
Gset
Hi
1)Given G, a group, and X a Gset What relationship exists between $\displaystyle G_x $ and $\displaystyle G_y $ if x and y belong to the same orbit?
2)
a)Let G be a group and X a Gset.Prove that there is a homomorphism
$\displaystyle p: G  Biy (X) $ determined by the action.
b)Now, given a homomorphism
$\displaystyle p: G  Biy (X) $, show that X is a Gset
I hope some help. thanks

Re: Gset
1) if x,y belong to the same orbit, then there is some g in G with g.x = y. thus g^{1}.y = g^{1}.(g.x) = (g^{1}g).x = e.x = x
now suppose that a stabilizes x (that is, a is in G_{x}), that is: a.x = x.
then for the g for which g.x = y, we have:
(gag^{1}).y = g.(a.(g^{1}.y)) = g.(a.x) = g.x = y, so gag^{1} stabilizes y.
on the other hand, suppose that b.y = y. note that b = ebe^{1} = g(g^{1}bg)g^{1}.
now (g^{1}bg).x = g^{1}.(b.(g.x)) = g^{1}.(b.y) = g^{1}.y = x.
this means that G_{y} = gG_{x}g^{1}, G_{x} and G_{y} are conjugate subgroups.
2) suppose X is a Gset.
then we have a mapping G x X → X given by (g,x) → g.x with (for all g,h in G, and x in X):
g.(h.x) = (gh).x
e.x = x
define φ:G → Sym(X) by: φ(g)(x) = g.x (that is φ(g) is the function that takes x to g.x, so φ(g):X → X).
we need to show that φ(g) is a bijection on X, and that φ(gh) = φ(g)○φ(h).
note that that (φ(g)○φ(g^{1}))(x) = φ(g)(φ(g^{1})(x)) = φ(g)(g^{1}.x) = g.(g^{1}.x) = (gg^{1}).x = e.x = x, and:
(φ(g^{1})○φ(g))(x) = φ(g^{1})(φ(g)(x)) = φ(g^{1})(g.x) = g^{1}.(g.x) = (g^{1}.g)(x) = e.x = x.
thus φ(g^{1}) is a twosided inverse for φ(g), so φ(g) is a bijection.
now φ(gh)(x) = gh.x = g.(h.x) = φ(g)[φ(h)(x)] = (φ(g)○φ(h))(x), for all x in X, thus φ(gh) = φ(g)○φ(h), φ is a homomorphism.
on the other hand, suppose we just have a homomorphism φ of G into Sym(X).
we can make X into a Gset as follows:
define g.x = φ(g)(x). all we have to do is verify this satisfies the rules for a Gset.
note that since φ is a homomorphism, it has to map the identity of G, to the identity map of X (since that is the identity of the group Sym(X)). this means that:
φ(e)(x) = x, for all x in X, so e.x = x.
also, since φ is a homomorphism, φ(gh) = φ(g)○φ(h), which means that:
(gh).x = φ(gh)(x) = (φ(g)○φ(h))(x) = φ(g)(φ(h)(x)) = g.(φ(h)(x)) = g.(h.x), so we have a Gset.