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Math Help - product of n-1 commutators

  1. #1
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    product of n-1 commutators

    I'm probably missing something very simple here, but I can't figure this one out.

    Let x_{1},...,x_{n} be elements in a group G.
    show that an element of the form x_{1}x_{2}...x_{n}x_{1}^{-1}x_{2}^{-1}...x_{n}^{-1} is a product of n-1 commutators.

    I tried this with induction on n, but that led nowhere and I tried showing how an element of the form abca^{-1}b^{-1}c^{-1} was a product of 2 commutators but i got stuck.

    Any hint would be appreciated

    SK
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  2. #2
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    Quote Originally Posted by skyking View Post
    I'm probably missing something very simple here, but I can't figure this one out.

    Let x_{1},...,x_{n} be elements in a group G.
    show that an element of the form x_{1}x_{2}...x_{n}x_{1}^{-1}x_{2}^{-1}...x_{n}^{-1} is a product of n-1 commutators.

    I tried this with induction on n, but that led nowhere and I tried showing how an element of the form abca^{-1}b^{-1}c^{-1} was a product of 2 commutators but i got stuck.

    Any hint would be appreciated

    SK


    abca^{-1}b^{-1}c^{-1}=[a,bc][b,c] (defining [x,y]:=xyx^{-1}y^{-1} ) .

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    abca^{-1}b^{-1}c^{-1}=[a,bc][b,c] (defining [x,y]:=xyx^{-1}y^{-1} ) .

    Tonio
    OK, so this is what I get then:

    [x_{1},x_{2}...x_{n}][x_{2},x_{3}...x_{n}]...[x_{n-1},x_{n}]=x_{1}...x_{n}x_{1}^{-1}...x_{n}^{-1}

    Thanks for the hint,

    SK
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