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Fintie Groups; Subgroups

  1. Feb 11, 2007 #1
    Show that whenever ab = ba, you have ba^(-1) = a^(-1)b.

    I don't know how to slove problem.

    pls help me..
     
  2. jcsd
  3. Feb 11, 2007 #2
    On the equality [tex]ba^{-1}=a^{-1}b[/tex] make a left multiplication and then a right multiplication by [tex]a[/tex] and see what you get.
     
  4. Feb 11, 2007 #3
    sorry.. I don't understand.. Please give me clear as that.. thanks.. smile..
     
  5. Feb 11, 2007 #4
    You don't know what a left or right multiplication is? It means that you multiply both the members of the equality, by the same quantity, on their left/right side.

    For example, a left multiplication by [tex]a[/tex]:

    [tex]ba^{-1}=a^{-1}b[/tex] => [tex]a{\cdot}ba^{-1}=a{\cdot}a^{-1}b[/tex]

    Now go on and make a right multiplication on the equation we obtained.
    Also consider what is the result of [tex]a{\cdot}a^{-1}[/tex] and [tex]a^{-1}{\cdot}a[/tex]
     
    Last edited: Feb 11, 2007
  6. Feb 11, 2007 #5
    Show that whenever ab = ba, you have ba^(-1) = a^(-1)b.

    then you said that

    just answers:
    For example, a left multiplication by LaTeX graphic is being generated. Reload this page in a moment.:

    LaTeX graphic is being generated. Reload this page in a moment. => LaTeX graphic is being generated. Reload this page in a moment.
     
  7. Feb 11, 2007 #6
    you mean that I have to make left and right..

    make left:
    ba^(-1) = a^(-1)b => aba^(-1) = aa^(-1)b

    make right:
    ba^(-1) = a^(-1)b => ba^(-1)a = a^(-1)ba

    then what??
     
  8. Feb 11, 2007 #7
    No. I mean make left, then, on the result of the left multiplication, make a right multiplication. Let's see what you get.
     
  9. Feb 11, 2007 #8
    i m sorry.. I lost and I dont understand what you talk about. Please help for that.
     
  10. Feb 11, 2007 #9

    berkeman

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    Staff: Mentor

    Thread moved from General Math to Homework forums. patelnjigar, you have received some very good help here so far from antonantal. You need to understand that we DO NOT do your homework problems for you. You need to show us more of your work on this problem, and stop trying to get it solved for you. Show us your work.
     
    Last edited: Feb 12, 2007
  11. Feb 11, 2007 #10
    Come on, you almost did it!

    You did a left multiplication by [tex]a[/tex] on

    [tex]ba^{-1}=a^{-1}b[/tex] and obtained

    [tex]a{\cdot}ba^{-1}=a{\cdot}a^{-1}b[/tex]

    Now if you do a right multiplication by [tex]a[/tex] on [tex]a{\cdot}ba^{-1}=a{\cdot}a^{-1}b[/tex] what do you obtain?
     
    Last edited: Feb 11, 2007
  12. Feb 12, 2007 #11
    ab=ba

    a^(-1).ab.a^(-1) = a^(-1).ba.a^(-1)

    (a^(-1).a) b^(-1) = a^(-1).b(a.a^(-1))

    e.b^(-1) = a^(-1).b.e

    b^(-1) = a^(-1).b

    is that right?? I hope that I made it...
     
  13. Feb 12, 2007 #12
    I can't figure out what is it that you don't understand. Look:
    I take [tex]ba^{-1}[/tex] ,I do a left multiplication by [tex]a[/tex] and get [tex]aba^{-1}[/tex] ,then I do a right multiplication by [tex]a[/tex] and get [tex]aba^{-1}a[/tex].

    You try it on the other member of the equation, [tex]a^{-1}b[/tex]
     
  14. Feb 12, 2007 #13
    ba^(-1) = a^(-1)b

    aba^(-1) = aa^(-1)b

    aba^(-1)a = aa^(-1)ba

    ab = ba

    i seem that it done answers.
     
  15. Feb 12, 2007 #14
    That's right. It wasn't that hard was it? :smile:
     
  16. Feb 12, 2007 #15

    Gokul43201

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    Staff Emeritus
    Science Advisor
    Gold Member

    You started off correctly but made a typo (bolded) in the second step. Else you would have got the correct answer.
     
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