I'm trying to prove that any product of two transpositions is in a normal subgroup $H$ of $S_n$.
I know that a normal subgroup of $S_n$ has index 2 and hence contains no transpositions. And but I don't think I understand what it means by any product of two transpositions is in $H$ when $H$ contains no transpositions. I thought of it as any permutations can be written as a product of transpositions but I am quite lost.
This proof is by @ViníciusNovelli and is given in the comments.
Since $A_n\lhd S_n$ and $\operatorname{sgn}:S_n\to \{1,-1\}$ with $\ker(\operatorname{sgn})=A_n$, the sign of two transpositions is $1$, so their product must be in $A_n$. So let $H=A_n$.