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Math Help - Automorphisms of a group G - Inn(G) and Aut(G)

  1. #1
    Super Member Bernhard's Avatar
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    Automorphisms of a group G - Inn(G) and Aut(G)

    Dummit and Foote Section 4.4 Automorphisms Exercise 1 reads as follows:

    Let \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)

    ================================================== ====

    A start to the proof is as follows:

    \sigma \phi_g \sigma^{-1}

    = \sigma (\phi_g (\sigma^{-1} (x)))

    = \sigma (g. \sigma^{-1} (x). g^{-1} )

    = \sigma (g). x . \sigma (g^{-1} )

    = \sigma (g). x . {[\sigma (g)]}^{-1}

    Thus \sigma \phi_g \sigma^{-1} = \phi_{\sigma  (g)}.

    My question is - how do we use this result to show Inn(G) is a normal subgroup of Aut(G)?

    Can someone please help?

    Peter
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Re: Automorphisms of a group G - Inn(G) and Aut(G)

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

    Let \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)

    ================================================== ====

    A start to the proof is as follows:

    \sigma \phi_g \sigma^{-1}

    = \sigma (\phi_g (\sigma^{-1} (x)))

    = \sigma (g. \sigma^{-1} (x). g^{-1} )

    = \sigma (g). x . \sigma (g^{-1} )

    = \sigma (g). x . {[\sigma (g)]}^{-1}

    Thus \sigma \phi_g \sigma^{-1} = \phi_{\sigma  (g)}.

    My question is - how do we use this result to show Inn(G) is a normal subgroup of Aut(G)?

    Can someone please help?

    Peter
    Well, you first need to show that Inn is a subgroup. I am assuming you have done this?

    Now, a normal subgroup, N\unlhd G, is a subgroup whhere xyx^{-1}\in N for all y\in N and x\in G. So...you took an arbitrary automorphism, \sigma, and an arbitrary inner automorphism, \phi_g, and showed that \sigma\phi_g\sigma^{-1}=\phi_{\simga(g)}\in \operatorname{Inn}(G) which is precisely what you want. Here, x:=\sigma, y:=\phi_g, G:=\operatorname{Aut}(G) and N:=\operatorname{Inn}(G).
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