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Proving a function is onto

  1. Nov 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Let G be a group and define λg:G→G to be λg(x)=g.x , x[itex]\in[/itex]G.

    Show that λg is onto and one-to-one.

    3. The attempt at a solution
    Suppose g.x=g.x' g-1.g.x=g-1.g.x' which means x=x'.

    How should I show that the function is onto?
     
  2. jcsd
  3. Nov 10, 2012 #2

    HallsofIvy

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    If y is any member of G, then [itex]\lambda(g^{-1}y)= y[/itex]
     
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