Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


$f$ 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?

share|cite|improve this question
Still no motivation nor explanation about what you tried... – Did Jul 15 '12 at 15:26

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?

share|cite|improve this answer
$\{+1,-1\}$, ahh I see, – Un Chien Andalou Jul 15 '12 at 15:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.