Can anyone help me with this problem please... thank you
Show that for every subgroup H of Sn for n ≥ 2, either all the permutations in H are even or exactly half of them are even.
I guess this approach might work:
Let $\displaystyle X$ be the set of all even permutations of $\displaystyle H$. Now if $\displaystyle X = H$ the proof is complete. Otherwise there exists a non-empty set of odd permutations $\displaystyle Y$.
Now you define a one-to-one onto map $\displaystyle \phi : X \mapsto Y$.
Define $\displaystyle \phi (\sigma) = (1,2)\sigma$ for $\displaystyle \sigma \in X$.
This shows that $\displaystyle |X| = |Y|$ thus there are exactly one half even permutations and one half odd permutations.