Proving Invariance of Subsets under Group Actions

[playa007]

Homework Statement

Let G be a group acting on a set X, and let g in G. Show that a subset Y of X is invariant under the action of the subgroup <g> of G iff gY=Y. When Y is finite, show that assuming gY is a subset of Y is enough.

Homework Equations

If Y is a subset of X, we write GY for the set { g·y : y Y and g G}. We call the subset Y invariant under G if GY = Y (which is equivalent to GY ⊆ Y)

The Attempt at a Solution

gY=Y implies that gy is in Y for all g in G and y in Y. G is a group so all powers of G would be in G as well. so <g>Y=Y must be true also and Y is invariant under action of <g>. If Y was invariant under the action of <g> but gY=/=Y. This would mean that there exists some y' in Y such that gy =/= y' so the set <g>Y would not contain the element y' and <g>Y would not equal to Y which contradicts that Y is invariant. I'm also wondering how to proceed for finite case. Any help would be much appreciated; thanks.

 

[Dick]

I really don't even understand your proof of the first part. Can you work on it? But here's what you should be thinking about. Take Z the set of integers. Take G to be the actions on Z defined by g_a(z)=z+a for all a in Z. G is a group, right? Now take Y to be the positive integers. g_1(Y) is contained in Y, right again? Y IS NOT invariant under <g_1>. Why not? But then Y is infinite. Why can't this happen if you have a group action leaving a FINITE set invariant?