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Math Help - Orbits of an element and the Stabilizer of the element

  1. #1
    Super Member Bernhard's Avatar
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    Orbits of an element and the Stabilizer of the element

    I am reading Dummit and Foote Section 4.1 on Group Actions and Permutation Representations.

    Proposition 2 reads:


    ================================================== =========
    "Let G be a group acting on the non-empty set A. The relation on A defined by

    a \sim b if and only if a = g.b for some g \in G

    is an equivalence relation.

    For each a \in A, the number of elements in the equivalence class containing a is |G: G_a| , the index of the stabilizer of a."

    ================================================== =========

    The proof that a = g.b is an equivalence relation is clear to me. I am interested (and struggling with) in some elements of D&F's proof regarding the second sentence. The second part of the proof proceeds as follows:

    "To prove the last statement of the proposition we exhibit a bijection between the left cosets of G_ a in G and the elements of the equivalence class of a. Let \C_a be the class of a, so

    C_a = {g.a | g \inG}

    Suppose b = g.a \in C_a. Thus g \G_a is a left coset of G_a in G.

    The map

    b = g.a \mapsto g G_a

    is a map from C_a to the set of left cosets of G_a in G. This map is surjective since for any g \in G the element g.a is an element of C_a. Since g.a = h.a if and only if h^{-1}g \in G_a if and only if g G_a = h G_a, the map is injective, hence is a bijection. This completes the proof."
    ================================================== ===

    The problems I am having is with the statements:

    a) This map is surjective since for any g \in G the element g.a is an element of C_a.

    b) Since g.a = h.a if and only if h^{-1}g \in G_a if and only if g G_a = h G_a, the map is injective

    Can anyone help by giving a very explicit proof or indication of why these statements are true?

    Would be appreciative of help.

    Peter
    Last edited by Bernhard; December 23rd 2011 at 03:57 PM.
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  2. #2
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    Re: Orbits of an element and the Stabilizer of the element

    ok, we have a map from C_a \to G/G_a from the orbit of a (that is all possible images g.a) to the set of left cosets of G_a.

    that is, given b in the set A, such that b is in the equivalence class of a, then b~a, so b = g.a (g permutes A in such a way as to take a to b).

    so we send b (=g.a) to the coset of G_a that contains g, which is gG_a.

    now, given an arbitrary coset of G/G_a, say xG_a, we know that x.a is some element of the orbit of a,

    and our mapping takes x.a to xG_a, so x.a is a pre-image of xG_a (pretty much because that's the way we defined the orbit of a).

    now, suppose two elements of [a], let's call them b and c, get mapped to the same coset. we can write b = g.a, and c = h.a, so we have:

    gG_a = hG_a. this means that h^{-1}g \in G_a.

    but what is G_a? it's the stabilizer of a; that is, the set of g in G such that g.a = a. so, h^{-1}g.a = a.

    therefore:

    h(h^{-1}g).a = h.a
    (hh^{-1})g.a = h.a
    eg.a = h.a
    g.a = h.a, that is, b = c.

    perhaps this will be more clear if we see it "in action" (excuse the terrible pun). let's have A = {1,2,3,4}, and let G be the group D4 (where we are considering A as the set of vertices (corners) of the square).

    now, just from looking at the rotations alone, we can see that the orbit of 1, is all of A. let's write the elements of D4 in permutation form:

    {e, (1 2 3 4), (1 3)(2 4), (1 4 3 2), (1 2)(3 4), (1 4)(2 3), (1 3), (2 4)}.

    what is Stab(1)?

    well clearly e stabilizes (or "fixes") 1, and the only other element to do so is (2 4).

    so Stab(1) = {e, (2 4)}. the left cosets are as follows:

    Stab(1) = {e, (2 4)}
    (1 2 3 4)Stab(1) = {(1 2 3 4), (1 2)(3 4)}
    (1 3)(2 4)Stab(1) = {(1 3)(2 4), (1 3)}
    (1 4 3 2)Stab(1) = {(1 4 3 2), (1 4)(2 3)}, and our bijection is:

    1<--->Stab(1)
    2<--->(1 2 3 4)Stab(1) (every element of (1 2 3 4)Stab(1) takes 1-->2)
    3<--->(1 3)(2 4)Stab(1) (every element of (1 3)(2 4)Stab(1) takes 1-->3)
    4<--->(1 4 3 2)Stab(1) (every element of (1 4 3 2)Stab(1) takes 1-->4).

    and indeed we see that |[1]| = |A| = [D4:Stab(1)] = 4.

    you can look at it this way: if g,h take a to different elements of A, then gG_a, hG_a give rise to different cosets of G_a.
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  3. #3
    Super Member Bernhard's Avatar
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    Re: Orbits of an element and the Stabilizer of the element

    Thanks for that, Deveno

    Working through your post now

    Peter
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    Super Member Bernhard's Avatar
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    Re: Orbits of an element and the Stabilizer of the element

    Quick question:

    You write:

    "and our mapping takes x.a to x G_a , so x.a is a pre-image of x G_a(pretty much because that's the way we defined the orbit of a)."

    Dummit and Foote define 'orbit' as follows:

    "The equivalence class {g.a | g \inG} is called the orbit of G contaning a."

    How does this definition relate to the fact that g.a is taken to g G_a?

    Peter
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    Re: Orbits of an element and the Stabilizer of the element

    if G = {g1,g2,g3,....} then the orbit of a is {g1.a,g2.a,g3.a,....} and the cosets of G_a are \{g_1G_a,g_2G_a,g_3G_a,\dots\}.

    it's obvious we can send the elements of the orbit of a to the corresponding coset, what isn't obvious is that the duplicates in each list correspond to each other.
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