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I am still working on this problem on radical of finite group:

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

I have managed to parse the problem into these followings, consisting of what I know and what I don't know:

(1) In my class note the radical of finite group $R(G)$ is defined as: $$R(G) := E(G)F(G),$$ where $E(G)$ and $F(G)$ are called respectively the Layer and Fitting of $G$. Unfortunately, the definitions of $E(G)$ and $F(G)$ are long, winding and arduous chains of sub-definitions, which I believe won't be useful for solving this problem. Fortunately, the same text has established that $E(G)$ and $F(G)$ are normal subgroup of $G.$ Hence it is easy to conclude that $R(G) \lhd G$.

(2) The problem states that $R(G)$ is simple, meaning that $R(G)$ does not have any non-trivial normal subgroup. It is further stated that $R(G)$ is non-ablien.

(3) Now, $Aut(R(G))$ is group of isomorphisms from $R(G)$ to itself. I learn from Wikipedia here that conjugation is a handy example of automorphism. Therefore I visualize the $Aut(R(G))$ as follow as an aid in solving this problem: $$\begin{align} \varphi_g &: R(G) \to R(G) \qquad &&g \in G\\ &: r \mapsto r^g &&r \in R(G) \\ &: r \mapsto grg^{-1} \end{align}$$ Please correct me if there is any misstep in this visualization.

(4) Now, here comes the biggest challenge for me: How to prove $G$ is a subgroup of $Aut(R(G))$? Especially, how to prove, first and foremost, that $G$ is a subset of $Aut(R(G))$? To me at least, this proposition is counter-intuitive since $R(G)$ is normal subgroup of $G$. Did I miss anything here?

Any help or take from you would be most appreciated. Thank you for your time and help.

POST SCRIPT AFTER RESPONSE FROM "mesel" ~~~~~~~~~~~~~~
Thanks to "mesel" for his deep analysis, and here is the line-by-line analysis as I understood it. Any misunderstanding herein, 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. $\blacksquare$

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Lemma1: $C_G(R(G))\leq R(G)$. (property of $R(G)$)

Lemma2: If $N$ is a normal subgroup of $G$ then $C_G(N)$ is also a normal subgroup of $G$.

By lemma1,2 we can say that $C_G(R(G))$ is a normal subgroup of $G$ which is contained in $R(G)$.

But since $R(G)$ is simple, $C_G(R(G))=1$ or $C_G(R(G))=R(G)$.

ıf $C_G(R(G))=R(G)$ then $R(G)$ is abelian which is not the case.

Thus we have $C_G(R(G))=e$.

Now, Let $G$ act on $R(G)$ by conjugation then you will get homomorphism $\phi$ from $G$ to $Aut(R(G))$ with kernel $C_G(R(G))=e$. Then $\phi$ is an embedding as kernel is trivial. We are done.

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  • $\begingroup$ Thanks for your response! I can pretty much digest your analysis albeit slowly, but I have difficulties understanding the last paragraph. Do you mind if you elaborate it into line-by-line? Especially: (1) How do you know that $C_G(R(G))$ is the kernel of $G$? (2) What is an "embedding"? And finally, (3) How do you go from proposition that $\phi$ is an embedding to conclude that $G$ is subgroup of $Aut(R(G))$? $\endgroup$ – Amanda.M Jan 7 '15 at 20:45
  • $\begingroup$ @A.Magnus: If a group $G$ act on a normal subgroup $N$ by conjugation then $ker=\{g\in G| gng^{-1}=n \ for \ all \ n\}$. Notic that $Ker$ is the exactly $C_G(N)$. If you take $R(G)=N$ then the result fallows. When the kernel is trivial, $\phi$ is one to one homomorphism. Thus, $\phi(G)\leq Aut(G)$ and $G\cong \phi(G)$ which conclude the result. $\endgroup$ – mesel Jan 7 '15 at 21:30
  • $\begingroup$ Many thanks to you! I parsed out your analysis line-by-line in the Post Script above. But anyway, don't you think that $ker(\phi)$ should be $\{g \in G \mid gng^{-1} = e, \forall n \in N \}$ instead? Also, shouldn't it be $\phi(G) \leq Aut(R(G))$ instead? Thanks again. $\endgroup$ – Amanda.M Jan 8 '15 at 16:59
  • $\begingroup$ @A.Magnus: The kernel is consisting of elements beheving likes identity function. Thus, $Ker(\phi)=\{g\in G|gng^{-1}=n \}$. (I think you confused the term behaving like identity function and sending elements to identity). By the way, You are welcome. $\endgroup$ – mesel Jan 8 '15 at 17:29

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