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 $\,A_n\,$
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 class of the identity map, size $1$;
- the conjugacy class of transpositions, size $6$;
- the conjugacy class of $(a~b)(c~d)$, size $3$;
- the conjugacy class of $3$-cycles, size $8$;
- the conjugacy class of $4$-cycles, size $6$.
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.
Best Answer
The elements of the Klein $4$-group sitting inside $A_4$ are precisely the identity, and all elements of $A_4$ of the form $(ij)(k\ell)$ (the product of two disjoint transpositions).
Since conjugation in $S_n$ (and therefore in $A_n$) does not change the cycle structure, it follows that this subgroup is a union of conjugacy classes, and therefore is normal.