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

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.

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## Answers and Replies

Is N(H) supposed to be the normalizer, and Z(R(G)) supposed to be the center of R(G)?

Is N(H) supposed to be the normalizer, and Z(R(G)) supposed to be the center of R(G)?
Yes, that is correct. ##Z## and ##N## stand for centralizer and normalizer. Thanks!

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)).

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)).

Oops! I made careless typo. Center ##Z(G) ## have the elements of ##G## that commute with every element in ##G##.

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.

jbunniii
Homework Helper
Gold Member
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)##?

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First let's address the normality of the various subgroups ...

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.