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Math Help - Show a,ba^-1b^-1 is an element of A_n

  1. #1
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    Show a,ba^-1b^-1 is an element of A_n

    Let a,b be in S_n.
    Show a,ba^-1b^-1 is an element of A_n
    aba^-1b^-1
    a(ba^-1)b^-1
    a(ab)b^-1
    aa(bb^-1)
    aa
    =e
    Have no clue if what I did was correct
    A_n is alternating group
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  2. #2
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    Quote Originally Posted by kathrynmath View Post
    Let a,b be in S_n.
    Show a,ba^-1b^-1 is an element of A_n
    aba^-1b^-1
    a(ba^-1)b^-1
    a(ab)b^-1
    aa(bb^-1)
    aa
    =e
    Have no clue if what I did was correct
    A_n is alternating group

    Even permutation times even permutation = ...? Odd permutation times odd permutation = ?

    Tonio

    Ps. And no, what you did isn't correct. There's no commutativity in S_n , for n >= 3
    Last edited by tonio; November 8th 2010 at 06:16 PM. Reason: Addition
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  3. #3
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    even times even=even
    odd times odd = odd
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  4. #4
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    Quote Originally Posted by kathrynmath View Post
    even times even=even
    odd times odd = odd

    Good, so now just remember that a permutation and its inverse have the same parity...and now solve your problem!

    Tonio
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  5. #5
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    Ok a permutation and it's inverse have the same parity.
    aba^-1b^-1
    So a*a^-1=even
    and b^-1*b=even
    or
    a*a^-1=even
    and b^-1*b=odd
    or
    a*a^-1=odd
    and b^-1*b=even
    or
    a*a^-1=odd
    and b^-1*b=odd
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  6. #6
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    I don't know if that was the right idea
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  7. #7
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    Quote Originally Posted by kathrynmath View Post
    I don't know if that was the right idea

    **** Sigh **** .... [a,b]=aba^{-1}b^{-1} = (ab)(ba)^{-1} ... and thus we only care whether ab (and ALSO ba) is even or odd...

    Tonio
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  8. #8
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    Oh, I hadn't thought of writing it that way. Since they are the same parity, they must be in A_n?
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  9. #9
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    No no no, not for permutations. The rules for parity for multiplying permutations are analagous to the rules for ADDING integers.
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  10. #10
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    Quote Originally Posted by kathrynmath View Post
    Oh, I hadn't thought of writing it that way. Since they are the same parity, they must be in A_n?

    No! No matter what parity ab, (ba)^{-1} have, when you multiply them the parity is even and thus the product is in A_n...

    Tonio
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