# Automorphisms

• Mar 28th 2011, 05:27 PM
topsquark
Automorphisms
This one is mostly a problem with the definitions, I think.

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

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

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

-Dan
• Mar 28th 2011, 06:14 PM
ojones
You need to show that these are the only automorphisms of $\displaystyle \mathbb{Z}$. Use the fact that automorphisms map generators to generators. Now, what are the generators of $\displaystyle \mathbb{Z}$?
• Mar 29th 2011, 05:23 AM
topsquark
Quote:

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

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 $\displaystyle \mathbb{Z}$ ( -1 and 1) there are only two automorphisms that are in $\displaystyle \text{Aut} \mathbb{Z}$. Since $\displaystyle \text{Aut} \mathbb{Z}$ is a group of two members it must be isomorphic to $\displaystyle \mathbb{Z}_2$.

-Dan
• Mar 29th 2011, 03:17 PM
ojones
Yes, this will do it provided you know how to fill in the details.