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If x=x^-1 for all x in G, then it's the same in H, for isomorphisms.

  1. Nov 13, 2011 #1
    1. The problem statement, all variables and given/known data

    G and H are isomorphic.
    Prove that if x-1=x for all x in G, then x-1=x for all x in H.


    2. Relevant equations

    G is isomorphic to H means there is an operation preserving bijection from G to H.

    3. The attempt at a solution



    I need a point in the right direction.
     
  2. jcsd
  3. Nov 13, 2011 #2

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    Pick an operation involving x and x^-1 and map it?

    Or perhaps you should look up the exact definition of an inverse?
     
  4. Nov 13, 2011 #3
    Thank goodness you're here!

    Okay, so um, [tex] \theta (x) = y [/tex] ?

    inverse means [tex] x*x^{-1} = e [/tex]
     
  5. Nov 13, 2011 #4

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    Thanks!
    It's nice to feel appreciated! :blushing:


    What about [tex] \theta (x*x^{-1}) ? [/tex]


    Inverse in a group G means according to the axioms:
    For each a in G, there exists an element b in G such that a • b = b • a = e​



    So let's pick an element a in H.
    Can you find an (the!) inverse b that satisfies the axiom, with the help of the isomorphism you have?
     
  6. Nov 13, 2011 #5
    [tex] \theta (x*x^{-1})=\theta (x)*\theta (x^{-1}) [/tex]
    [tex] \theta (e_G)=e_H=\theta (x)*\theta (x^{-1}) [/tex]
    let [itex] \theta (x)=a [/itex]
    then [itex] e_H=a*\theta (x^{-1}) [/itex]
    thus, the inverse of [itex] \theta (x)=a [/itex] in H is [itex] \theta (x^{-1}) [/itex]

    Yes?
     
  7. Nov 13, 2011 #6
    Oh wait, but since [itex] x=x^{-1} , \theta (x^{-1})=\theta (x)= a [/itex]
    Thus, each a in H, a=a-1

    Did I prove it? Did I prove it? :D
     
  8. Nov 13, 2011 #7

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  9. Nov 13, 2011 #8
    Hooray! Thank you very much. That was the most painless problem ever :)
     
  10. Nov 13, 2011 #9

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    Ah, but then, when things *click*, most of them are painless.
    Perhaps you should revisit older problems in time...
     
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