Dummit and Foote 12.2.16: Determining all $2 \times 2$ matrices with entries from $\mathbb F _{19}$ of order $2$ This is exercise 12.2.16 of Abstract Algebra by Dummit and Foote.

Show that $x^5-1 =(x-1)(x^2-4x+1)(x^2+5x+1)$ in $\mathbb F_{19}[x]$. Use this to determine, up to similarity, all $2 \times 2$ matrices with entries from $\mathbb F_{19}$ of (multiplicative order) $5$.


*

*First of all, the multiplicative group of $\mathbb F_{19}$ is $C_{18}$, the cyclic group of order 18, with generator $2$, so I presume they mean that the matrices are of order $5$.


*Secondly, I prove the decomposition result as stated and then see that due to Cayley-Hamilton we know that any $2 \times 2$ matrix $A$ with minimal polynomial that divides $x^5-1$ must also satisfy $A^5-I=0  \implies A^5= I$.
The candidate polynomials are:
$x-1$, $x^2-4x+1$ and $x^2+5x+1$.
This question is similar to what they do on page 487, but I am not quite sure how one goes from these candidates to the list of permissible invariant factors, and after that, the matrices.

We looks at the divisors of $x^5-1$ and use the decomposition as given. Notice for a $2 \times 2$ matrix it is only possible to get $2$ blocks of size $1$ or $1$ block of size $2\times 2$.
Recall that for a polynomial of the form $b_0 +b_1x +x^2$, we get the following companion matrix according to page 475:
$$ \begin{pmatrix}
    0 & -b_0 \\
    1 & -b_1 
\end{pmatrix}$$
A $1\times 1$ block simply has the value of $-b_0$ for $b_0 + x$.

*

*$x-1$, $x-1$ give us:
$$\begin{pmatrix}
    -(-1) & 0 \\
    0 & -(-1) 
    \end{pmatrix}=\begin{pmatrix}
    1 & 0 \\
    0 & 1 
    \end{pmatrix}=I.$$
This matrix is of course not of order $5$, but of order $1$, therefore it satisfies $I^5=I$.


*$x^2-4x+1$ gives us:
$$A= \begin{pmatrix}
0 & -1 \\
1 & 4 
\end{pmatrix}\equiv \begin{pmatrix}
0 & 18 \\
1 & 4 
\end{pmatrix} \bmod {19}$$


*$x^2+5x+1$ gives us:
$$B=\begin{pmatrix}
0 & -1 \\
1 & -5 
\end{pmatrix}\equiv \begin{pmatrix}
0 & 18 \\
1 & 14 
\end{pmatrix} \bmod {19}$$
One can easily verify that
$$ A^5 \equiv B^5 \equiv 1$$
Since $5$ ia prime and the matrices are nonidentity, this is the order.
