The following may help you:
1) A subgroup of a group is normal in it iff it is a union of conjugacy classes
2) Two permutations in $\,S_n\,$ are conjugate iff they have the same cycle decomposition (i.e., the same lengths of cycles and the same ammount of cycles of each length)
3) A conjugacy class of an even permutation in $\,S_n\,$ remains exactly the same class in $\,A_n\,$ unless all the disjoint cycles in the cyclic decomposition of the permutation are of odd and different lengths, in which case the equivalence class splits in two classes in $\,A_n\,$
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 :
- for $n$ arbitrary, $\mathfrak S(n)$ acts on two tokens with the following rule: even permutations do nothing, odd ones swap the two tokens (you have to admit this action is pretty boring).
- one exceptional case: $\mathfrak S(4)$ acts on three tokens, as we just saw.
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.)
Best Answer
We can see that $(gx)H=xH$ if and only if $x^{-1}gx=x^{-1}(gx)\in H$.
In other words, $g\in G$ is contained in $K$ if and only if all of its conjugates are in $H$, i.e., the conjugacy class containing $g$ is completely covered by $H$.
As for $H$, observe that $(1~2~3~4)$ is of order $4$ and $(2~4)$ is of order $2$, where $$(1~2~3~4)(2~4)=(1~2)(3~4),\quad(2~4)(1~2~3~4)^{-1}=(2~4)(1~4~3~2)=(1~2)(3~4);$$ that is, $$(1~2~3~4)(2~4)=(2~4)(1~2~3~4)^{-1}.$$ It follows that $H\cong D_8$, the dihedral group of squares. We can then easily write the elements in $H$ by $$H=\{\operatorname{id},(1~3),(2~4),(1~2)(3~4),(1~3)(2~4),(1~4)(2~3),(1~2~3~4),(1~4~3~2)\}.$$
Consider the conjugacy classes in $S_4$ now:
The conjugacy classes totally covered by $H$ are the class of the identity map, and the class of $(a~b)(c~d)$. Therefore, $$K=\{\operatorname{id},(1~2)(3~4),(1~3)(2~4),(1~4)(2~3)\}.$$ Apparently, $K\cong V_4$, the Klein's four group.