Subgroup of a symmetric group Sn

Homework Statement

Show that if G is a subgroup of a symmetric group Sn, then either every element of G is an even permutation or else exactly half the elements of G are even permutations.

The Attempt at a Solution

We have a hint for the problem. If all the elements of G are even, then there is nothing to prove. That is clear because even times even will yield and even number of permutations so there is closure under multiplication.
And if you have even transposition the inverse of it will be in the set as well so clearly if we have an even number of permutations its a subgroup.

If it is not we are to let e, o(1), 0(2),...o(k-1) be the even elements of G..

I am stuck so i looked at S3 a subgroup of S9. S3 has 6 elements.
(1 3 2) (1 2 3) and e are even
(1 2) (1 3) and (2 3) are odd
So exactly half like the theorem states but i am stuck.

So, let G be a subgroup of Sn. Let Godd be the subset of G consisting of all odd permutations, and let Geven be the subset of G consisting of all even permutation.

Try to show first that Geven is a subgroup of G (this is somewhat analogous with what you've shown so far).

There are two possibilities:
1) Godd is empty: then G consists only out of even permutation, this is what we wanted to show.

2) There exists an $$\sigma\in G_{odd}$$. Now try to show that $$G_{odd}=\{\sigma\tau~\vert~\tau \in G_{even}\}$$. Or in more highbrow terminology: Geven acts on Godd, and the orbit of an element in Godd is entire Godd.

Dick