Let $H\le S_4$ be a subgroup such that $H \not\subset A_4$. Show that $|H \cap A_4|=|H\cap (12)A_4|$. I would like to have a hint for the cinstruction of the bijection. Here is my attempt:
Each element of $A_4$ is of sign $1$. If we look at $(12)A_4$, then clearly sign of it's elements is $-1$ and $|(12)A_4|=12$. We observe first that $A_4\cap (12)A_4=\emptyset$ and that $S_4=A_4\sqcup (12)A_4$.
Then, as $H\not\subset A_4$, clearly $H\cap(12)A_4 \neq \emptyset$ (probably $H\subseteq (12)A_4 (?) )$.
Now, I have to show that $|H\cap A_4|=|H\cap (12)A_4|$. To show it's order equality, I have to find a bijection between $H\cap A_4$ and $H\cap (12)A_4$ (or injection in 2 different "directions" and apply Shroeder Bernstein's theorem). But, I dont't really see how to define it properly as $H\cap A_4 =\emptyset$ or it has an intersection. If someone could give a hint, I would really appreciate it. Thank you in advance for help!
$A_4$ is a normal subgroup of $S_4$ with index 2. Proof this for $A_4\cap H\subseteq H$.