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Proving that Aut(K), where K is cyclic, is abelian
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[QUOTE="Mr Davis 97, post: 6072354, member: 515461"] So, we know that ##G## can act on ##K## by conjugation since ##K## is normal. Consider the permutation representation of this action, ##\varphi : G \to S_K##. Note that $$\ker (\varphi) = \{g\in G \mid \varphi (g) = \operatorname{id}\} = \{g\in G \mid \varphi (g)(k) = k,\forall k\in K\} = \{g\in G \mid gkg^{-1} = k,\forall k\in K\} = C_G(K).$$ If I can establish that the image of ##\varphi## is ##\operatorname{Inn}(K) \le \operatorname{Aut}(K)##, then by the first isomorphism theorem ##G/C_G(K) \cong \operatorname{Inn}(K) \le \operatorname{Aut}(K)##. But ##\operatorname{Aut}(K)## is abelian, so ##G/C_G(K) \cong \operatorname{Inn}(K) ## is also abelian, which means we're done I think. I'm just having a bit of trouble seeing that the image of ##\varphi## is ##\operatorname{Inn}(K)## for some reason. I see that ##\varphi (G) = \{\varphi (g) \mid g\in G\}##, but why is this set ##\operatorname{Inn}(K)##? It looks like this set is ##\operatorname{Inn}(G)##. EDIT: Actually, maybe the image of ##\varphi## doesn't have to be ##\operatorname{Inn}(K)##. Maybe it's just an unnamed subgroup of ##\operatorname{Aut}(K)##, in which case I would still be done. [/QUOTE]
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Proving that Aut(K), where K is cyclic, is abelian
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