1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Inner Automorphisms as a Normal Subgroup

  1. Feb 26, 2008 #1
    1. The problem statement, all variables and given/known data

    Let G be a group. We showed in class that the permutations of G which send products to products form a subgroup Aut(G) inside all the permutations. Furthermore, the mappings of the form [itex]\sigma_b(g)=bgb^{-1}[/itex] form a subgroup inside Aut(G) called the inner automorphisms and denoted Inn(g).

    Prove that the inner automorphisms form a normal subgroup of Aut(G).

    2. Relevant equations

    None

    3. The attempt at a solution

    I attempted this problem one way, and my professor said I was going about it the wrong way - I did multiplications of the permutations and apparently I'm supposed to use function composition of the permutations. So this is what I have: I must show that, given a permutation [itex]\tau[/itex] that [itex]\tau \sigma_b \tau^{-1}[/itex] is in the set of inner automorphisms (at least, that's how we've been showing normality in class and my professor told me that this is at least correct).

    So, I have
    [tex]\tau\left(\sigma_b\left(\tau^{-1}\left(g\right)\right)\right)&=&\tau\left(b\tau^{-1}(g)b^{-1}\right)\\
    &=&\tau(b)\tau\left(\tau^{-1}(g)\right)\tau(b^{-1})\\
    &=&\tau(b)g\tau(b)^{-1}\\
    &=&\sigma_{\tau(b)}(g)[/tex]

    My question is, is [itex]\tau[/itex] equal to [itex]\tau^{-1}(g)[/itex] or is it [itex]\left(\tau(g)\right)^{-1}[/itex], or are these the same thing? I've done the problem assuming that [itex]\tau^{-1}[/itex] means [itex]\tau^{-1}(g)[/itex] and it seems to work, but I'm a bit uneasy about this. Can anybody confirm whether I've done this properly or not?
     
  2. jcsd
  3. Feb 26, 2008 #2
    This is good
    [itex]\tau[/itex] is a mapping and [itex]\tau^{-1}(g)[/itex] is an element of the group on which the mapping is defined. Your question is not making sense. You have successfully shown that that the inner automorphisms are normal in the group of automorphisms. You can stop there.
     
    Last edited: Feb 26, 2008
  4. Feb 26, 2008 #3
    Okay, so I worded my question a bit strangely. Sorry :redface:

    What I meant to ask was whether [itex]\tau^{-1}(g)=\left(\tau(g)\right)^{-1}[/itex] in general. I have a feeling that it's not and that's what I was uneasy about.
     
  5. Feb 26, 2008 #4

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Have you tried any special cases? What happens when you apply tau to both sides?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Inner Automorphisms as a Normal Subgroup
Loading...