1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Is that mapping the homomorphism?

  1. Apr 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Let [itex]G[/itex] be a group and let [itex]Aut(G)[/itex] be the group of automorphisms of [itex]G[/itex].

    (a) For any [itex]g \in G[/itex], define [itex]\phi_{g}(x) = g^{-1}xg[/itex]. Check that [itex]\phi_{g}(x)[/itex] is an automorphism.

    (b) Consider the map:

    [itex]\Phi:G \rightarrow Aut(G)[/itex]
    [itex]g \mapsto \phi_{g}[/itex]

    Check that [itex]\Phi[/itex] is a homomorphism.

    2. The attempt at a solution

    Evaluate [itex]\Phi[/itex] at gh with arbitrary x, so:

    [itex]\Phi(gh) = \phi_{gh}(x) = (gh)^{-1}x(gh) = h^{-1}g^{-1}xgh = \phi_{h}(\phi_{g}(x))[/itex]
    [itex]= \phi_{h} \circ \phi_{g} (x) = \Phi(h) \circ \Phi(g)[/itex]

    But the operation is reversed for this situation. So this is considered to be an antihomomorphism.

    Any comments?
  2. jcsd
  3. Apr 27, 2013 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    In question (a), you probably meant "check that ##\phi_g## is an automorphism". Do you need help with (a)? You have only shown your work for (b).

    At the far left, you need to write ##\Phi(gh)(x)=\phi_{gh}(x)=##. The x appears out of nowhere in your calculation, and then it magically disappears at the end. So you need to do something similar at the end.

    I agree that we end up with ##\Phi(gh)=\Phi(h)\circ\Phi(g)##.
  4. Apr 27, 2013 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The multiplication on the Automorphism group is sometimes written backwards to accomodate this (for example on the wikipedia page about this homomorphism) - where
    [tex] \left(\phi_{g} \phi_{h}\right)(x) = \phi_h \left( \phi_g(x) \right) [/tex]
    so you should check to make sure that multiplication is composition read from right to left

    If this isn't what it's supposed to be then probably there's a type and the homomorphism is supposed to be
    [tex] g \mapsto \phi_{g^{-1}} [/tex]
  5. Apr 27, 2013 #4
    Nope, don't need help with part (a). Sorry to indicate that beforehand.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted