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$f$ is an automorphism of an infinite cyclic group $G$ then

1.$f^n\neq \Id_G$



if $f^n=\Id_G$ then every element of $G$ will have finite order but in an infinite cyclic group only identity element has finite order, same for 2, so $3$ is correct?

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Still no motivation nor explanation about what you tried... –  Did Jul 15 '12 at 15:26

1 Answer 1

Why would $f^n=\text{id}_G$ imply that every element of $G$ will have finite order? Just because $$f^n(a)=\underbrace{f(f(\cdots f}_{n\text{ times}}(a)))=a$$ does not mean that $a^n=a$.

Hint: An infinite cyclic group is isomorphic to $\mathbb{Z}$. Which individual elements of $\mathbb{Z}$ generate $\mathbb{Z}$? Any group homomorphism $f:\mathbb{Z}\to G$ is determined by what it does to generators. Where can a homomorphism $f:\mathbb{Z}\to\mathbb{Z}$ send a generator if it is to be an automorphism?

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$\{+1,-1\}$, ahh I see, –  El Angel Exterminador Jul 15 '12 at 15:26

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