# Integer matrices which are not a power

$$\DeclareMathOperator\SL{SL}\DeclareMathOperator\Sp{Sp}$$In a group $$G$$, an element $$g$$ is said to be primitive if there is no $$h \in G$$ and integer $$n >1$$ such that $$g = h^n$$. (For clarification, I consider finite order elements to be not primitive)

I was wondering, in the case $$G$$ is $$\SL_n(\mathbb{Z})$$ or $$\Sp_{2n}(\mathbb{Z})$$, if there exists a criterion for primitivity of matrices. I actually even struggle to find examples of primitives matrices in these groups.

In $$\Sp_{2}(\mathbb{Z})=\SL_{2}(\mathbb{Z})$$, the matrix $$\pmatrix{1 & 1 \\ 0 & 1}$$ is primitive. (This can be shown by considering its action on $$\mathbb{H^2}$$, for example.)

But this does not generalize (easily at least) to higher dimension. For example, and quite surprisingly maybe $$\pmatrix {1 & 0 & 0& 1\\ 0 & 1 & 0 & 0\\ 0 & 1 & -1 & 1\\ 0 & 1& -1 & 0 }^3 = \pmatrix{1 & 1 & 0& 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0& 0 & 1}$$

Anyway, it seems like some things should be known, but it is very hard to find anything on google since primitive matrix usually means something else …. I would appreciate any input.

• The MO post "Condition for a matrix to be a perfect power of an integer matrix" seems relevant, although it doesn't have a complete answer to your question: mathoverflow.net/questions/375584/… – Joe Silverman Jan 21 at 16:35
• @GeoffRobinson wait, if $a^2=1$ then $a=a^3$? – Fedor Petrov Jan 21 at 17:15
• @DenisT I follow his definition in the first paragraph of the post. – Fedor Petrov Jan 21 at 17:24
• The problem in $\mathrm{GL}_2(\mathbf{Z})$ or $\mathrm{SL}_2(\mathbf{Z})$ is easily solvable using the amalgam decomposition. For $\mathrm{SL}_d(\mathbf{Z})$, $d\ge 3$ I don't even see in an obvious way that it's algorithmically solvable, but I'm pretty sure it is; how efficiently, I'm less sure. – YCor Jan 21 at 19:17
• About primitive: it can be shown that every element of infinite order in $\mathrm{GL}_n(\mathbf{Z})$ (or in any of its subgroups) of infinite order is a power of a primitive element: indeed every abelian subgroup in such a group is finitely generated (use that it's a discrete subgroup of its Zariski closure, which is a virtually connected abelian Lie group). – YCor Jan 21 at 23:47

Here is a relatively easy sufficient condition. If $$M \in SL_n(\mathbb{Z})$$ is the $$k^{th}$$ power of some other matrix $$N$$ then every eigenvalue $$\lambda$$ of $$N$$ has the property that $$\lambda^k$$ is an eigenvalue of $$M$$, and conversely. If we can choose $$M$$ and an eigenvalue $$\mu$$ of $$M$$ such that every $$k^{th}$$ root of $$\mu$$ has the property that its minimal polynomial has degree larger than $$n$$, then $$N$$ cannot exist.
For this to work $$\mu$$ can't be a root of unity. As an explicit example we'll take $$M = \left[ \begin{array}{ccc} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1 \end{array} \right]$$, the Fibonacci matrix augmented by a $$-1$$ to live in $$SL_3(\mathbb{Z})$$, whose eigenvalues are $$-1$$ and the golden ratios $$\phi, \varphi = \frac{1 \pm \sqrt{5}}{2}$$. Any $$\lambda$$ such that $$\lambda^k = \phi, \varphi$$ for $$k \ge 2$$ generates an extension of $$\mathbb{Q}$$ which contains $$\mathbb{Q}(\sqrt{5})$$ and hence is either equal to $$\mathbb{Q}(\sqrt{5})$$ or else has even degree greater than $$2$$ and so at least $$4$$. Moreover $$\lambda$$ is an algebraic integer and a unit.
But $$\phi, \varphi$$ each generate the unit group of $$\mathcal{O}_{\mathbb{Q}(\sqrt{5})} = \mathbb{Z}[\phi]$$ (up to signs). So if $$\mathbb{Q}(\lambda) = \mathbb{Q}(\sqrt{5})$$ then $$\lambda$$ must be $$\pm \phi$$ or $$\pm \varphi$$, but this is ruled out by taking the absolute value. So the minimal polynomial of $$\lambda$$ must have degree at least $$4$$, which means $$\lambda$$ can't be an eigenvalue of a matrix in $$SL_3(\mathbb{Z})$$.