Subgroup of a symmetric group Sn

[leonamccauley]

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.

Homework Equations

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.

 

[micromass]

So, let G be a subgroup of S_n. Let G_odd be the subset of G consisting of all odd permutations, and let G_even be the subset of G consisting of all even permutation.

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

There are two possibilities:
1) G_odd 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: G_even acts on G_odd, and the orbit of an element in G_odd is entire G_odd.

 

[Dick]

Or maybe just note that if o is an odd member of G and G is a subgroup then oG=G. How does multiplication by o affect the oddness and evenness of members of G?