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Thread: G-set

  1. #1
    Aug 2012



    1)Given G, a group, and X a G-set What relationship exists between $\displaystyle G_x $ and $\displaystyle G_y $ if x and y belong to the same orbit?


    a)Let G be a group and X a G-set.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 G-set

    I hope some help. thanks
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  2. #2
    MHF Contributor

    Mar 2011

    Re: G-set

    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-1g).x = e.x = x

    now suppose that a stabilizes x (that is, a is in Gx), 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-1bg)g-1.

    now (g-1bg).x = g-1.(b.(g.x)) = g-1.(b.y) = g-1.y = x.

    this means that Gy = gGxg-1, Gx and Gy 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.
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