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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.

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  • $\begingroup$ This might help. $\endgroup$
    – Shaun
    Commented Aug 14, 2020 at 10:43
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    $\begingroup$ $A$ has order three. There are seven non-zero elements in $\Bbb{Z}_2^3$, so $A$ must have a fixed point. That gives you the first column (w.r.t. to a family of bases). If you have seen Maschke's theorem (from rep theory), (one of) its proof (by averaging argument) implies that $A$ preserves "an inner product", so $A$ also acts on a carefully chosen complementary 2-dimensional subspace. That has three non-zero vectors and we are basically done. $\endgroup$ Commented Aug 14, 2020 at 15:40
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    $\begingroup$ Anyway, that fixed point gives you the central factor, and basically you are left with $\Bbb{Z}_2^2\rtimes\Bbb{Z}_3$. I'm sure you can show that must be isomorphic to $A_4$. $\endgroup$ Commented Aug 14, 2020 at 15:46

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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.

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  • $\begingroup$ Thank you for your helping me so many times. I still don't understand this: "if the minimal polynomial matrix is $x^2+x+1$, since this is irreducible over $\mathbb F_2$, the matrix would be similar to a sum of $2×2$ blocks". Is there a general theorem or theory for matrix on a field $F$ where the characteristic polynomial of the matrix doesn't split on $F$? Thanks. $\endgroup$
    – Andrews
    Commented Aug 15, 2020 at 11:03
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    $\begingroup$ Yes, try searching for rational canonical form. (I learnt that before the Jordan canonical form.) $\endgroup$
    – Derek Holt
    Commented Aug 15, 2020 at 11:54
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    $\begingroup$ An alternative approach is to use Maschke's theorem in representation theory to deduce that an ${\mathbb F}_2H$-module, with $H=C_3$, is a direct sum of irreducible modules, and that there are two such irreducibles, of dimensions 1 and 2. $\endgroup$
    – Derek Holt
    Commented Aug 15, 2020 at 12:00

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