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January 7th, 2015, 11:20 AM   #1
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Finite Group is Subgroup of Its Radical's Automorphism

I have been working on this problem on radical of finite group:

Quote:
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 would like to 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.
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January 8th, 2015, 09:02 AM   #2
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Solution to the problem

I received helps from a member of Math Stack Exchange "mesel" here. Many many thanks to "mesel"! Here is the line-by-line analysis as I understood it, hope it will be useful to some of you. 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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