In order for a group $G$ to be isomorphic to a semidirect product $G\cong H\rtimes K$, you must have:
- A subgroup $\mathcal{H}$ of $G$ that is isomorphic to $H$;
- A subgroup $\mathcal{K}$ of $G$ that is isomorphic to $K$;
- Normality of $\mathcal{H}$, $\mathcal{H}\triangleleft G$;
- Trivial intersection of $\mathcal{H}$ and $\mathcal{K}$; and
- $G = \mathcal{H}\mathcal{K}$.
In order to show that $S_4$ is isomorphic to $V_4\rtimes S_3$, you need, inter alia, to produce a subgroup of $S_4$ that is isomorphic to $S_3$; formally, $S_3$ is not a subgroup of $S_4$, because the elements of $S_4$ are bijections from $\{1,2,3,4\}$ to itself, while the elements of $S_3$ are bijections of $\{1,2,3\}$ to itself. The map $j$ is used to find a subgroup $\mathcal{K}$ that is isomorphic to $S_3$, but is inside $S_4$.
First of all let me point out that $D_{24}$ cannot be isomorphic to $S_4$. $S_4$ does not have an element of order 12, where as $D_{24}$ does. This is because $D_{24}$ as a set is
$$\{1,r,r^2 ,\ldots r^{11}, sr, sr^2, \ldots sr^{11}\}$$
where the $r_i's$ are the rotations and the $sr^j$ the reflections; so that $r$ is your element of order 12. For the sake of knowledge though $S_4$ does have a subgroup of order 12 sitting inside it that is just $A_4$. Just to tell you though the subgroups of $S_4$ are (up to isomorphism):
$$\{e\}, \hspace{2mm} C_2, \hspace{2mm} C_3, \hspace{2mm} V_4, \hspace{2mm} C_4, \hspace{2mm} S_3, \hspace{2mm} D_8, \hspace{2mm} A_4, \hspace{2mm} S_4.$$
Actually it is not so hard to see directly that $S_3$ and $D_6$ are isomorphic, just list down the elements explicitly and if they obey exactly the same relations, their Cayley Tables are the same so you have your isomorphism.
By the way I think for your last line you want a surjective homomorphism from $S_4$ to $S_3$ to show that $S_3$ is solvable only if $S_4$ is. The way you do that is like this:
We get a homomorphism $\varphi$ from $S_4$ to $S_3$ by considering the action of $S_4$ on a finite set $Y = \{\Pi_1, \Pi_2, \Pi_3\}$ where $\Pi_1 = \left\{\{1,2\}, \{3,4\}\right\}$, $\Pi_2 = \left\{\{1,3\}, \{2,4\}\right\}$ and $\Pi_3 =\left\{ \{1,4\}, \{2,3\}\right\}$. Each $\Pi_i$ is one way of partitioning a set of 4 elements into two subsets of equal cardinality. $S_4$ acts on a $\Pi_i$ by permuting the numbers in each partition, so for example the cycle $(1234)$ interchanges $\Pi_1$ and $\Pi_3$ while keeping $\Pi_2$ fixed, so that $\varphi( (1234) ) = (13)(2)$.
It is easy to see that $\ker \varphi$ is the Klein four- group $V_4= \{e, (12)(34), (14)(23), (13)(24)\}$ where $e$ denotes the identity in $S_4$. If you go further you should know that by the first isomorphism theorem the quotient group $S_4/V_4$ is isomorphic to the image of $\varphi$.
Now to see the surjectivity of $\varphi$, recall we have the counting formula
$|S_4| = |\ker \varphi||\operatorname{Im} \varphi|$. Therefore the order of the image is $24/4 = 6$. Now recall the image is a subgroup of $S_3$ so if we have a subgroup of order 6 sitting inside $S_3$, it must be the whole of $S_3$.
To answer your question on why $S_4$ is solvable, note that the only normal subgroups in $S_4$ are the trivial group, the Klein 4-group $V_4$, $A_4$ and $S_4$ itself.
So we have a subnormal series
$$\{e\} \triangleleft V_4 \triangleleft A_4 \triangleleft S_4.$$
Now you should know that $V_4/\{e\} \cong V_4$ that is abelian (all groups of order 4 are). You also know that $|A_4/V_4| = 3$ so that $A_4/V_4 \cong C_3$ which is abelian. Finally $|S_4/A_4| = 2$ so that $S_4/A_4 \cong C_2$ that is also abelian. So it remains to check that $A_4$ is normal in $S_4$ and $V_4$ is normal in $A_4$. The former you already know for the index of $A_4$ in $S_4$ is two. For the second I suggest listing out the conjugacy classes of $A_4$ and show that $V_4$ is a union of conjugacy classes. It should not be so hard to compute all the conjugacy classes of $A_4$ since there are only 12 elements. Once you've done all of this, you've proved that $S_4$ is solvable!
Best Answer
I'm guessing that by $\,Z_4\,$ you actually mean the Klein viergrupp
$$C_2\times C_2\cong \{(1)\,,\,(12)(34)\,,\,(13)(24)\,,\,(14)(23)\}$$
since the above one is the only normal subgroup of order $\,4\,$ of $\,S_4\,$ , but then, since the cyclic group of order $\,6\,$ is obviously abelian ,we get
$$S_4/Z_4\cong C_6\Longleftrightarrow S_4^{'}\leq Z_4$$
and this is false since $\,(123)\notin Z_4\,$ , but $\,(123)\in A_4=S_4^{'}\,$.
Thus, as the only other group of order $\,6\,$, up to isomorphism, is $\,S_3\,$ ,we get that $\,S_4/Z_4\cong S_3\,$