Prove that if G is a finite p-group acting on a finite set S with p not dividing |S|, then G has at least one orbit which contains only one element.

I am completely lost. Please help.

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- April 13th 2011, 08:14 AMpage929Prove G has at least one orbit
Prove that if G is a finite p-group acting on a finite set S with p not dividing |S|, then G has at least one orbit which contains only one element.

I am completely lost. Please help. - April 13th 2011, 08:31 AMSwlabr
- April 13th 2011, 09:48 AMDeveno
to amplify, consider a fixed element x of a set S that G acts on. if we set Gx =

{a in G: a(x) = x}, the stabilizer of x, we can form a map φ from G/Gx (the set of left-cosets of Gx) to the orbit of x by:

φ(bGx) = b(x) (if a is in the stablizer Gx, ba(x) = b(a(x)) = b(x), since the elements of Gx don't do anything to x).

now suppose b' is in bGx, so b' = ba, where a is in Gx. then b'(x) = ba(x) = b(x), so this map is well-defined.

suppose that two cosets of Gx, give us the same result: φ(bGx) = φ(cBx), so b(x) = c(x).

then bc^-1(x) = cc^-1(x) = e(x) = x, so bc^-1 is in Gx, so bGx = cGx, the two cosets are equal. this means φ is injective.

but if y is in the orbit of x, this means that for SOME g in G, g(x) = y, in which case φ(gGx) = g(x) = y,

so φ is surjective as well. so if G is finite, then the size of the orbit of x is the number of cosets of Gx, which is the index [G:Gx].

now Gx is a subgroup of G, so |Gx| divides |G|, and since |G| = |Gx|*[G:Gx], [G:Gx] divides |G| as well.

what does this all mean if |G| = p^k?

1) possibility A- [G:Gx] = 1. this happens if all of G fixes x, which is what we hope might happen.

2) possibility B- [G:Gx] = p^r, this happens if the orbit has more than one element.

(stuff for you to prove in the middle: two orbits are either the same, or have no elements in common).

so divide S into all its orbits. we have |S| = k + p(other stuff). we want to prove k isn't 0.

suppose it was. then 0 = |S| - p(other stuff). since p divides 0, and p divides p(other stuff), then wouldn't p divide |S|?