If $H$ is a subgroup of $S_n$ and if H is not contained in $A_n$ prove that precisely one half of the elements of H are even permutations.

Question

If $H$ is a subgroup of $S_n$ and if H is not contained in $A_n$ prove that precisely one half of the elements of H are even permutations. I know that multiplying two odd permutations gives a even, and that one even and one odd return an odd permutation but i have no idea where to go from there.

Answer 1

Hint:

Let $\;\sigma\in H\;$ be an odd permutation, and now define

$$\;f:A_N\cap H\to H\setminus(A_n\cap H)\;,\;\;f(x):=\sigma x\;$$

Prove $\;f\;$ is a bijection.

Answer 2

Every subgroup of $S_n$ contains an even permutation (which one?). Then the hypothesis says that $H$ contains an odd permutation also. Then consider the map $$\sigma:H\rightarrow \{1,-1\}, h\mapsto \mbox{sign}(h).$$ Can you complete proof?