Claim: Non-trivial semidirect product $(\mathbb Z_2 \oplus \mathbb Z_2 \oplus\mathbb Z_2) \rtimes_\varphi \mathbb Z_3 \cong A_4 \oplus \mathbb Z_2$.
I'm classifying groups of order $24$, and this is the case when $\mathbb Z_2 \oplus \mathbb Z_2 \oplus\mathbb Z_2$ is the Sylow-$2$ subgroup and $\mathbb Z_3$ acts non-trivially on it, which yields a homomorphism $\varphi: \mathbb Z_3 \to \text{Aut}(\mathbb Z_2 \oplus \mathbb Z_2 \oplus\mathbb Z_2) = \text{GL}_3(\mathbb F_2)$.
Let $A = \varphi(\bar{1})$.
It is of order $3$ in $\text{GL}_3(\mathbb F_2)$ with minimal polynomial $x^2+x+1=0$ (wrong. see the answer by Derek Holt).
Some suggest that $A$ can be quasi-diagonalized to $\left(\begin{smallmatrix} 1 & 0 & 0 \\0 & 1 & 1 \\ 0 & 0 & 1\end{smallmatrix}\right)$, so for non-trivial $\varphi$, we have
$(\mathbb Z_2 \oplus \mathbb Z_2 \oplus\mathbb Z_2) \rtimes_\varphi \mathbb Z_3 \cong ((\mathbb Z_2 \oplus \mathbb Z_2) \rtimes \mathbb Z_3) \oplus \mathbb Z_2 \cong A_4 \oplus \mathbb Z_2$.
Such diagonalization method works well for groups of order $18$.
However, Jordan normal form only works in algebraically closed field, and $\mathbb F_2$ is not algebraically closed.
Especially, $x^2+x+1=0$ has no root in $\mathbb F_2$.
So is this diagonalization method correct?
And if not, How can we prove the claim rigorously?
Thanks for your time and effort.
Best Answer
You wrote: it is of order 3 in ${\rm GL}_3(\mathbb F_2)$ with minimal polynomial $x^2+x+1$, but that is wrong.
You know only that its minimal polynomial divides $x^3-1 = (x-1)(x^2+x+1)$.
Since we are assuming that the action is non-trivial, the minimal polynomial cannot be $x-1$. If it was $x^2+x+1$ then, since this is irreducible over ${\mathbb F}_2$, the matrix would be similar to a sum of $2 \times 2$ blocks. But that would imply that the dimension was even, which it is not.
So the minimal polynomial must be $x^3-1$, and the matrix is the sum of a 1-dimensional block, the identity, and a $2 \times 2$-block, which you can take to be the companion matrix of the polynomial. You are not using the Jordan Canonical Form Theorem here.