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Showing that a map from factor group to another set bijective

  1. Oct 19, 2011 #1
    Let G have a transitive left action on a set X and set [itex] H = G_x [/itex] to be the stabilizer of any point x. Show that the map defined by [itex] f: G/H \rightarrow X [/itex] 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 [itex] x = g_1^{-1}g_2 x [/itex] which implies that [itex] g_1^{-1}g_2 \in H [/itex] and this shows [itex] g_1 H = g_2 H [/itex].

    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?
  2. jcsd
  3. Oct 21, 2011 #2
    [itex]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)[/itex].

    This is OK.

    You need to start with an arbitrary [itex]y[/itex] in [itex]X[/itex]. Since [itex]G[/itex] is transitive there exist [itex]g\in G[/itex] such that [itex]g \cdot x = y[/itex]. So [itex]y = f(gH)[/itex] i.e. [itex]f[/itex] is onto.
  4. Oct 25, 2011 #3
    thanks for clearing that up for me! and sorry for the late reply.
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