Here is a recipe for finding all subgroups of Sn that are isomorphic to a fixed and completely understood group G, like G = S3.
First, find all conjugacy classes of subgroups H of G, and label them by their index $[G:H] = |G| / |H|$. For example, for G = S3:
- "1" is the subgroup { 1, (123), (132), (12), (13), (23) }
- "2" is the subgroup { 1, (123), (132) }
- "3" is the subgroup { 1, (12) }, or one of its conjugates: { 1, (13) } or { 1, (23) }
- "6" is the subgroup { 1 }
Each of these defines a way for G to act (multiplication on the cosets of H in G). For instance, for G = S3:
- "1" is the action on 1 point { A }, where every element of G leaves A alone
- "2" is the action on 2 points { B, C }, where the 2-cycles { (12), (13), (23) } all switch B and C, but everyone else { 1, (123), (132) } leaves A and B alone
- "3" is the action on 3 points { 1, 2, 3 }, where everybody does the natural thing
- "6" is the action on 6 points { E = 1, F = (123), G = (132), H = (12), I = (13), J = (23) }, where everybody acts as multiplication
Now write n as an unordered sum of the indices. For instance, if n = 4, then there is only one way:
- "4 = 3 + 1" acts on 1,2,3, and A=4. The combined action is { 1, (123), (132), (12), (13), (23) }, where nobody moves 4.
A better example is n = 5, where there are two ways:
- "5 = 3 + 1 + 1" acts on 1,2,3 and leaves 4 and 5 alone: { 1, (123), (132), (12), (13), (23) }
- "5 = 3 + 2" acts on 1,2,3 and B=4, C=5: { 1, (123), (132), (12)(45), (13)(45), (23)(45) }
Maybe even better is n = 6, where there are four ways:
- "6 = 3 + 1 + 1 +1" is { 1, (123), (132), (12), (13), (23) }
- "6 = 3 + 2 + 1" is { 1, (123), (132), (12)(45), (13)(45), (23)(45) }
- "6 = 3 + 3" needs {4,5,6} to be a second copy of {1,2,3}, and so it is: { 1, (123)(456), (132)(465), (12)(45), (13)(46), (23)(56) }
- "6 = 6" is just { 1, (123)(456), (132)(465), (14)(26)(35), (15)(24)(36), (16)(25)(34) } = { 1, (EFG)(HIJ), (EGF)(HJI), (EH)(FI)(GJ), etc.}
In each case, we are writing a general action of G as a sum of simple "transitive" actions of G on the cosets of a subgroup H. This is called the orbit stabilizer theorem; your course should mention this at least superficially. The idea of literally adding them up is one way to view "permutation characters" in character theory; that will be in a later course, most likely.
This only works if (a) you understand G very well, and (b) the "big" group is Sn. A similar thing works if the "big" group is a general linear group, but then the H are replaced by modules and this is called representation theory of finite groups.
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
Just a “visual” construction of the isomorphism between $\mathfrak S(4)/V_4$ and $\mathfrak S(3)$...
It's quite well known that $4 = 2 +2$. Concretely, it means that if you have four things, you can always group them into two pairs. What's remarkable is that there are exactly three such groupings: $$ \{1, 2, 3, 4\} = \{1, 2\} \sqcup \{3,4\} = \{1, 3\} \sqcup \{2,4\} = \{1, 4\} \sqcup \{2,3\}.$$
If I call these three groupings $\mathbf{Order}$, $\mathbf{Parity}$ and $\mathbf{5sum}$, every permutation of $\{1,2,3,4\}$ induces a permutation of these three things. For example, the $4$-cycle $(1234)$ induces the transposition $(\mathbf{Order}\ \mathbf{5sum})$. So I get a homomorphism $$ \mathfrak S(4) \to \mathfrak S(\{\mathbf{Order}, \mathbf{Parity}, \mathbf{5sum}\})\simeq \mathfrak S(3).$$ It's quite easy to show that this morphism is surjective (we already found the transposition $(\mathbf{Order}\ \mathbf{5sum})$ in its image, it's not hard to find another transposition or a $3$-cycle, and that will be enough to generate the whole of $\mathfrak S(3)$.)
Let's give a proof that the kernel is exactly $V_4$: let $\sigma$ be a nontrivial permutation in it. So for example, there are two elements $a \neq b$ such that $\sigma(a) = b$ (let's call $c$ and $d$ the two remaining elements) In that case, because $\sigma$ preserves the grouping $\{a,b\} \sqcup \{c,d\}$, it must send $b$ back to $a$. And because it preserves $\{a,c\} \sqcup \{b,d\}$ and it exchanges $a$ and $b$, it must exchange $c$ and $d$ as well. So $\sigma = (a\, b)(c\, d) \in V_4$.
Therefore, the factorisation theorem gives you an isomorphism $\mathfrak S(4)/V_4 \to \mathfrak S(3)$.
Of course, this proof is not the shortest (well, it certainly isn't the shortest to write, but if you're allowed to make a lot of pictures or to play with four actual tokens, essentially all the arguments become self-evident). But it really makes the isomorphism concrete. There's an isomorphism between $\mathfrak S(4)/V_4$ and $\mathfrak S(3)$ because $4$ equals $2+2$ in $3$ different ways...
A final remark: if $n \neq 4$, the only proper normal subgroup of $\mathfrak S(n)$ is $\mathfrak A(n)$. That means that the situation I described here is pretty exceptional. If you want $\mathfrak S(n)$ to act on fewer than $n$ objects, the only interesting actions are :
I may be overenthusiastic, but this simple remark gives much cachet to this innocent-looking isomorphism. (Another peculiarity of the same kind is that the only interesting action of $\mathfrak S(n)$ on $n+1$ tokens is an action of $\mathfrak S(5)$ on 6 tokens, coming from another exceptional behaviour of the symmetric group.)