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I have seen this question in I.N.Herstein, that if |G|>2 then G has a non-trivial automorphism.

The converse seems to be true. How to answer this question: That is G has a non-trivial automorphism, then prove that |G|>2.

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The contrapositive statement is "If |G| <= 2, then G has no automorphisms." From this perspective, it should be obvious. –  David Speyer Nov 4 '09 at 20:37
    
This is elementary enough that I don't think there should be any more discussion about it than there already has been (the question has been completely answered in both directions), so I'm closing the question. –  Anton Geraschenko Nov 4 '09 at 21:41
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closed as too localized by Anton Geraschenko Nov 4 '09 at 21:35

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1 Answer

up vote 5 down vote accepted

Which direction are you asking for a proof for?

The converse is very easy, as one only has to show that Z/2 (the only group of order 2) has no non-trivial automorphisms. This is obvious.

It's not too hard to prove the other direction either, but I can't tell whether you're interested in that or not.

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To clarify, a quick sketch of the other direction: Non-abelian groups are easily seen to have to have a non-trivial inner automorphism; abelian groups have the automorphism which takes x to -x. The latter is trivial iff your group is a vector space over Z/2Z, in which case you have a bunch of other automorphisms if it has dimension > 1. –  Harrison Brown Nov 4 '09 at 20:58
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