Automorphisms

**topsquark** · March 28th 2011, 05:27 PM

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

**ojones** · March 28th 2011, 06:14 PM

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 $\mathbb{Z}$ ?

**topsquark** · March 29th 2011, 05:23 AM

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

**ojones** · March 29th 2011, 03:17 PM

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

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