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Let G be a finite group of order n. Must every automorphism of G have order less than n?

The question is a copy from this question. However, I do not have access to the reference given there as an answer so I would be grateful if someone can post a proof for this question here.

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A proof is also given as corollary 3.3 in Isaacs's Finite Group Theory: books.google.com/… . The non-trivial fact you need is theorem 2.20, which says if $A\lneq G$ is cyclic, then $[A:core(A)] < [G:A]$. – Steve D Aug 4 '12 at 14:41
It seems you can view the proof of Lucchini's theorem as well (it is long!): books.google.com/… – Steve D Aug 4 '12 at 14:45
@SteveD: Thanks. I got the book. I will study the proof and then write it down. As an aside optional part, it seems from the comments on the mathoverflow site that the proof by M V Horoševskiĭ uses very elementary arguments and I am guessing it is in the spirit of the Herstein's proof about existence of Sylow subgroups. Lets see if I see find that kind of proof. :-) – Jayesh Badwaik Aug 4 '12 at 15:32
Maybe I should clarify. The argument given by Isaacs is very elementary, it's just somewhat long. And I mean long by his standards, so basically a little more than a page. – Steve D Aug 4 '12 at 15:48
Okay! Thanks :-) I thought very elementary meant combinatorial style elementary. Thanks again. – Jayesh Badwaik Aug 4 '12 at 15:49

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