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Bijection of a group

  1. Sep 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Let (G,*) be a group, and denote the inverse of an element x
    by x'. Show that f: G to G defi ned by f(x) = x' is a bijection,
    by explicitly writing down an inverse. Given x, y in G, what is
    f(x *y)?

    2. Relevant equations



    3. The attempt at a solution

    Okay, I think I'm just over thinking this... because it seems obvious that f is a bijection, since any function that has an inverse is bijective, and f obviously has an inverse where f-1(x')=x. And since x' and x are both in G, and G is a group then f-1 is well defined. But I keep having trouble with how to write proofs like this. Could someone help me out? Thanks!
     
  2. jcsd
  3. Sep 28, 2012 #2

    jgens

    User Avatar
    Gold Member

    Try writing something like: "Notice that [itex]f \circ f = \mathrm{id}_G[/itex] and this implies that [itex]f[/itex] is a bijection."
     
  4. Sep 28, 2012 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, since the inverse of the inverse of a group element is the group element itself so this function is its own inverse.
     
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