1. Automorphisms

This one is mostly a problem with the definitions, I think.

I am asked to prove that $\text{Aut} \mathbb{Z}$ is isomorphic to $\mathbb{Z}_2$, where $\text{Aut} \mathbb{Z}$ is the group of all automorphisms of $\mathbb{Z}$.

Obviously the identity function $I: \mathbb{Z} \to \mathbb{Z}: x \mapsto x$ is in $\text{Aut} \mathbb{Z}$. The only other automorphism I can come up with is $f: \mathbb{Z} \to \mathbb{Z}: x \mapsto -x$. etc, etc. to finish showing the isomorphism between $\text{Aut} \mathbb{Z}$ and $\mathbb{Z}_2$.

Are there truly only two members in $\text{Aut} \mathbb{Z}$? It seems there should be more, but I can't find any.

-Dan

2. You need to show that these are the only automorphisms of $\mathbb{Z}$. Use the fact that automorphisms map generators to generators. Now, what are the generators of $\mathbb{Z}$?

3. Originally Posted by ojones
You need to show that these are the only automorphisms of $\mathbb{Z}$. Use the fact that automorphisms map generators to generators. Now, what are the generators of $\mathbb{Z}$?
Thank you. I was unaware of that fact. (Edit: It's pretty obvious, actually, now that I've had some time to think about it.) To finish the argument then, since there are only two generators of $\mathbb{Z}$ ( -1 and 1) there are only two automorphisms that are in $\text{Aut} \mathbb{Z}$. Since $\text{Aut} \mathbb{Z}$ is a group of two members it must be isomorphic to $\mathbb{Z}_2$.

-Dan

4. Yes, this will do it provided you know how to fill in the details.