I recently came across this linear algebra problem:
Let $ A $ be a $ 2 \times 2 $ matrix with integer entries satisfying $ A^n = I_2 $ for a natural $ n $. We are asked to show $ A^{12} = I $.
All I know is det$ (A) = \pm 1 $ but cannot figure out how to relate $ n $ and 12. Thanks for helping.
(At first for no good reason I thought we should get $A^2=I$. But no, if $A=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ then $A^2=-I$.
Below we do obtain the nominally stronger result that $A^6=I$ or $A^4=I$.)
The Cayley-Hamilton theorem shows that there is a monic quadratic $p$ with integer coefficients that annihilates $A$: There exist $b,c\in \Bbb Z$ with $$p(A)=A^2+bA+cI=0.$$
(Of course CH is a fairly large gun, but CH restricted to $2\times 2$ matrices is trivial; just write out what $\det(A-\lambda I)$ is in terms of $a_{j,k}$.)
Now say $m$ is the minimal polynomial for $A$. If $\deg(m)=1$ then $A=\lambda I$: hence $\lambda\in\Bbb Z$ and $\lambda^n=1$, so $\lambda=\pm1$ and so $A^2=I$.
If $\deg(m)>1$ then since $m|p$ we must have $m=p$. Hence $p|q$, where $q(t)=t^n-1$.
The (complex) zeroes of $p$ are roots of unity, in particular they have modulus $1$. And the zeroes of $p$ must be simple.
So if $p$ has real zeroes then $p(t)=t^2-1$ and we're done. If the roots of $p$ are complex then there exists $\alpha\in\Bbb C\setminus \Bbb R$ with $$t^2+bt+c=(t-\alpha)(t-\overline\alpha).$$
In particular $\alpha+\overline\alpha\in\Bbb Z$; since $|\alpha|=1$ and $\alpha$ is not real this shows that the real part of $\alpha$ is $0$ or $\pm1/2$. Since $\alpha$ is a root of unity it follows that $\alpha^2=1$ or $\alpha^6=1$. But if $\alpha=\pm i$ then $p(t)=t^2+1$, hence $A^2=-I$ and $A^4=I$.
If otoh $\alpha^6=1$ then $p(t)|(t^6-1)$; hence $A^6=I$.
Detail: Why does $\Re\alpha=\pm1/2$ imply that $\alpha^6=1$? Because $\Re\alpha=\pm1/2$ and $|\alpha|=1$ imply that $\alpha=\pm e^{\pm 2\pi i/3}$.