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Math Help - Automorphisms and Conjugation

  1. #1
    Super Member Bernhard's Avatar
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    Automorphisms and Conjugation

    Exercise 1 in Dummit and Foote section 4.4 Automorphisms reads as follows:

    If \sigma \in Aut(G) and {\phi}_g is conjugation by g prove that \sigma {\phi}_g {\sigma}^{-1} = {\phi}_{\sigma{(g)}.

    Deduce that Inn(G) is a normal subgroup of Aut(G).

    Can anyone help me get started on this problem?

    Notes:

    (1) I have attached Dummit and Foote section 4.4 in case readers need to view the terminology and notation used. This upload includes the text of the exercise.

    (2) I am assuming that the mapping {\phi}_g is h \longrightarrow gh g^{-1}

    (3) I am not sure what Dummit and Foote mean by {\sigma{(g)} - unless they mean the set of automorphisms that map g into g???
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    MHF Contributor Drexel28's Avatar
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    Re: Automorphisms and Conjugation

    Quote Originally Posted by Bernhard View Post
    Exercise 1 in Dummit and Foote section 4.4 Automorphisms reads as follows:

    If \sigma \in Aut(G) and {\phi}_g is conjugation by g prove that \sigma {\phi}_g {\sigma}^{-1} = {\phi}_{\sigma{(g)}.

    Deduce that Inn(G) is a normal subgroup of Aut(G).

    Can anyone help me get started on this problem?

    Notes:

    (1) I have attached Dummit and Foote section 4.4 in case readers need to view the terminology and notation used. This upload includes the text of the exercise.

    (2) I am assuming that the mapping {\phi}_g is h \longrightarrow gh g^{-1}

    (3) I am not sure what Dummit and Foote mean by {\sigma{(g)} - unless they mean the set of automorphisms that map g into g???
    What they mean is this. There is a natural map \Phi:G\to \text{Aut}(G) defined by the rule (\Phi(g))(x)=gxg^{-1} (where x\in G--note that since \Phi(g) is supposed to be a map that this makes sense). They then define \text{im }\Phi to be \text{Inn}(G). They then ask you to show that \sigma\circ\Phi(g)\circ\sigma^{-1}=\Phi(\sigma(g)) (where, note again, that \sigma(g)\in G) and so one can deduce that \text{Inn}(G)\unlhd \text{Aut}(G).
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    Super Member Bernhard's Avatar
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    Re: Automorphisms and Conjugation

    I am still struggling with both the meaning and the difference between \phi_g and \phi_{\sigma{(g)}}?

    Can anyone clarify this for me and go a bit further with the proof?

    Peter
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    MHF Contributor Drexel28's Avatar
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    Re: Automorphisms and Conjugation

    Quote Originally Posted by Bernhard View Post
    I am still struggling with both the meaning and the difference between \phi_g and \phi_{\sigma{(g)}}?

    Can anyone clarify this for me and go a bit further with the proof?

    Peter

    Think about this. Take G=\mathbb{Z}, \sigma(x)=2x, and g=3. Then, they're trying to have you prove that \sigma\circ\Phi(3)\circ\sigma^{-1}=\Phi(\sigma(3))=\Phi(6).
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    Super Member Bernhard's Avatar
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    Re: Automorphisms and Conjugation

    Thanks

    The example was very helpful in getting an understanding of the problem.

    Basically he equivalence between your notation and D&Fs is as follows:

    \phi_g  \equiv \Phi (g)

    and

    \phi_{\sigma (g)}  \equiv \Phi (\sigma (g)) where \sigma (g) (as is usual) is the map of g under \sigma

    I am still struggling with the proof of \sigma \phi_g \sigma^{-1} = \phi_{\phi (g)} and would appreciate some further help.

    Regarding Inn(G) \unlhd Aut(G) it appears to me that \phi_g is essentially conjugation of G by g and hence \phi_g is an inner automorphism of G and hence \phi_g \in Inn(G).

    But where to from here to show Inn(G) \unlhd Aut(G)

    Peter
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    Re: Automorphisms and Conjugation

    two maps are equal if they take on equal values for every element in their domain.

    let's say we have any old automorphism of G, which we call \sigma. this is a bijective homomorphism, and let's say \sigma:x \to y. clearly, \sigma^{-1}:y \to x.

    now by \sigma_g, we mean the particular automorphism that is conjugation by g: \sigma_g:x \to gxg^{-1} for any x. so:

    \sigma \circ \sigma_g \circ \sigma^{-1}(y) = \sigma(\sigma_g(\sigma^{-1}(y)))

     = \sigma(\sigma_g(x)) = \sigma(gxg^{-1})

     = \sigma(g)\sigma(x)\sigma(g^{-1}) (since \sigma is a homomorphism)

     = \sigma(g)y(\sigma(g))^{-1} (again, since \sigma is a homomorphism, and by our definition of y)

     = \sigma_{\sigma(g)}(y).

    so when we conjugate the inner automorphism corresponding to conjugation by g, by another automorphism \sigma, we get the same map as conjugation by the element \sigma(g).

    this shows that for any automorphism \sigma, that \sigma\mathrm{Inn}(G)\sigma^{-1} \subseteq \mathrm{Inn}(G)
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    Super Member Bernhard's Avatar
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    Re: Automorphisms and Conjugation

    Thanks for that really helpful post

    I will now work through the proof

    Peter
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    Re: Automorphisms and Conjugation

    note: i just noticed i use \sigma_g where you use \phi_g, which is probably preferrable to avoid confusion.
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    Super Member Bernhard's Avatar
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    Re: Automorphisms and Conjugation

    Thanks for that helpful clarification

    Peter
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