Automorphism Group of Radical of Finite Group

In summary, the conversation discusses a problem regarding the automorphism group of the radical of a finite group. It is known that the automorphism group is a group of isomorphic mappings from the radical to itself. The radical is defined as a long chain of subgroups, including the Fitting subgroup and the Layer of the group. The problem is to prove that the group is a subgroup of the automorphism group, and this is achieved through a monomorphism from the group to the automorphism group. The proof involves showing that the centralizer of the radical is trivial, and then constructing an injective homomorphism from the group to the automorphism group.
  • #1
A.Magnus
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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.
 
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  • #2
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##
 

Related to Automorphism Group of Radical of Finite Group

What is the Automorphism Group of Radical of Finite Group?

The Automorphism Group of Radical of Finite Group is a group of automorphisms (isomorphisms from a group to itself) that preserve the radical of a finite group. The radical of a group is the intersection of all of its maximal normal subgroups. This group is often denoted as Aut(R(G)), where G is the finite group and R(G) is its radical.

What is the significance of the Automorphism Group of Radical of Finite Group?

The Automorphism Group of Radical of Finite Group is important in the study of group theory because it provides a way to analyze the structure of a finite group and its normal subgroups. It can also be used to classify groups and understand their properties.

How is the Automorphism Group of Radical of Finite Group related to the center of a group?

The center of a group is a subgroup that contains all the elements that commute with every element in the group. The Automorphism Group of Radical of Finite Group is closely related to the center of a group because it is a subgroup of the center and contains all the automorphisms that preserve the radical of the group.

What is the order of the Automorphism Group of Radical of Finite Group?

The order of the Automorphism Group of Radical of Finite Group is equal to the order of the radical of the finite group. This is because the group consists of automorphisms that preserve the radical, so the number of elements in the group will be equal to the number of elements in the radical.

How can the Automorphism Group of Radical of Finite Group be calculated?

The Automorphism Group of Radical of Finite Group can be calculated by finding all the automorphisms that preserve the radical of the finite group. This can be done by analyzing the structure of the group and its normal subgroups. It can also be calculated using computational methods such as group theory software.

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