Let $σ ∈ Aut(G)$, where $G$ is a nonotrivial finite group. Then show that the order $o(σ)$ of $σ$ is less than $|G|$.
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1What've you tried? – Arpit Kansal Sep 19 '15 at 11:54
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2This is Horosevskii's theorem, and is not trivial. There is a proof in Finite Group Theory by I. Martin Isaacs. See http://math.stackexchange.com/a/190199/589 and http://mathoverflow.net/questions/39294/non-trivial-consequences-of-baers-theorem-and-lucchinis-theorem-in-subnormalit. – lhf Sep 19 '15 at 12:03
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You should give more detaisl about question and what you have tried. The proof is not so straightforward. Sketch of the proof;
Let $\Gamma$ be semidirect product of $A=<\sigma>$ and $G$.
Claim1: $Core_{\Gamma}(A)=e$.
Notice that $G\cap Core_{\Gamma}(A)=e$ and they are both normal in $\Gamma$. Hence they commutes with each other. But if $e\neq a\in Core_{\Gamma}(A)$ then $a$ commutes with elements of $G$, which means $a$ acts trivially on $G$, which is not possible as $a\in Aut(G)$.
Theorem: Let $\Gamma$ be a finite group and $A$ be a cyclic subgroup of $G$. Then $|A:Core(A)|<|{\Gamma}:A|$.
Now, By appliying the theorem, we get $|A|<|G|$.
