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Math Help - Projection function of some kind?

  1. #1
    Forum Admin topsquark's Avatar
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    Projection function of some kind?

    My problem is in proving the following:
    Let H be a subgroup of a group G. Then |Hg| = |H| = |gH| for all g in G.
    The text suggests that I construct the function
    f: Hx \to H : hx \mapsto h

    for all x where h \in H and x \in G. All the text says is that f is "easily shown to be a bijection."

    First, this looks suspiciously like a projection map of some kind. I'll tackle that later...I should be able to show that f is a surjection.

    The problem I am having is this: f maps a set of cosets onto a single element of H. Okay, but looking at a couple of examples of the function acting explicitly on a given group:

    Let
    H = \{e, a, a^2, a^3 \} and H < G then
    f(e \cdot a) = e \text{ and } f(a \cdot e) = a This implies that f is multiple valued, ie f(a) is equal to both e and a. Which can't be right.

    Should I be taking the function of a coset, Hx = \{e, a, a^2, a^3 \}x , in which case hx is a representative of that coset? I'm obviously looking at something wrong, but I don't know what.

    Thanks!
    -Dan
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  2. #2
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    Quote Originally Posted by topsquark View Post
    My problem is in proving the following:
    The text suggests that I construct the function
    f: Hx \to H : hx \mapsto h
    for all x where h \in H and x \in G. All the text says is that f is "easily shown to be a bijection."
    Look f(h_1g)=f(h_2g) be definition means h_1=h_2.
    Operate by g and get h_1g=h_2g, injective.

    If h\in H then f(hg)=h, surjective.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Plato View Post
    Look f(h_1g)=f(h_2g) be definition means h_1=h_2.
    Operate by g and get h_1g=h_2g, injective.

    If h\in H then f(hg)=h, surjective.
    Thank you. I see what I was doing wrong earlier. And thanks for the bijection proof as well.

    -Dan
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