<s>[INST] Element of order $7$ in $GL(4,2)$ Find an element of order $7$ in $GL(4,2)$, the group of all invertible $4 \times 4$ matrices with entries in $\mathbb{F}_2$. 
I'd like a more a constructive way to find the required element in $GL(4,2)$ in lieu of explicity going through elements of $GL(4,2)$ and computing their orders aimlessly. 
First, I know that if $A$ is the desired element, we know $A$ satisfies $A^7 = I$ $\Rightarrow$ $A$ satisfies the polynomial $p(x) = x^7 - 1$ $\Rightarrow$ $p(x) = x^7-1$ is an annihilating polynomial for $A$. This means that the minimal polynomial of $A$ must divide $p(x) = x^7 - 1$. Over $\mathbb{F}_2$, I found that this polynomial factors completely as $p(x) = (x+1)(x^3 + x^2 + 1)(x^3 + x + 1)$. 
Now, the minimal polynomial of $A$ cannot equal $p(x)$, since $p(x)$ has degree greater than $4$. Similarly, the minimal polynomial of $A$ cannot involve both factors $(x^3 + x^2 + 1)$ and $(x^3 + x + 1)$. Furthermore, the minimal polynomial cannot simply be $(x+1)$ alone, as this would mean that $A$ is equal to the identity matrix multiplied by $-1$, which has order $2$. Thus, for degree reasons, it must be that the minimal polynomial is either $m_A(x) = (x+1)(x^3 + x^2 + 1)$ or $m_A(x) = (x+1)(x^3 + x + 1)$. In either case, the minimal polynomial coincides with the characteristic polynomial, since the characteristic polynomial must have degree $4$. 
Thus, the characteristic polynomial of $A$ is either $p_A(x)  = (x+1)(x^3 + x + 1)$ or $p_A(x) = (x+1)(x^3 + x^2 + 1)$. Then, once can build the companion matrices corresponding to each of these characteristic polynomials, and check if either of them has order $7$. It seems, a prior, one of them must be of order $7$. 
However, neither of the corresponding companion matrices is working for me as a solution. Where did I make a mis-step in my logic ? Is there an even better method to do this ? 
Thanks! 
 [/INST] If $\lambda\in\mathbb{F}_{2^3}$ is such that $\lambda^3=\lambda+1 $ then $\lambda^6=\lambda^2+1 $ and $\lambda^7 = \lambda^3+\lambda = 1 $, so the companion matrix of $x^3-x-1$ (as an element of $\text{GL}(3,\mathbb{F}_2)$) is a $3\times 3$ matrix with order $7$, which can be easily completed to a $4\times 4$ matrix with order $7$. Given $M=\left(\begin{smallmatrix}0&0&1\\1&0&1\\0&1&0\end{smallmatrix}\right)$ we have
$M^2=\left(\begin{smallmatrix}0&0&1\\1&0&1\\0&1&0\end{smallmatrix}\right)\quad$ $M^3=\left(\begin{smallmatrix}1&0&1\\1&1&1\\0&1&1\end{smallmatrix}\right)\quad$
$M^4=\left(\begin{smallmatrix}0&1&1\\1&1&2\\1&1&1\end{smallmatrix}\right)\quad$
$M^5=\left(\begin{smallmatrix}1&1&1\\1&2&2\\1&1&2\end{smallmatrix}\right)\quad$ $M^6=\left(\begin{smallmatrix}1&1&2\\2&2&3\\1&2&2\end{smallmatrix}\right)\quad$
$M^7=\left(\begin{smallmatrix}1&2&2\\2&3&4\\2&2&3\end{smallmatrix}\right)\equiv I\pmod{2}.$
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