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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. 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 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. $\blacksquare$ Tags automorphism, finite, group, group theory, radical, subgroup Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Archykhain Abstract Algebra 0 November 23rd, 2011 04:29 PM johnmath Abstract Algebra 1 March 1st, 2011 08:34 AM envision Abstract Algebra 3 October 4th, 2009 10:37 PM envision Abstract Algebra 1 October 4th, 2009 03:24 AM mlchai Abstract Algebra 1 August 21st, 2009 01:06 PM

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