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Math Help - Automorphisms on S7

  1. #1
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    Automorphisms on S7

    Hi guys,

    I am looking an automorphism (isomorphism onto itself) of S7 (symmetric group on 7 letters). The extra condition I have been given is that no element is allowed to remain fixed after applying said isomorphism (eg 1 should not be sent back to 1). To make this clearer, I am looking for a function, f, on 7 letters that:

    1. Is a bijection
    2. Satisfies f(xy)=f(x)f(y)
    3. Leaves no element of S7 fixed

    I have tried fiddling around with a lot of modular arithmetic and conjugacy classes etc, but it is hard to find such a function. Perhaps such a function does not exist?
    Thanks in advance
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  2. #2
    Member ModusPonens's Avatar
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    Re: Automorphisms on S7

    The identity has to remain fixed. So, either you didn't mention that, or the automorphism does not exist.
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  3. #3
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    Re: Automorphisms on S7

    i think you are confusing the concepts of set-automorphism (bijection from a set to itself) and group-automorphism (isomorphism from a group to itself).

    an element of Aut(S7) maps permutations to permutations.

    only elements of S7 map numbers to numbers (or letters to letters..whatever we decide the call the elements of our 7-element set).

    however, it is true that every automorphism of S7 is inner, so Aut(S7) is isomorphic to S7

    (that is every element of Aut(S7) is a map: τ→στσ-1, for some σ in S7.

    if we call this map fσ, then σ→fσ is a group isomorphism).

    and there DO exist elements of S7 that do not leave any letter fixed, for example: 7-cycles.

    however, conjugation by a 7-cycle will still be the identity map on all powers of that 7-cycle:

    σ(σk-1 = σ1+k-1 = σk.

    perhaps this will convince you that every automorphism of S7 fixes some element of S7 (which is the same as saying every σ in S7 is conjugate to itself, which should not be surprising).
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