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An earlier question today motivates this slight variant:

For what natural numbers $n$ does there exist a non-identity integer $2\times 2$ matrix $A$, such that $A^n = I$? (And let's say $A^k \ne I$ for $|k| < |n|$, too.)

Clearly there are solutions for $n=2,3$, because $(-I)^2 = I$, and for $n=3$ we have $\left( \begin{smallmatrix} -2 & 1\\ -3 & 1 \\\end{smallmatrix} \right)$

The solution for $n = 3$ suggests to me that something involving the euclidean algorithm might come into play...and the fact the the determinant is $\pm 1$ suggests that this is really an $SL(2, \mathbb Z)$ (or $PSL(2, \mathbb Z)$)problem...but that's a group I'm woefully ignorant about.

For $n = 4$, there's $\left( \begin{smallmatrix} 0 & -1\\ 1 & 0 \\\end{smallmatrix} \right)$. And right about there I run out of ideas.

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John Hughes
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3 Answers3

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For $n=6$ there's $\left( \begin{smallmatrix} 2 & -1\\ 3 & -1 \\\end{smallmatrix} \right)$, not surprisingly. And these are the only such $n$, even if you allow rational entries instead of just integers.

Suppose that $A^n=I$ but $A^k\ne I$ for $k<n$. Then the minimal polynomial of $A$ divides $x^n-1$ but doesn't divide $x^k-1$ for $k<n$. It follows that the minimal polynomial of $A$ divides the $n$th cyclotomic polynomial $\Phi_n(x)$. But that polynomial is irreducible over the rationals, and so the minimal polynomial of $A$ equals $\Phi_n(x)$. Finally, the minimal polynomial of $A$ has degree at most $2$, since it divides the characteristic polynomial of $A$ by the Cayley-Hamilton theorem. Therefore the degre of $\Phi_n$ is at most $2$, which means that $\phi(n)$ is at most $2$, and so $n\in\{1,2,3,4,6\}$.

Greg Martin
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    Nice answer! And it used only things that I already (sort of) knew, showing that I should have been able to get there myself. – John Hughes Jan 11 '15 at 22:24
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    Everything's easier in hindsight :) I do find it a good instinct, whenever I see something about a poynomial expression a matrix, to translate it into a statement about the matrix's minimal polynomial - that tends to trigger the stuff-I-know part of my brain better than the original formulation. – Greg Martin Jan 12 '15 at 01:58
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$n=6$ is another possibility: let $A$ be any matrix with characteristic polynomial $X^2 - X + 1$.

There are no other possibilities, because the characteristic polynomial $p$ of $A$ has degree $2$. If $A$ has multiplicative order $n$, then $p$ must be a cyclotomic polynomial, which has degree $2$ only for $n\in \{1,2,3,4,6\}$.

Andrew Dudzik
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$$R_{\theta}=\begin{bmatrix} cos \theta & -sin \theta \\ sin \theta& cos \theta \end{bmatrix}\\$$if $$R_{\theta+2k\pi}=R_{\theta}\\(R_{\theta})^n=R_{n\theta}\\\theta=\frac{2k\pi}{n} \rightarrow (R_{\theta})^n=R_{n\theta}=R_{n\frac{2k\pi}{n}}=R_{2k\pi}=I $$

Khosrotash
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    Perhaps you missed the part about the entries being integers. This is a good answer, however, to the question I linked, which was about real matrices rather than integer ones. – John Hughes Jan 11 '15 at 22:23
  • Matrices with integer ,similar to R(theta) $$A^n=I\A=B^{-1}R_{\theta}B $$ – Khosrotash Jan 11 '15 at 22:28
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    Sure...if you can find such a $B$, then this works. Let's take $n = 8$, so $\theta = \pi/4$. Can you find a matrix $B$ such that $B^{-1} R_\theta B$ actually has all integer entries? The other two answers claim that you cannot, and I think I believe them. – John Hughes Jan 12 '15 at 01:50