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This question already has an answer here:

Prove that every finite group having more than two elements has a nontrivial Automorphism.

It is from Topics in Algebra by Herstein. I am not able to solve.

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marked as duplicate by Douglas S. Stones, Thomas Andrews, lhf, Jack Schmidt, Start wearing purple May 14 '13 at 14:50

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  • $\begingroup$ try $x\mapsto a^{-1}xa$. $\endgroup$ – user59671 May 14 '13 at 14:29
  • $\begingroup$ Do you see how to solve this when some element has order $>2$? $\endgroup$ – Ted Shifrin May 14 '13 at 14:30
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    $\begingroup$ @CutieKrait, great unless $G$ is abelian ... :) $\endgroup$ – Ted Shifrin May 14 '13 at 14:32
  • $\begingroup$ @TedShifrin: then try $x\mapsto x^p$ for an appropriate $p$. $\endgroup$ – user59671 May 14 '13 at 14:42
  • $\begingroup$ @CutieKrait, great unless every element of $G$ has order $2$ ... :) $\endgroup$ – user1729 May 14 '13 at 17:31
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Suppose that $G$ is abelian with exponent greater than 2 (i.e. not every element has order 2). Then mapping $g\in G$ to $g^{-1}$ is an automorphism. It's clearly bijective, and since $G$ is abelian it is indeed a homomorphism.

If $G$ is abelian with exponent 2, then it is a vector space over $\mathbb Z/2$. Choose a basis for this vector space, and pick your favourite two basis vectors to interchange. This is a nontrivial automorphism.

If $G$ is not abelian, then the center of $G$, $Z(G)\neq G$. Now, let $G$ act on itself via conjugation. These are all automorphisms. Convince yourself that if $Z(G)\neq G$, there is a nontrivial one.

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