Characterization of parity in the left regular representation of a finite group

Let G be a finite group and let \pi : G \rightarrow S_G be the left regular representation of G. Prove that if x \in G is an element of order n and if |G| = nm, then \pi(x) is a product of m disjoint n-cycles. Deduce that \pi(x) is an odd permutation if and only if |x| is even and |G|/|x| is odd.


This action of G is faithful, so that the induced action of H = \langle x \rangle on G is faithful. Note that under this action, for all g \in G we have \mathsf{stab}_H(g) = 1, so that [H : \mathsf{stab}_H(g)] = n. Thus each H-orbit of G has order n; the H-orbit of an element g \in G is precisely the cycle containing g in the decomposition of \pi(x) since H is cyclic with generator x. Because |G| = mn, there are m distinct orbits. Thus \pi(x) is a product of m disjoint n-cycles.

Clearly if |x| is even and |G|/|x| is odd, then \pi(x) is an odd permutation by Proposition 25.

Now if \pi(x) is an odd permutation and |x| is odd, then the number of cycles in \pi(x) of even length is zero, which is even, so that \pi(x) is an even permutation- a contradiction. Thus |x| is even. Now by Proposition 25, |G|/|x| must be odd.

Post a comment or leave a trackback: Trackback URL.

Comments

  • Gobi Ree  On December 7, 2011 at 9:23 pm

    Faithfulness does not imply that the stabilizer of any point is 1. So the reasoning in the first paragraph is a little misleading. It will be better to say that:

    This action of G satisfies that the stabilizer of any point is the identity subgroup, so does the induced action.

    • nbloomf  On December 9, 2011 at 1:43 am

      Thank you; that was poorly worded.

  • Gobi Ree  On December 7, 2011 at 9:33 pm

    typo: \mathsf{stab}_H(a)\mathsf{stab}_H(g)
    the cycle containing G – the cycle containing g

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: