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?

  • $\begingroup$ It might help to realize $S_3$ as $\Bbb Z_3\rtimes\Bbb Z_2$ so that you're left with showing $\Bbb Z_3\rtimes(\Bbb Z_2 \times \Bbb Z_2) \cong (\Bbb Z_3\rtimes\Bbb Z_2)\times\Bbb Z_2$. $\endgroup$ Nov 15, 2019 at 14:16
  • $\begingroup$ Is there a good way to show that? $\endgroup$
    – Vasting
    Nov 18, 2019 at 21:53
  • $\begingroup$ This question might help. $\endgroup$
    – Andrews
    Nov 28, 2019 at 17:24

3 Answers 3


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. $(12)(123)(12) \neq (123)$.

  • $\begingroup$ (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.) $\endgroup$
    – hunter
    Nov 28, 2019 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 $(y_1,1)$.

  • $\begingroup$ $(y_1,0 )$ 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 $(y_1,0)$? $\endgroup$ Nov 12, 2019 at 4:44

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