# Automorphism Group of Radical of Finite Group

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.

Last edited:

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##