an element of $\text{SL}(3, \mathbf Z)$ of order 5

Question

It is known that the orders of elements of $\text{SL}(2, \mathbf Z)$ are restricted: for example, there is no element of order 5.

What about for $\text{SL}(3,\mathbf Z)$? Does there exist an element of order 5? Of order 7? Of order any integer?

How would one go about constructing such a matrix?

There are some known results over finite fields, but I am not at all familiar with the $\mathbb{Z}$ case. — Randall, Feb 06 '24 at 01:04
Note that Tahara has some typos and errors, as is worked out here: https://math.stackexchange.com/a/4219995 — colt_browning, Feb 06 '24 at 01:29

score 0 · Accepted Answer · answered Feb 06 '24 at 01:27

An element of order $n$ would have an eigenvalue of that order, necessarily together with the other $\phi(n)-1$ primitive roots of unity of that order, which means that $\phi(n)\leqslant3$, which gives the same crystallographic restriction as in dimension $2$: only orders $1,2,3,4,6$, as Mariano said in the comment.

an element of $\text{SL}(3, \mathbf Z)$ of order 5

1 Answers1