TITLE: Proving $\mathrm{Aut} \;\Bbb{Z}\cong \Bbb{Z_2}$
QUESTION [5 upvotes]: I have been asked to prove $\mathrm{Aut}\; \Bbb{Z}\cong \Bbb{Z_2}$. 
I figured that if I prove that $\mathrm{Aut}\; \Bbb{Z}$ has two elements, then I'm done. Clearly, $f(a)=a$ and $g(a)=-a$ can be those two automorphisms. Based on some rough calculations, I know there can't be any more automorphisms. Does this form a proof by any stretch of imagination? 
Thanks in advance!

REPLY [5 votes]: Indeed, if we set $G=\mathbb Z=\langle x\rangle$ so $G$ be infinite and if $f\in \operatorname{Aut}(\mathbb Z)$ then $f(x)$ must be a generator of $G$ itself. Now think of how many generators $G$ has? This is very interesting to me because an infinite group has a finite automorphism group.