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Thread: Projection function of some kind?

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    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
    $\displaystyle f: Hx \to H : hx \mapsto h$

    for all x where $\displaystyle h \in H$ and $\displaystyle x \in G$. 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
    $\displaystyle H = \{e, a, a^2, a^3 \}$ and H < G then
    $\displaystyle 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, $\displaystyle Hx = \{e, a, a^2, a^3 \}x$ , 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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    Quote Originally Posted by topsquark View Post
    My problem is in proving the following:
    The text suggests that I construct the function
    $\displaystyle f: Hx \to H : hx \mapsto h$
    for all x where $\displaystyle h \in H$ and $\displaystyle x \in G$. All the text says is that f is "easily shown to be a bijection."
    Look $\displaystyle f(h_1g)=f(h_2g)$ be definition means $\displaystyle h_1=h_2$.
    Operate by $\displaystyle g$ and get $\displaystyle h_1g=h_2g$, injective.

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

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

    -Dan
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