Answer to Question #178275 in Abstract Algebra for 123

Question

Prove that for every subgroup H of S_n , either all of the permutations in H are even, or exactly half are even.

Answer 1

Suppose that there is at least one odd permutation $\tau$ (if there's no odd permutations, we conclude that all permutations are even). Then, for every even permutation $\alpha \in H$ we obtain distinct odd permutations $\alpha \cdot \tau \in H$ (as if $\alpha_1 \cdot \tau = \alpha_2\cdot \tau$, then by multiplying by $\tau^{-1}$ on the right we get that $\alpha_1=\alpha_2$) and therefore we conclude that there is at least as many odd permutations as even permutations. However, for every odd permutation $\beta$ (including the case $\beta=\tau$) we can apply the same argument: $\beta \cdot \tau$ are distinct even permutation. Therefore there is at least as many even permutations odd permutations. By double inequality we conclude that there the same quantity of even and odd permutations, i.e., exactly half of permutations are even.