If $f:\mathbb{Z} \to \mathbb{Z}$ is an isomorphism, prove that $f$ is the identity map.

Question

I am a little baffled by this question. Is it safe to assume that since $f$ is an isomorphism, $f (1) = 1$ ? And, if it is safe to assume this, could I construct a proof by induction, by using the fact that $f(1) =1$,$f(-1)=-f(1)$, $f(1+1) = f(1) +f(1)$, .... etc. to prove the claim? Thanks for your time as always guys!

Answer 1

I will assume that you mean an automorphism in the additive group of the integers. In this case what you are trying to prove is false.

You can prove by induction (first with the case $n\ge 0$ and then use $f(-n)=-f(n)$) that $f(n)=nf(1)$ for all integer $n$. Thus if we want a surjective map we need $nf(1)=1$ for some integers $n$ and $f(1)$ hence $f(1) \in \{-1,+1\}$. It is easy to see that in these two cases we indeed get an isomorphism so it is not unique.