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Sorry if this is an easy one, I'm a little rusty on my group theory. My first guess was that it's simply the inverse limit of the Aut($\mathbb{Z}/p^i\mathbb{Z})$, with the map when $i\leq j$ given by taking $\sigma\in$ Aut$(\mathbb{Z}/p^j\mathbb{Z})$ to the map $\tilde{\sigma}:\mathbb{Z}/p^i\mathbb{Z}\rightarrow\mathbb{Z}/p^i\mathbb{Z}$ defined by solving $\phi\circ\sigma=\tilde{\sigma}\circ\phi$ where $\phi:\mathbb{Z}/p^j\mathbb{Z}\rightarrow\mathbb{Z}/p^i\mathbb{Z}$ is the reduction map, but that seems too optimistic - I couldn't think of any reason $\tilde{\sigma}$ would be well-defined, much less be an automorphism of $\mathbb{Z}/p^i\mathbb{Z}$.

Also, barring a full answer to my question, I would be interested in whether Aut$(\mathbb{Z}_p)$ is a $p$-group. If not, what can we say about the elements $\sigma\in$ Aut$(\mathbb{Z}_p)$ with order a power of $p$?

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2 Answers 2

up vote 8 down vote accepted

First the $p$-adic integers are finitely generated (actually cyclic) pro-$p$ group therefore from a result of Serre all automorphisms are continuous. Now as it cyclic it is enough to see what happens to $1$. It has to go to another generator, i.e. any element of the form $a_0+a_1p+a_2p^2+\cdots$, where $0 \leq a_i < p$ for all $i$ and $a_0 \ne 0$. Hence, every element in $\mathbb{Z}_p$ is just multiplied by $a_0+a_1p+a_2p^2+\cdots$. Thus, the automorphism group is the multiplicative group of $\mathbb{Z}_p$.

It is of course not a $p$-group, but it contains a subgroup of index $p-1$ which is pro-$p$ and is actually again a cyclic pro-$p$ group, i.e. isomorphic to $\mathbb{Z}_p$ and thus have no elements of finite order.

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I'm not sure exactly which result of Serre you're referring to, but this is an elementary question so let's be more explicit. Let $G$ be a profinite group and let $f$ be an automorphism of the underlying abstract group. Then $f$ is a homeomorphism iff it is continuous iff $f^{-1}$ is continuous iff $f$ maps open subgroups to open subgroups. Since every group automorphism preserves the index of subgroups and any open subgroup has finite index, a suficient condition for every group automorphism of $G$ to be a homeo is that every finite index subgroup of $G$ is open. This is the case here... –  Pete L. Clark Jul 5 '10 at 3:09
Continued: it was rather recently proven that any finitely generated profinite group has the property that every finite index subgroup is open. I believe this has been known for pro-$p$-groups for a long time: that's probably related to the result of Serre that you're referring to. –  Pete L. Clark Jul 5 '10 at 3:11
Indeed, Serre proved that for a finitely generated pro-$p$ group a subgroup of finite index is open. Thus, the topology is determined by the algebra. Hence, every automorphism must be continuous. The questions whether every subgroup of finite index of a finitely generated profinite group is open was known as Serre's problem. It was fairly recently proved to have an affirmative answer by Nik Nikolov and Dan Segal. –  Yiftach Barnea Jul 5 '10 at 7:21
Surely the continuity of group automorphisms is obvious for $\mathbb Z_p$: the subgroups $p^n \mathbb Z_p$ are basic open neighbourhoods of the identity, and they are preserved by group homomorphisms. –  fherzig Jul 5 '10 at 8:16
fherzig, yes you are right the case of $\mathbb{Z}_p$ is triviel. I am not sure what is the origin of Serre's argument, I think it was in some letter, but I have a terrible memory so don't trust me too much. Luis Ribes knows the history of the problem, so if you are interested you can ask him. –  Yiftach Barnea Jul 5 '10 at 9:47

Any automorphism of $\mathbb{Z}_p$ preserves whether an element is divisible by $p^k$, so it is Lipschitz (in particular, continuous) with respect to the $p$-adic norm. On the other hand, any automorphism must preserve $\mathbb{Z}$, which is dense in $\mathbb{Z}_p$.

What I should've said is that any automorphism is determined by its behavior on $\mathbb{Z}$, hence by its behavior on $1$. To make up for that mistake, let me offer a sketch of the description of the structure of $\mathbb{Z}_p^{\ast}$. This group clearly splits up as the direct product of $(\mathbb{Z}/p\mathbb{Z})^{*}$ and the multiplicative group $U = 1 + p \mathbb{Z}_p$. It is now an interesting exercise to show that the exponential map $x \mapsto u^x, x \in \mathbb{Z}, u \in U$ extends to a map from $\mathbb{Z}_p$ to $U$ which, given the right choice of $u$, is an isomorphism for $p > 2$. For $p = 2$, Yiftach Barnea's otherwise excellent answer is slightly wrong and $U$ is in fact isomorphic to $\{ \pm 1 \} \times \mathbb{Z}_2$.

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The part of the answer about preserving Z seems to refer to automorphisms of the ring of p-adic integers, whereas the (title of the) question asks about automorphisms of the additive group. It seems to me that multiplication by any unit in Z_p is an automorphism of the additive group, and that these will be the only automorphisms. –  Andreas Blass Jul 5 '10 at 0:28
Thanks for pointing out my mistake. I always disliked even primes they are just troubles! –  Yiftach Barnea Jul 5 '10 at 9:39
Speaking of which: mathoverflow.net/questions/915/… –  Qiaochu Yuan Jul 5 '10 at 10:04

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