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How to show G/Z(R(G)) is isomorphic to Aut(R(G))?

  1. Jan 31, 2015 #1
    I am working on this problem with lots and lots of nesting definitions like this following, and I have been trying to get help from here as well as http://www.quora.com/How-do-I-prove-G-Z-R-G-is-isomorphic-to-Aut-R-G [Broken], but none gave me complete help:

    Show that ##G/Z(R(G))## is isomorphic to a subgroup of ##Aut(R(G))##.​

    For your info, ##R(G)## is called the Radical of ##G##, defined as ##R(G) := E(G)F(G)##, where ##E(G)## is called the Layer of ##G##, and ##F(G)## is called the Fitting of ##G##. Long story short, it suffices (I think) to say that since both ##E(G)## and ##F(G)## are normal according to a lemma, therefore ##R(G)## is normal too.

    And then the problem comes with a hint: Use Exercise 3.4. This exercise has the following mappings:

    ##\begin{align}
    \alpha \ &: \ N_G(H) \to Aut(H), \quad g \mapsto \alpha_g \tag{1}\\
    \text{where} \ \alpha_g \ &: \ H \to H, \quad h \mapsto h^g, \tag{2}
    \end{align}##​

    therefore I think that this hint is directing me to use conjugate automorphism.

    I was thinking about letting ##\varphi : G/Z(R(G)) \to Aut(R(G))##, and thenstructuring this ##\varphi## into mappings like (1) and (2), so that at least I have a visual idea. But after getting limited responses from my earlier postings, I am not so sure that will be the right direction. Any help or hints would be very, very much appreciated.

    Thank you for your time and help.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Jan 31, 2015 #2
    Is N(H) supposed to be the normalizer, and Z(R(G)) supposed to be the center of R(G)?
     
  4. Jan 31, 2015 #3
    Yes, that is correct. ##Z## and ##N## stand for centralizer and normalizer. Thanks!
     
  5. Jan 31, 2015 #4
    Wait, the center or the centralizer?

    Either way, I'm not familiar with the Radical, but I would try proving that the map that they gave is an automorphism(replacing H with R(G), and then show that the G/(Z(R(G)) is isomorphic to N(R(G)).
     
  6. Jan 31, 2015 #5
    Oops! I made careless typo. Center ##Z(G) ## have the elements of ##G## that commute with every element in ##G##.
     
  7. Jan 31, 2015 #6
    Ah ok, well thats about as much as I can offer on this one. I would start by redfining that map with R(G) in place of H and prove that it is an automorphism.
     
  8. Jan 31, 2015 #7

    jbunniii

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    First let's address the normality of the various subgroups. Even without knowing the definitions of ##E##, ##F##, and ##R##, it's a solid bet that all three are not only normal but characteristic subgroups. As Isaacs points out in his Algebra text, if a subgroup is prefixed with "the", then it's generally characteristic. E.g. the center of ##G##, the layer of ##G##, the radical of ##G##, the Fitting subgroup of ##G## are all characteristic subgroups of ##G##. (One can, of course, show this rigorously in each specific case.) Using this reasoning, ##Z(R(G))## is the center of ##R(G)##, which is the radical of ##G##, so ##Z(R(G))## is characteristic in ##R(G)##, which is characteristic in ##G##. So ##Z(R(G))## is normal in ##G## and ##G/Z(R(G))## is a group.

    The natural mapping to consider is ##\phi : G \to Aut(R(G))## such that ##\phi(g) : R(G) \to R(G)## is conjugation by ##g##. This mapping makes sense because ##R(G) \lhd G##, so the restriction of conjugation to ##R(G)## is an automorphism of ##R(G)##. What is the kernel of ##\phi##? By definition, it is the set of all ##g\in G## such that ##\phi(g)## is the identity map on ##R(G)##. What does that mean in this case? (Hint: it's a centralizer.)

    By the first isomorphism theorem, ##G/\text{ker}(\phi)## is isomorphic to ##\phi(G)##, which is a subgroup of ##Aut(R(G))##.

    Now what's the relation between ##Z(R(G))## and ##\text{ker}(\phi)##?
     
    Last edited: Jan 31, 2015
  9. Jan 31, 2015 #8
    I am posting this message just to acknowledge your response. Thank you. Give me sometime to digest and I will get back with you shortly. Thanks again as always.
     
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