Re: Normalizers, Centralizers and orbits

mareg_at_mimosa.csv.warwick.ac.uk
Date: 11/20/04 

Date: Sat, 20 Nov 2004 14:35:48 +0000 (UTC)

In article <20041120033840.06323.00000583@mb-m26.aol.com>,
        warren065@aol.com (Warren065) writes:
>I have the following problem:
>
>Let P be a subgroup of S_n (the symmetric group) where P is of prime order and
>suppose x belongs to S_n normalizes but DOES NOT centralize P. Show that x
>fixes at most one point in each orbit of P.
>
>[My proof is more of a wordy explanation. I just can't think of how to prove
>this explicitly]
>
>Proof: Let P be a subgroup of S_n of prime order. So P=(1,.....,p). Suppose x
>belongs to S_n fixes 1. Now if x fixes any other point in (1,....,p) and
>normalizes (1,2,....,p) then it fixes all points 1,2,....,p. Then x would be
>an element of the centralizer. So if x fixes 2 points in an orbit of P, then
>it centralizes P. Therefore x fixes at most one piont in each orbit of P.

You need a bit more detail than that. Let P = < g >. Let {1,...,p} be
an orbit of P where g = (1,2,...,p) ..., and suppose x normalizes P
and fixes the two points 1 and t of {1,...,p}. Let h = g^(t-1),
so h(1) = t. Then x h x^-1 (1) = x h (1) = x(t) = t. So x h x^-1 is
a power of h that maps 1 to t, and hence x h x^-1 = h, and x centralizes
h and hence also g, because P = < h > = < g >.

Derek Holt.