# How to prove $\mathbb Z_3\rtimes(\mathbb Z_2\mathbb \times\mathbb Z_2) \cong S_3\times\mathbb Z_2$?

I know that there is a unique semidirect product $$\mathbb Z_3\rtimes(\mathbb Z_2\times \mathbb Z_2)$$, defined by mapping two of the order two generators of $$\mathbb Z_2\times \mathbb Z_2$$ to the inversion automorphism of $$\mathbb Z_3$$.

However, I am not exactly sure how to proceed to show that $$\mathbb Z_3\rtimes(\mathbb Z_2\mathbb \times\mathbb Z_2) \cong S_3\times\mathbb Z_2$$.

What exactly is the map I could construct?

• It might help to realize $S_3$ as $\Bbb Z_3\rtimes\Bbb Z_2$ so that you're left with showing $\Bbb Z_3\rtimes(\Bbb Z_2 \times \Bbb Z_2) \cong (\Bbb Z_3\rtimes\Bbb Z_2)\times\Bbb Z_2$. – Cameron Williams Nov 15 '19 at 14:16
• Is there a good way to show that? – Vasting Nov 18 '19 at 21:53
• This question might help. – Andrews Nov 28 '19 at 17:24

Let $$G = S_3 \times \{\pm 1\}$$ (writing the cyclic group multiplicatively).

Now map $$G \to \{\pm 1\} \times \{\pm 1\}$$ by the rule $$(\sigma, \epsilon) \mapsto (\epsilon \cdot \text{sign}(\sigma), \epsilon).$$ The kernel is $$A_3 \times \{1\}$$, so we have an exact sequence $$1 \to A_3 \to G \to \{\pm 1\} \times \{\pm 1\} \to 1,$$ and there is a splitting $$\{\pm 1\} \times \{\pm 1\} \to G$$ given by $$(-1, 1) \mapsto ((12), 1)$$ and $$(1, -1) \mapsto (\text{id}, -1)$$.

It follows that $$G$$ is some semidirect product of $$A_3$$ by $$\{\pm 1\} \times \{\pm 1\}$$ as desired (using $$A_3 = \mathbb{Z}/3\mathbb{Z}$$). We still must check that it is the non-trivial semi-direct product, i.e. that the lift $$((12), 1)$$ acts non-trivially on $$A_3$$ by conjugation. This is indeed true since e.g. $$(12)(123)(12) \neq (123)$$.

• (We can say the same without talking about exact sequences: $G$ is generated by the subgroups $A_3 \times \{ 1\}$ and $\langle (12), \{\pm 1\}$; the former is normal and the intersection is trivial, so it's a semidirect product, the conjugation action is non-trivial so it's the non-trivial semidirect product.) – hunter Nov 28 '19 at 17:19

Informally, you are looking for a central direct factor $$C_2$$. So how do you find that?

Name the generators, so write $$Z_3 \rtimes (Z_2 \times Z_2)= \langle x \rangle \rtimes (\langle a \rangle \times \langle b \rangle)$$

The semidirect product $$\rtimes_\Psi$$ you defined is given by mapping $$\Psi: a \mapsto \phi$$ and $$b \mapsto \phi$$ where $$\phi \in \operatorname{Aut}(Z_3)$$ is the map defined by $$\phi(x) = x^2$$. Observe now that $$\Psi(ab) = \phi \phi$$ and that $$\phi(\phi(x)) = \phi(x^2) = \phi(x)^2 = x^4 = x$$ that is, $$\Psi(ab)=\operatorname{Id}$$.

This means that $$ab$$ is in the kernel of $$\Psi$$. Having realized that, it is just a matter of changing generators for $$C_2 \times C_2$$, so write $$Z_3 \rtimes (Z_2 \times Z_2)= \langle x \rangle \rtimes_\Psi (\langle a \rangle \times \langle ab \rangle)$$ and now it is clear that this is the same thing as $$(\langle x \rangle \rtimes_\Psi \langle a \rangle) \times \langle ab \rangle$$

which is $$S_3 \times C_2$$ (note that $$Z_3 \rtimes Z_2 = S_3$$).

If $$G$$ is the group defined by the semidirect product you defined and $$H$$ is the direct product of $$S_3\times \mathbb Z_2$$, then all elements of $$G$$ take the form $$g=(x_1,x_2,x_3)$$ where $$x_1\in\mathbb Z_3$$ and $$x_2,x_3\in\mathbb Z_2$$. All elements of $$H$$ take the form $$h=(y_1,y_2)$$ where $$y_1\in S_3$$ and $$y_2\in\mathbb Z_2$$.

If I understand your definition correctly, then for the group $$G$$, the following should hold: $$(x_1,0,0)(a,b,c)=(x_1+a,b,c)$$ $$(x_1,1,1)(a,b,c)=(x_1+a,b+1,c+1)$$ $$(x_1,0,1)(a,b,c)=(-x_1+a,b,c+1)$$ $$(x_1,1,0)(a,b,c)=(-x_1+a,b+1,c)$$

Then you should be able to find an isomorphism between $$G$$ and $$H$$ by mapping the elements of $$G$$ that take the form $$(x_1,0,0)$$ and $$(x_1,1,1)$$ to elements of $$H$$ that take the form $$(y_1,0)$$ and elements of $$G$$ in the form $$(x_1,0,1)$$ and $$(x_1,1,0)$$ to elements of $$H$$ in the form $$(y_1,1)$$.

• $(y_1,0 )$ has elements of order 2, while neither $(x_1,0,0)$ or $(x_1,1,1)$ is order 2. Should it be that $(x_1,0,0)$ and $(x_1,0,1)$ take the form $(y_1,0)$? – Salmonella mayonnaise Nov 12 '19 at 4:44