Let $p$ be a prime number. Can you give me a few examples of automorphisms of $\Bbb Z_{p^\infty}$ other than the identity function? I'm looking for an elemetary way to construct them.
It can be proved that any homomorphism $f:\Bbb Z_{p^n}\to \Bbb Z_{p^m}$ can be extended to a homomorphism $f:\Bbb Z_{p^\infty}\to \Bbb Z_{p^\infty}$ but the proof uses axiom of choice and does not give any information about contructing a homomorphism.
Multiplication by $p$ is an endomorphism of $G = \mathbb{Z}_{p^{\infty}}$ which is locally nilpotent in the sense that for all $g \in G$ we have $p^n g = 0$ for some $n$ (possibly depending on $g$). In general, given a locally nilpotent endomorphism of an abelian group, we can make sense of a formal power series in that endomorphism acting on the abelian group since when applying such a formal power series to any particular $g \in G$, all but finitely many terms vanish. Explicitly, if $a_n, n \ge 0$ is a sequence of integers, then
$$g \mapsto \sum_{n \ge 0} a_n p^n g$$
is a well-defined endomorphism of $G$.
This means that $G$ (and more generally any abelian group all of whose elements are $p$-power torsion) admits a natural action of the $p$-adic integers. It follows that any unit in the $p$-adic integers gives an automorphism of $G$: more concretely, for any sequence $a_n, n \ge 0$ of integers such that $a_0 \neq 0 \bmod p$,
$$g \mapsto \sum_{n \ge 0} a_n p^n g$$
is a well-defined automorphism of $G$. In fact every automorphism has this form; one way to see this is to use the fact that the $p$-adic integers are the Pontryagin dual of $G$.
Here is another very familiar example of a locally nilpotent endomorphism: the endomorphism $\frac{d}{dx}$ acting on $k[x]$. It follows that we can apply arbitrary power series in $\frac{d}{dx}$ over $k$ to $k[x]$, and in particular if $k$ has characteristic zero we can apply the power series $e^{t \frac{d}{dx}}, t \in k$, which turns out to be translation by $t$ as you would expect.
This blog post is indirectly related.
