External semi direct product Isomorphism excercise

Question

Would anyone please explain how to show $$\Bbb Z_3 \rtimes _\vartheta \Bbb Z_2\cong S_3?$$

I understand that $\Bbb Z_2 \cong{\rm Aut}\Bbb Z_3$, but how can it help me construct an isomorphism to $S_3$?

My train of thought is finding a normal subgroup $$H \trianglelefteq S_3 $$ such that $H \cong\Bbb Z_3$, and to find another subgroup $K \le S_3$ which is $K \cong \Bbb Z_2$ and $K \cap H = {e}$ , and $KH = S_3$

but I have no methodical way to actually find these subgroups.

thanks

Answer 1

Hint: Notice that $A_3 \simeq \mathbb{Z}_3$ and $A_3 \triangleleft S_3$. Now you can proceed in two ways. Either explicitly find a subgroup $H$ of order 2 (the embedding of $S_2$ for example) or just consider a Sylow $2$-subgroup. Now use the fact that $|HK| = \frac{|H||K|}{|H\cap K|}$ for subgroups $H$ and $K$ to show that $S_3 = A_3H$.

Answer 2

Your sought $H$ and possible $K$'s pop up quite naturally from the following, I think.

By definition: $$\Bbb Z_3 \rtimes_\vartheta \Bbb Z_2:=(\Bbb Z_3\times\Bbb Z_2,\circ_\vartheta)$$ where, for every $(n_1,h_1),(n_2,h_2)\in \Bbb Z_3\times\Bbb Z_2$: $$(n_1,h_1)\circ_\vartheta(n_2,h_2):=(n_1+\vartheta_{h_1}(n_2),h_1+h_2)$$ The only nontrivial action of $\Bbb Z_2$ on $\Bbb Z_3$ is the following: \begin{alignat*}{1} \vartheta:\Bbb Z_2&\longrightarrow& {\rm{Aut}(\Bbb Z_3)} \\ 0&\longmapsto& (i\longmapsto i) \\ 1&\longmapsto& (i\longmapsto -i) \\ \end{alignat*} whence: $$(n_1,0)\circ_\vartheta(n_2,h_2)=(n_1+n_2,h_2) \tag 1$$ and $$(n_1,1)\circ_\vartheta(n_2,h_2)=(n_1-n_2,1+h_2) \tag 2$$ From $(1)$ and $(2)$ follow: $$(n_1,0)^2=(2n_1,0)\stackrel{(\text{for}\space n_1= 1,2)}{\ne}(0,0), \space\space\space (n_1,0)^3=(0,0), \space\space\space (n_1,1)^2=(0,0)$$ Therefore, $\Bbb Z_3 \rtimes_\vartheta \Bbb Z_2$ has two elements of order $3$ (namely $(1,0)$ and $(2,0)$) and three elements of order $2$ (namely $(0,1)$, $(1,1)$ and $(2,1)$), just like $S_3$ has. Now, define $\varphi\colon \Bbb Z_3 \rtimes_\vartheta \Bbb Z_2\longrightarrow S_3$ by$^\dagger$: \begin{alignat}{1} &(0,0)\longmapsto () \\ &(1,0)\longmapsto (123) \\ &(2,0)\longmapsto (132) \\ &(0,1)\longmapsto (12) \\ &(1,1)\longmapsto (13) \\ &(2,1)\longmapsto (23) \\ \end{alignat}

You can prove that this map is indeed operation-preserving (i.e. an isomorphism).

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$^\dagger$Candidate isomorphisms must preserve elements' order.