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Inner Automorphism

  1. Oct 31, 2012 #1
    I'm working some homework where we're introduced to inner automorphisms. I have that G is a group. Given a fixed g in G, and fg:G → G, fg(x)= gxg-1.

    I'm just a little confused about the notation. Are both g and x in G? I've been reading through the wiki http://en.wikipedia.org/wiki/Inner_automorphism and that seems to be the case, however when I'm working on proofs I wasn't sure which variable I was supposed to be working with.

    For instance, if I'm proving surjectivity, (or injectivity or bijectivity) I would want to show f(x1)=f(x2), however I would assume g stays fixed, correct? But when I'm proving f has an inverse, I would prove that (fg)-1=fg-1, which is using a different fixed g in G? Similarly, The wiki says that when G is abelian, the inner automorphism contains only the identity, but if I wanted to prove that, would I work with two different g's or two different x's?

    I have a vague idea of how this relates to permutations, but it's not sinking in and my hang ups about the notation is preventing me from understanding a lot of what I'm reading. f is a permutation function? Or is G a set of permutations? Actually, now I'm confusing myself even worse. Could someone clarify for me?
     
  2. jcsd
  3. Oct 31, 2012 #2

    micromass

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    Yes.

    Correct. Every g induces an inner automorphism [itex]f_g[/itex]. The function [itex]f_g[/itex] is a function of x, so only x is variable.

    Yes.

    You need to prove that each function [itex]f_g[/itex] is equal to the identity.

    A permutation of a set X is by definition a bijection [itex]g:X\rightarrow X[/itex]. Any inner automorphism [itex]f_g[/itex] is a permutation on G since it is bijective.
     
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