If the left regular representation of a finite group G contains an odd permutation, then G has a subgroup of index 2

Let G be a finite group and \pi : G \rightarrow S_G the left regular representation. Prove that if \pi[G] contains an odd permutation, then G has a subgroup of index 2.


If \pi[G] \leq S_G contains an odd permutation, then \pi[G] \not\leq A_G. Now A_G \vartriangleleft S_G has index 2, which is prime. By this previous exercise, we have [ \pi[G] : A_G \cap \pi[G] ] = 2. Since the action of G on itself by left multiplication is faithful, G \leq S_G can be identified with \pi[G]. Thus G has a subgroup of index 2; namely the preimage of A_G \cap \pi[G].

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: