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Math Help - Automorphism Proof

  1. #1
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    Automorphism Proof



    Attempt:

    I don't know how to show that \pi (g) = g^{-1}, using the given hint. Any help to get me started is really appreciated. But I have tried to show that G is abelian and that it has an odd order:

    If \pi is an automorphism then for all g,h \in G



    Hence G is Abelian. (Right?)

    Now to show that the group is of an odd order. We can split the whole set G\{1} into disjoint pairs .

    Then G\{1} has even number of elements and thus G (including 1) has an odd number of elements.

    Is this right?

    P.S. Latex wasn't working properly today.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by demode View Post


    Attempt:

    I don't know how to show that \pi (g) = g^{-1}, using the given hint. Any help to get me started is really appreciated. But I have tried to show that G is abelian and that it has an odd order:

    If \pi is an automorphism then for all g,h \in G



    Hence G is Abelian. (Right?)

    Now to show that the group is of an odd order. We can split the whole set G\{1} into disjoint pairs .

    Then G\{1} has even number of elements and thus G (including 1) has an odd number of elements.

    Is this right?

    P.S. Latex wasn't working properly today.
    I would actually do it in reverse thought (I've done this problem before, Dummit and Foote--right?). Namely, the key thing to note is that to show "a=b" here you just need to show that "pi(ab')=ab' " where ' denotes inverse. For example, fiddle around to show that pi(aba'b')=aba'b' and conclude that aba'b'=e. Then, using abelianess note that pi(pi(g)g)=pi(pi(g))pi(g)=gpi(g)=pi(g)g and so conclude that pi(g)g=e for every g in G. The rest of your work looks fine.
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  3. #3
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    if π(a)a^-1 = π(b)b^-1, then π(b^-1a) = b^-1a, whence b^-1a = e, so a = b.

    this shows we can write any element of G uniquely as π(a)a^-1 for some a.

    so for any g in G, gπ(g) = (π(a)a^-1)(π(π(a)a^-1)) = (π(a)a^-1)(aπ(a^-1)) = π(a)π(a^-1) = π(e) = e.

    hence π(g) = g^-1, for all g in G.

    but if g-->g^-1 is an automorphism: then (gh)^-1 = h^-1g^-1 = g^-1h^-1, so hg = gh.

    if G has an element x of order 2, then π(x) = x^-1 = x, and x is a fixed point of π, so G has no elements of order 2,

    and hence G must be of odd order.
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