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Prove that (a^-1)^-1=a

  1. Nov 3, 2013 #1
    1. The problem statement, all variables and given/known data

    Let (G,[itex]\circ[/itex]) be a group. Show that [itex]\forall[/itex]a[itex]\in[/itex]G [tex](a^{-1})^{-1} = a[/tex]
    3. The attempt at a solution

    I came up with the following. [itex]a^{-1}[/itex] is the inverse of [itex](a^{-1})^{-1}[/itex], therefore we have:

    [tex]a^{-1} \circ (a^{-1})^{-1} = e[/tex]
    But for [itex]a^{-1} \circ (a^{-1})^{-1}[/itex] to be equal to e, it has to be the case that:
    [tex]a^{-1} \circ a = e[/tex]
    , therefore [itex](a^{-1})^{-1}[/itex] has to be equal to a.

    Now I think the proof is incorrect, but I'm not sure. I think the mistake is trying to imply that [itex](a^{-1})^{-1}[/itex] is equal to a, just because both [itex]a^{-1} \circ (a^{-1})^{-1}[/itex] and [itex]a^{-1} \circ a[/itex] are equal to a. Is this a valid proof technique?
     
  2. jcsd
  3. Nov 3, 2013 #2

    arildno

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    Well, isn't a group associative?
    Thus,

    We have:
    a^-1(ring)(a^-1)^-1=e

    so that:
    a(ring)(a^-1(ring)(a^-1)^-1))=a(ring)e (*)

    Invoking associativity on LHS in (*) should do the trick.
     
  4. Nov 3, 2013 #3
    Ah yes, didn't think of that option. Thanks a lot for your help.

    In general if you try to prove equalities like this, what should you think about? Should you just consider all definitions that are valid and then just try some things out until you come up with the desired equality or what should the process of a finding a proof for a statement like this look like?
     
  5. Nov 3, 2013 #4

    arildno

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    There isn't any foolproof method to kill all the problems you might encounter.

    But, as you do more of such problems, your brain sort of figures out the structure, and clever ideas might start popping up by themselves.
     
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