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Math Help - Inverse element proving

  1. #1
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    Inverse element proving

    Hello there,

    Given a, b, c elements of the set G, and G being a group relating to the * operator.
    (a * b) is the inverse element of itself.

    How can I prove that (b * a) is inverse to itself as well?

    Tried a lot of methods but all of them seem very far fetched (with all ending at (a * b) = (b * a)).

    Any help would be much appreciated.
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  2. #2
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    Re: Inverse element proving

    Quote Originally Posted by BscStud View Post
    Given a, b, c elements of the set G, and G being a group relating to the * operator.
    (a * b) is the inverse element of itself.

    How can I prove that (b * a) is inverse to itself as well?
    If you were given only a and b, then there is a way to prove this, but since you are given c as well, I am not sure.

    Given: abab = e where e is the unit.
    b(abab) = b
    b(abab)b^(-1) = bb^(-1)
    (baba)(bb^(-1)) = e
    baba = e

    P.S. Questions about groups should be posted to the University Algebra forum.
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  3. #3
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    Re: Inverse element proving

    Quote Originally Posted by emakarov View Post
    If you were given only a and b, then there is a way to prove this, but since you are given c as well, I am not sure.

    Given: abab = e where e is the unit.
    b(abab) = b
    b(abab)b^(-1) = bb^(-1)
    (baba)(bb^(-1)) = e
    baba = e

    P.S. Questions about groups should be posted to the University Algebra forum.
    Thanks for your response, I will make sure to post it next time in the correct forum. Wasn't aware.

    As for your response - I'm not sure how you reached the (baba)(bb^(-1)) = e ?
    As if you say (baba)(bb^(-1)) = e, you need to have somehow to prove baba = e in the first place since bb^(1) is the unit, or am I missing something?
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  4. #4
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    Re: Inverse element proving

    Quote Originally Posted by BscStud View Post
    I'm not sure how you reached the (baba)(bb^(-1)) = e ?
    This particular line I am getting from the previous line "b(abab)b^(-1) = bb^(-1)" by replacing bb^(-1) with e in the right-hand side and by using associativity to rewrite b(abab)b^(-1) as (baba)(bb^(-1)) in the left-hand side.
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  5. #5
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    Re: Inverse element proving

    Quote Originally Posted by emakarov View Post
    This particular line I am getting from the previous line "b(abab)b^(-1) = bb^(-1)" by replacing bb^(-1) with e in the right-hand side and by using associativity to rewrite b(abab)b^(-1) as (baba)(bb^(-1)) in the left-hand side.
    Got it, but isn't making (ab)(ab) to (ba)(ba) relying on the group being commutative (it hasn't been given to be commutative)?

    Thanks again.
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  6. #6
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    Re: Inverse element proving

    Quote Originally Posted by BscStud View Post
    As for your response - I'm not sure how you reached the (baba)(bb^(-1)) = e ?
    As if you say (baba)(bb^(-1)) = e, you need to have somehow to prove baba = e in the first place since bb^(1) is the unit, or am I missing something?
    Here is a different way:

     \begin{align*}  (ab)(ab) &= e \\ ba&=a^{-1} b^{-1}\\&= (ba)^{-1}\end{align*}
    Thanks from BscStud
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  7. #7
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    Re: Inverse element proving

    Quote Originally Posted by BscStud View Post
    Got it, but isn't making (ab)(ab) to (ba)(ba) relying on the group being commutative (it hasn't been given to be commutative)?
    I am not replacing (ab)(ab) by (ba)(ba); I am replacing b(abab)b^(-1) by (baba)(bb^(-1)). Note that the order of group elements is not changed: it's bababb^(-1) in both cases. Therefore, commutativity is not used, only associativity.
    Thanks from BscStud
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  8. #8
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    Re: Inverse element proving

    Think I got it, thanks Plato and emakarov!
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