Find $\left|\left\{A\in(\Bbb Z/3\Bbb Z)_{2\times2}:A^{50}\equiv I_2\pmod3\right\}\right|$

Question

Find the number of $2\times2$ reversible matrices $A$ whose elements range in $\Bbb Z/3\Bbb Z$ such that $A^{50}\equiv I_2\pmod3$.

I used python to find the answer.

Here is the result: $$\begin{bmatrix} -1 & -1 \\ 0 & 1 \\ \end{bmatrix} \begin{bmatrix} -1 & 0 \\ -1 & 1 \\ \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & 1 \\ \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 1 & 1 \\ \end{bmatrix} \begin{bmatrix} -1 & 1 \\ 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 0 & -1 \\ -1 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 0 & -1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & -1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & -1 \\ \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & -1 \\ \end{bmatrix}$$ How do we solve this by hand? (Group theory may or may not be relevant)

Answer 1

Fleshing out the hint in the comments. This is mostly linear algebra.

A starting point is the observation that $G=GL_2(K), K=\Bbb{Z}/3\Bbb{Z}$, is a group of order $48$. Therefore every element $A\in G$ satisfies the equation $A^{48}=1_G=I_2$. Consequently, you are asking for a list of elements $A\in G$ satisfying $A^2=I_2$.

Linear algebra tells us that the eigenvalues (necessarily in $K$ or its quadratic extension field as we are dealing with 2x2 matrices) of such a matrix $A$ can only be $\pm1$. We deal with the resulting three cases separately, but I begin by showing that $A$ is necessarily diagonalizable. Namely, the identity $$ \vec{x}=\frac12(I_2+A)\vec{x}+\frac12(I_2-A)\vec{x} $$ makes sense for every (column) vector $\vec{x}\in K^2$ because we can divide by $2$. Furthermore, the two vectors on the right hand side are eigenvectors of $A$ belonging to $\lambda=\pm1$ respectively. Hence every vector is a linear combination of eigenvectors, and the first claim follows from this (you may have known this piece already).

On with the cases:

1. If $\lambda=+1$ is a double eigenvalue, then diagonalizability forces $A=I_2$.
2. Similarly, if $\lambda=-1$ is a double eigenvalue, then we must have $A=-I_2$.
3. If both $\pm1$ are eigenvalues, then we know that the characteristic polynomial of $A$ is $\chi_A(x)=(x-1)(x+1)$. For $2\times2$-matrices we have the useful formula $$\chi_A(x)=\det(xI_2-A)=x^2-tr(A)x+\det A.$$ This means that the trace (=the sum of diagonal entries) of $A$ must vanish, and that the determinant must be equal to $-1$. What's more, these necessary conditions are also sufficient, as any matrix satisfying them shares the characteristic polynomial. So all you need to is to list the matrices with zero trace and determinant $-1$. In particular, the diagonal must be either $(0,0)$, $(1,-1)$ or $(-1,1)$, and the determinant then constrains the off-diagonal entries in a way that makes writing down the list by hand rather straight forward.

Answer 2

HINT:

Any $n\times n$ matrix $A$ satisfies an equation of degree $n$ ( its characteristic equation) : $P(A)=0$. If it satisfies another equation $Q(A)=0$, then $D(A)=0$, where $D = \gcd(P,Q)$. But note that $D$ is of degree $\le n$ and divides $Q$.

Now, $x^{50}-1$ has only one factor of degree $\le 2$, $x^2-1$. Reduce to $A^2-1=0$, a problem that was seen before