# Non-trivial semidirect product $(\mathbb Z_2 \oplus \mathbb Z_2 \oplus\mathbb Z_2) \rtimes_\varphi \mathbb Z_3 \cong A_4 \oplus \mathbb Z_2$

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.

• This might help. Commented Aug 14, 2020 at 10:43
• $A$ 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. Commented Aug 14, 2020 at 15:40
• 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 $A_4$. Commented Aug 14, 2020 at 15:46

## 1 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.

• 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. Commented Aug 15, 2020 at 11:03
• Yes, try searching for rational canonical form. (I learnt that before the Jordan canonical form.) Commented Aug 15, 2020 at 11:54
• 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. Commented Aug 15, 2020 at 12:00