Group homomorphisms

Discussion in 'Physics & Math' started by arfa brane, Jun 18, 2012.

  1. arfa brane call me arf Valued Senior Member

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    The problem is:

    Prove that the function \( \theta_x : g \mapsto g^x = x^{-1}gx \) on a group G is an isomorphism from G to G.
    Here's my attempt so far:

    Let \( g_1,g_2 \in G \) such that \( \forall x \in G \):

    \( {g_1}^x{g_2}^x \;=\; \( x^{-1}g_1x\)\(x^{-1}g_2x) \;=\; x^{-1}g_1\(xx^{-1}\)g_2x \;=\; x^{-1}g_1g_2x \;=\; \(g_1g_2\)^x \)​
    ,
    then g[sup]x[/sup] is a homomorphism from G to G.

    Suppose \( {g_1}^x \;=\; {g_2}^x \), then \( x^{-1}g_1x \;=\; x^{-1}g_2x \) so g[sub]1[/sub] = g[sub]2[/sub] by cancellation.
    So g[sup]x[/sup] is 1:1.

    Now to show g[sup]x[/sup] is onto, I need to show that each (or some arbitrary) element in the codomain of g[sup]x[/sup] has at least one element in its domain.

    Is the codomain the image of g[sup]x[/sup], and is x the element in the domain of g[sup]x[/sup]?
    Just, you know, making sure.
     
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  3. arfa brane call me arf Valued Senior Member

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    I think what I need to show to prove g[sup]x[/sup] is surjective is that its image is the whole of G.
    If it maps all x in G to G, maybe it's "clearly" onto.

    The question doesn't specifically say "for all x". But I guess I can assume it means "some arbitrary x".
     
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  5. temur man of no words Registered Senior Member

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    You seem to sometimes confuse the parameter \(x\) of \(\theta_x\) with its argument \(g\). The parameter \(x\) is just a fixed element of \(G\), and we have to show that \(\theta_x:G\to G\) is an isomorphism. You have done most of the work, what remains is to show surjectivity, which means that for any \(h\in G\) there exists \(g\in G\) such that \(g^x=h\). This equation is easily solved: \(g=xhx^{-1}\equiv h^{x^{-1}}\). In other words, \(\theta_x^{-1}=\theta_{x^{-1}}\).
     
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  7. arfa brane call me arf Valued Senior Member

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    Ok, thanks. So \( \theta_x \) is a function of g, not a function of x. Or rather it maps some g in G to the ccnjugate of x?
     
  8. arfa brane call me arf Valued Senior Member

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    No that's inaccurate; if \(g^x=h\), then g is conjugate to h. So x conjugates g.
    We haven't covered conjugation as such, but when we did G-sets, we got this example:

    For H ≤ G, G is an H-set if we define h·g = hgh[sup]-1[/sup], ∀h ∈ H, ∀g ∈ G.
    Then we have · : G x H → G; e·h = h; (g[sub]1[/sub]g[sub]2[/sub])h = g[sub]1[/sub](g[sub]2[/sub]h).
     

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