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 G-set.

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 two-sided 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 G-set as follows:

define g.x = φ(g)(x). all we have to do is verify this satisfies the rules for a G-set.

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 G-set.