Number of matrices $A$ in $M_2(\mathbf{Z}/3\mathbf{Z})$ satisfying $A^{50}=I.$ 
Find the number of $2$x$2$ matrices $A$ with entries in $\mathbf{Z}/3\mathbf{Z}$ satisfying $A^{50}=I$.

Propositions as far as I know:

*

*The number of $2$x$2$ matrices with entries in $\mathbf{Z}/3\mathbf{Z}$ is $81$.

*$A^{50}=I$ implies that the eigenvalues of $A$ are roots of $\lambda^{50}=1.$

*In $\mathbf{Z}/3\mathbf{Z}$, if $A$ is diagonalizable and $\det(A)\neq 0$, then $A^{50}=I.$

*$A^{50}=I$ implies $\det(A)\neq 0$
Using prop. $3$ I get the following $14$ matrices satisfying $A^{50}=I$ (at least $14$ matrices, maybe
more).


But I'm not sure whether the condition in prop. 3 is necessary? Note that $\textrm{char} (\mathbf{Z}/3\mathbf{Z})\neq 0$.
 A: Consider: A matrix $A=\begin{pmatrix}a & b \\ c & d \end{pmatrix}$ is not a unit exactly if $ad - bc=0$. How many possibilities does this have?

*

*$a=0,b=0$ (9 possibilies)

*$d=0,c=0$ and $a\neq 0$ or $b\neq 0$ (8 possibilities)

*$a=0,c=0$ and $b\neq0, d\neq 0$ (4 possibilities)

*$d=0,b=0$ and $a\neq0,c\neq 0$ (4 possibilities)

*$ad=1,bc=1$ (4 possibilities)

*$ad=2,bc=2$ (4 possibilities)

for a total of $9+8+4+4+4+4=33$ possibilities.
Thus we have $3^4 - 33 = 48$ matrices that are units. The multiplicative order of such a matrix thus has to divide $48$, so it divides $50$ if and only if it divides the $\gcd(48,50)=2$.
Thus we only need to consider matrices with multiplicative order 1 or 2.
Order 1 is trivially only possible for $E$. So we only need to solve:
$$ A^2 = E$$
(this also includes $A=E$).
This is equivalent to
$$ a^2+bc = 1 = cb+d^2 $$
$$ ab+bd = 0 = ca+dc $$
Thus if $a=0$ then $bc=1$ (thus $b=c=\pm1$), $d=0$. So we get
$\begin{pmatrix} 0 & \pm 1 \\ \pm 1& 0 \end{pmatrix}$ ($\pm$ linked, 2 possibilites)
If $a=\pm1$ then $a^2=1$, so $bc=0$, thus $d=\pm1$. Then either $b=0$ or $c=0$. Assume $b=0$ ($c=0$ analogue). So $0=ca+dc$ if either $c=0$ or $a+d=0$. Thus either $c=0$ or $a=-d$. This gives us

*

*$\begin{pmatrix} \pm1 & 0 \\ 0& \pm1 \end{pmatrix}$ ($\pm$ not linked, 4 possibilites)

*$\begin{pmatrix} \pm 1 & 0 \\ c & \mp 1 \end{pmatrix}$ ($c=1,2$, $\pm$ linked, 4 possibilities)

*$\begin{pmatrix} \pm 1 & b \\ 0& \mp 1 \end{pmatrix}$ (similar 4 possibilities)

So we have a total of $2+4+4+4=14$ such matrices.
A: $\mathbb Z/3\mathbb Z$ is isomorphic to the field $\mathbb F_3$ of characteristic $3$. Suppose first that the characteristic polynomial of $A$ is irreducible over $\mathbb F_3$. Then there are only three possible choices:
\begin{cases}
x^2+1,\\
x^2+x-1,\text{ or}\\
x^2-x-1.\\
\end{cases}
The first choice is impossible because it implies that $A^2=-I$ and $A^{50}=-I$. The second choice is impossible too, otherwise $A^2=I-A$ and $A^4=(I-A)^2=A^2-2A+I=2I=-I$, but then the condition $A^{50}=I$ will imply that $A^2=I$, which is a contradiction because $\pm1$ are not roots of $x^2+x-1=0$. The third choice is also impossible, because it reduces to the second choice by considering $B=-A$.
Hence the characteristic polynomial of $A$ must split over $\mathbb F_3$ and the eigenvalues of $A$ lie inside the set $\{1,-1\}$ (because $A$ is invertible). Since $50=-1\ne0$ in $\mathbb F_3$, if $A^{50}=I_2$, the Jordan normal form of $A$ over the algebraic closure of $\mathbb F_3$ cannot have any non-trivial Jordan block. Therefore $A$ is diagonalizable, and either $A=\pm I$ or $A$ is similar to $\operatorname{diag}(1,-1)$. In the latter case, $A$ is uniquely determined by the two eigenspaces corresponding to the eigenvalues $1$ and $-1$. Since $\mathbb F_3^2$ has $(3^2-1)/2=4$ different one-dimensional subspaces, there are $4(4-1)=12$ matrices similar to $\operatorname{diag}(1,-1)$. Counting also the two solutions $A=\pm I$, there are $12+2=14$ solutions in total.
