Showing that a map from factor group to another set bijective

[demonelite123]

Let G have a transitive left action on a set X and set H = G_x to be the stabilizer of any point x. Show that the map defined by f: G/H \rightarrow X where f(gH) = gx is well defined, one to one, and onto.

i think i know how to show well defined. letting g1 H = g2 H, if i multiply on the right both sides with x, then since H is the stabilizer of x, Hx = x so i get g1 x = g2 x which is what i wanted. i am having some trouble though on showing that this function is one to one and onto most likely because i am still getting acquainted with the ideas of factor groups and cosets.

So for one to one i want to show that if g1 x = g2 x then g1 H = g2 H. what i did was x = g_1^{-1}g_2 x which implies that g_1^{-1}g_2 \in H and this shows g_1 H = g_2 H.

For onto i tried saying that given any gx in X, you can then find gH in G/H which maps to it. but I'm not sure if this is valid or not. when proving onto do i need to start with an arbitrary x in X? or do i start with some gx in X?

 

[zoek]

Let G have a transitive left action on a set X and set H = G_x to be the stabilizer of any point x. Show that the map defined by f: G/H \rightarrow X where f(gH) = gx is well defined, one to one, and onto.

i think i know how to show well defined. letting g1 H = g2 H, if i multiply on the right both sides with x, then since H is the stabilizer of x, Hx = x so i get g1 x = g2 x which is what i wanted. i am having some trouble though on showing that this function is one to one and onto most likely because i am still getting acquainted with the ideas of factor groups and cosets.

g_1 H = g_2 H \implies g_2^{-1}g_1 H = H \implies g_2^{-1}g_1 \in H \implies g_2^{-1}g_1 \cdot x=x \implies g_1 \cdot x = g_2 \cdot x \implies f(g_1 H) = f(g_2 H).

So for one to one i want to show that if g1 x = g2 x then g1 H = g2 H. what i did was x = g_1^{-1}g_2 x which implies that g_1^{-1}g_2 \in H and this shows g_1 H = g_2 H.

This is OK.

For onto i tried saying that given any gx in X, you can then find gH in G/H which maps to it. but I'm not sure if this is valid or not. when proving onto do i need to start with an arbitrary x in X? or do i start with some gx in X?

You need to start with an arbitrary y in X. Since G is transitive there exist g\in G such that g \cdot x = y. So y = f(gH) i.e. f is onto.

 

[demonelite123]

thanks for clearing that up for me! and sorry for the late reply.