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Automorphism Group of Radical of Finite Group
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[QUOTE="A.Magnus, post: 4969559, member: 531393"] I received a generous help from a member of Math Stack Exchange "mesel" [URL='http://math.stackexchange.com/questions/1095533/finite-group-is-subgroup-of-its-radicals-automorphism'][I]here[/I][/URL], 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 [CENTER]##\begin{align} gr &= rg \\ g &= rgr^{-1} \\ &= e. \\ \end{align}##[/CENTER] (5) Let ##\phi## be a homomorphism from ##G## to ##Aut(R(G))##, where the automorphism is a conjugation: [CENTER]##\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}##[/CENTER] 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## [/QUOTE]
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Automorphism Group of Radical of Finite Group
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