External semi direct product Isomorphism excercise 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
 A: 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).

$^\dagger$Candidate isomorphisms must preserve elements' order.
A: 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 $S_3 = A_3H$.
