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January 7th, 2015, 11:20 AM  #1  
Newbie Joined: Jan 2015 From: Texas Posts: 4 Thanks: 0  Finite Group is Subgroup of Its Radical's Automorphism
I have been working on this problem on radical of finite group: Quote:
(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 subdefinitions, 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 nontrivial normal subgroup. It is further stated that $R(G)$ is nonablien. (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 counterintuitive 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.  
January 8th, 2015, 09:02 AM  #2 
Newbie Joined: Jan 2015 From: Texas Posts: 4 Thanks: 0  Solution to the problem
I received helps from a member of Math Stack Exchange "mesel" here. Many many thanks to "mesel"! Here is the linebyline 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. $\blacksquare$ 

Tags 
automorphism, finite, group, group theory, radical, subgroup 
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