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

[A.Magnus]

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 , 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.

 

[MostlyHarmless]

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

 

[A.Magnus]

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!

 

[MostlyHarmless]

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)).

 

[A.Magnus]

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 $G$.

 

[MostlyHarmless]

Ah ok, well that's 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]

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 $\text{ker}(\phi)$?

 

[A.Magnus]

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.