Conjugacy classes and orders of matrices. The following are prime decompositions in $\Bbb{Z}_7[x]$:
$x^8+1= (x^2-x-1)(x^2+x-1)(x^2+3x-1)(x^2+4x-1)$
$x^4+1= (x^2+3x+1)(x^2+4x+1)$
(a) Give representatives for the conjugacy classes of elements of order dividing 16 in $Gl_2(\Bbb{Z}_7)$ and give the order of each. 
(b) Show that there are no elements of order 32 in $Gl_2(\Bbb{Z}_7)$. 
(a) We can write the rational canonical forms as representatives. 
$x^2-x-1 \rightarrow \begin{bmatrix} 0 & 1 \\1 & 1 \end{bmatrix}$
$x^2+x-1  \rightarrow  \begin{bmatrix} 0 & 1 \\1 & -1 \end{bmatrix}$
$x^2+3x-1 \rightarrow \begin{bmatrix} 0 &  1 \\1 & -3 \end{bmatrix}$
$x^2+4x-1 \rightarrow \begin{bmatrix} 0 &  1 \\1 & -4 \end{bmatrix}$
$x^3+3x+1 \rightarrow \begin{bmatrix} 0 &  -1 \\1 & -3 \end{bmatrix}$
$x^2+4x+1 \rightarrow \begin{bmatrix} 0 &  -1 \\1 & -4 \end{bmatrix}$
$x^2+1 \rightarrow \begin{bmatrix} 0 &  -1 \\1 & 0 \end{bmatrix}$. 
The first four matrices have order 16, since they satisfy $x^8+1=0$ so $x^8=-1 \implies (x^8)^2 = 1$. The last two have order 8, since $x^4=-1 \implies x^8=1$. 
(b)We don’t know the factorization of $x^{16}+1$, so we can’t check whether or not the matrices that satisfy it actually belong to $Gl_2(\Bbb{Z}_7)$. I wasn’t able to tell whether this was even reducible in $Z_7[x]$ to begin with, so I was kind of stuck here, and I was wondering if anybody could give me a hint...
Thanks in advance

Part (a) Let $A\in\operatorname{GL}_2(\mathbb Z_7)$ be a matrix whose order divides $16$.
Then $A^{16} - I = 0$, so the minimal polynomial $f$ of $A$ is a divisor of $x^{16} - 1$.
We have
$$x^{16} - 1
= (x^8 + 1)(x^8 - 1)
= (x^8 + 1)(x^4 + 1)(x^2 + 1)(x+1)(x - 1)
$$
Since $A$ is $2\times 2$, the degree of $f$ is at most $2$.
Using the fact that $x^2 + 1$ is irreducible over $\mathbb Z_7$ (since $-1$ is not a square mod $7$) and the given factorizations of $x^4 + 1$ and $x^8+ 1$, we can enumerate all the possibilities for $f$.
Furthermore, in all the cases we can immediately give the exact order of $A$, which is given by the smallest exponent $k$ such that $f$ divides $x^k - 1$. Representatives for $A$ are given by the possible rational normal forms (which happen to be uniquely determined by $f$ in all cases). We get the following list:
$$
\begin{array}{ccc}
f & \operatorname{ord}(A) & \text{representatives} \\
\hline
x-1 & 1 & I \\
x+1 & 2 & -I \\
(x-1)(x+1) & 2 & \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix} \\
x^2 + 1 & 4 & \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix} \\
x^2 + 3x + 1 & 8 & \begin{pmatrix}0 & -1 \\ 1 & -3\end{pmatrix} \\
x^2 + 4x + 1 & 8 & \begin{pmatrix}0 & -1 \\ 1 & 3\end{pmatrix} \\
x^2 - x - 1 & 16 & \begin{pmatrix}0 & 1 \\ 1 & 1\end{pmatrix} \\
x^2 + x - 1 & 16 & \begin{pmatrix}0 & 1 \\ 1 & -1\end{pmatrix} \\
x^2 + 3x - 1 & 16 & \begin{pmatrix}0 & 1 \\ 1 & -3\end{pmatrix} \\
x^2 + 4x - 1 & 16 & \begin{pmatrix}0 & 1 \\ 1 & 3\end{pmatrix} \\
\end{array}
$$
Part (b)
Assume that that there is a matrix $A\in\operatorname{GL}_2(\mathbb Z_7)$ of order $32$.
Then the minimal polynomial $f$ of $A$ divides $x^{32} - 1$, but not $x^{16} - 1$ and is of degree $1$ or $2$.
In the algebraic closure of $\mathbb{Z}_7$ there is a zero $a$ of $f$ of multiplicative order $32$. (Otherwise, all zeros would have an order dividing $16$. Since $x^{32} - 1$ doesn't have multiple zeros, $f$ does not have multiple roots and so $f$ would divide $x^{16} - 1$.)
From $f(a) = 0$ we get $[\mathbb{Z}_7(a) : \mathbb{Z}_7] \leq \deg(f) = 2$.
So $\mathbb{Z}_7(a)$ is a field of size $7^1$ or $7^2$. Because up to isomorphisms, finite fields are uniquely determined by their order, and because $\mathbb{F}_7$ is a subfield of $\mathbb{F}_{49}$, we get $a\in\mathbb{F}_{49}^\times$.
Since $\mathbb{F}_{49}^\times \cong \mathbb{Z}_{48}$ does not contain an element of order $32$, this is a contradiction.
