Automorphism Group of Radical of Finite Group

[A.Magnus]

I am working on a problem on automorphism group of radical of finite group like this one:

Assume that $R(G)$ is simple and not commutative, show that $G$ is a subgroup of $Aut(R(G)).$

Here are what I know and what I don't know:

$Aut(R(G))$ is an automorphism group, whose elements consist of isomorphic mappings from $R(G)$ to itself. For visualization purpose, I envision the following example, where $g \in G, r \in R(G)$:

$\begin{align} (\phi_g) \in Aut(R(G)), \quad \phi_g &: R(G) \to R(G), \\ &: r \mapsto r^g \\ &: r \mapsto grg^{-1} \end{align}$
​

Please correct me if I was wrong. But after this, I am stuck on how to prove that $G$ is subgroup of $Aut(R(G)),$ for to me it is like relating apple to orange. Perhaps this is because I failed to understand what the radical of finite group $R(G)$ stands for. For your info, in the class note the definition of $R(G)$ is a long and winding chain that goes like these:

(a) $R(G) := E(G)F(G);$
(b) Where $F(G)$ is defined to be the (complex) product of all subgroups $O_p(G)$ with $p$ a prime number, this group $F(G)$ is called the Fitting subgroup of $G;$
(c) And $E(G)$ is defined to be the subgroup of $G$ generated by all components of $G,$ this group $E(G)$ is called the Layer of $G$. And then another chain of definitions: A subnormal subgroup of $G$ is called a component of $G$ if it is quasisimple; the group $G$ is called quasisimple if $G′ = G$ and $G/Z(G)$ is simple.​

I think these are too long to be useful in solving this problem, what I am looking for is a working definition, or the salient property of $R(G)$ to solve this problem. I would therefore appreciate any help or hints in solving this problem. Thanks for your time and help.

 

[A.Magnus]

I received a generous help from a member of Math Stack Exchange "mesel" here, many many thanks to "mesel" for his deep analysis! And here is the line-by-line analysis as I understood it. Any misunderstanding that I bring in, though, will completely be mine.

(1) From the Fundamental Lemma of Finite Group Theory, we have $C_G(R(G)) \subseteq R(G)$, and from that $C_G(R(G)) \leq R(G)$ easily follows. Since $C_G(R(G))$ is normal, therefore it is the normal subgroup of $R(G)$.

(2) But the question states that $R(G)$ is simple, meaning $C_G(R(G))$ is either $R(G)$ itself or $e$.

(3) If $R(G) = C_G(R(G))$, then $R(G)$ must be abelian which violates the premise given by the problem, therefore $C_G(R(G)) = e$.

(4) Notice that $C_G(R(G)) = \{g \in G \mid gr = rg, \forall r \in R(G) \}$, implying that

$\begin{align} gr &= rg \\ g &= rgr^{-1} \\ &= e. \\ \end{align}$​

(5) Let $\phi$ be a homomorphism from $G$ to $Aut(R(G))$, where the automorphism is a conjugation:

$\begin{align} \phi &: G \to \underbrace{(R(G) \to R(G))}_{Aut(R(G)} \\ &: \underbrace {g}_{= \ e} \mapsto (r \mapsto \underbrace {rgr^{-1}}_{= \ e}) \qquad \qquad \forall r \in R(G), \\ \end{align}$​

which implies that $\phi$ is monomorphism.

(6) Because of the injective homomorphism above, $G \cong \phi (G)$, and since $\phi(G) \leq Aut(R(G))$, therefore we conclude that $G$ is subgroup of $Aut(R(G))$ as required. $\blacksquare$