Why does the set of all automorphisms on the additive group of integers contain only 2 elements?

Question

I believe that $f(x)=x$ and $f(x)=-x$ are the only two automorphisms.

But the set of automorphisms on $\mathbb{Z}_n$ is isomorphic to $U_n$. So as $n$ increases, the number of automorphisms on $\mathbb{Z}_n$ increases. However, for the entire set of integers, there are only two automorphisms.

I want to understand why this changes at infinity. Why isn't there a limiting property?

Answer 1

Jose Carlos Santos has explained why the automorphism group of $\mathbb{Z}$ is very constrained, while that of $\mathbb{Z}_n$ is not as restricted, leading to more possibilities.

Paraphrasing US Supreme Court justices, I write separately to emphasize that I don’t think it is conceptually a good idea to think of $\mathbb{Z}$ as being a sort of “limit” of $\mathbb{Z}_n$ as $n$ gets larger. It just doesn’t.

First, if you think of $\mathbb{Z}_n$ as $\mathbb{Z}/n\mathbb{Z}$, then $\mathbb{Z}\cong\mathbb{Z}/0\mathbb{Z}$, so in fact $\mathbb{Z}$ isn’t what happens when “$n$ gets larger”, it’s what happened when $n$ was a small as possible.

Second, you don’t have (nontrivial) maps from $\mathbb{Z}_n$ into $\mathbb{Z}$: you only have maps going the other way. If you think of group morphisms as going “left to right”, like most of our function arrows, the $\mathbb{Z}$ is the leftmost group among the cyclic groups, not the rightmost. So there is not good way to think of $\mathbb{Z}$ as the “limiting group” of the $\mathbb{Z}_n$.

Thirdly, you want to be careful with how you think of the $\mathbb{Z}_n$ “fitting inside each other”. Note that $\mathbb{Z}_m$ has a (unique) subgroup isomorphic to $\mathbb{Z}_n$ if and only if $n|m$. So it’s not a nice straight progression, but rather a somewhat complicated arrangement of these groups as they fit inside each other; what you get is what is called a “directed partially ordered set”.

Now, there is a way to try to figure out what the “limiting group” of the $\mathbb{Z}_n$ is; it’s called a direct limit. If you do that to the finite cyclic groups, you don’t get $\mathbb{Z}$; instead, you get $\mathbb{Q}/\mathbb{Z}$, a very different group. And this group does have lots of automorphisms! You can decompose it into a direct product of Prüfer $p$-groups, and each of those has at least one automorphism for each unit of the $p$-adic integers (see here).

Answer 2

If $f$ is an automorphism of $\Bbb Z$, then there is some $m\in\Bbb Z$ such that $f(m)=1$, and therefore $mf(1)=1$. But this is possibly only if $m=\pm1$. So, either you have $f(1)=1$ or $f(-1)=1(\iff f(1)=-1)$.

But if you are working with $\Bbb Z_n$, then there is no this restriction that $mf(1)=1\iff m=\pm1$.

Answer 3

All these groups are 1-generated. Thus, the image of a generator must be a generator. (Isomorphisms preserve such things.) But $\Bbb Z$ has only two generators, whereas $\Bbb Z_n$ can have many. In fact, the number of automorphisms of $\Bbb Z_n$ equals the number of its generators.