Order $4$ subgroup of alternating group $A_4$

Question

I ran into the following problem:

Let $H$ be the subgroup $H = \{e, (1\,2)(3\,4), (1\,3)(2\,4), (1\,4)(2\,3)\}$ in $G = A_4 = H \cup \{(1\, 2\, 3), (1\, 3\, 2), (1\, 2\, 4), (1\,4\,2), (1\,3\,4),(1\,4\,3),(2\,3\,4),(2\,4\,3)\}$. Show that $H$ is normal.

Part of the solution mentions that "since every element of $G-H$ has order $3$, none of these elements can be contained in a subgroup of order $4$." I don't understand why this is the case.

Answer 1

If H was not normal then for some h$\in$H and some g$\in$(G-H), g$^{-1}$hg must end in G-H and have order 3 which will imply that,

g$^{-1}$h$^{3}$g = e i.e h$^{3}$g=g i.e. h$^{3}$=e contradiction to the Lagrange Theorem that h being an element of H of order '4' can't have order 3.