leq n \text{ if } m\neq 2\pmod{4}$$

Any finite subgroup of ${\rm GL}(n,{\mathbb Q})$ is conjugate to a subgroup of ${\rm GL}(n,{\mathbb Z})$, so extending to rational entries will not yield any additional finite orders, or even isomorphism types of finite subgroups.

On the other hand ${\rm GL}(1,{\mathbb C})$ clearly has elements of all possible finite orders.