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$\newcommand{\Id}{\operatorname{Id}}$

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

1.$f^n\neq \Id_G$

2.$f^2=\Id_G$

3.$f=\Id_G$

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...2012-07-15

1 Answers 1

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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,2012-07-15