# Showing $G/Z(R(G))$ isomorphic to $Aut(R(G))$

I am working on this problem with lots of nesting definitions:

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. Here is what I would like to get help from you:

Let $\varphi : G/Z(R(G)) \to Aut(R(G))$, could you please structure this $\varphi$ into mappings like (1) and (2), so that at least I have a visual idea?

Thank you for your time and help.

• Are you aware of the theorem that $C_G(R(G)) \le Z(R(G))$? You need that, because $C_G(R(G))$ is exactly the kernel of the map $\alpha:N_G(R(G)) \to {\rm Aut}(R(G))$. Jan 23, 2015 at 9:01
• The result you need is that if $f:G \to H$ is a group homomorphism, then $f$ induces a homomorphism $\bar{f}:G/\ker (f) \to H$. (That's essentially the 1st Isomorphism Theorem.) Jan 23, 2015 at 9:04