A cyclic group of order $p^n$, $n\geq 1$, always has an automorphism of order $p-1$ (well known).

Let $C_{p^n}=\langle x\colon x^{p^n}=1\rangle$, and $H=\langle x^p\rangle$, the unique subgroup of index $p$. I wanted to know how an automorphism of order $p-1$ permutes the elements of $C_{p^n}$. My guess was following:

$H$ has $p$ cosets in $G$: $$H, xH, x^2H, \cdots, x^{p-1}H.$$ Every automorphism will take trivial coset $H$ to itself (sine it is characteristic subgroup). Hence any automorphism will permute the remaining cosets. Then, the behavior of automorphism of order $p-1$ would be permuting remaining $p-1$ cosets cyclically.

Question: Does there exists an automorphism $\sigma$ of order $p-1$ which permutes the cosets in the manner

$$ xH\mapsto x^2H \mapsto \cdots \mapsto x^{p-1}H \mapsto xH$$


Your claim is already not true for $n=1$. In that we think of $C_p$ as the additive group of the field $\mathbb{F}_p$ with $p$ elements and the automorphisms are just multiplication by nonzero scalars. The orbit of an automorphism will then be $1\mapsto \zeta\mapsto \zeta^2 \mapsto \zeta^3 \mapsto \cdots \mapsto \zeta^{p-2}\mapsto 1$, which will not be $1\mapsto 2\mapsto 3 \cdots$ unless $p=3$.

You do however get an orbit of size $p-1$, and this can be lifted to $p^n$. In general, the automorphisms of $\mathbb{Z}/p^n$ are identified with $(\mathbb{Z}/p^n)^\times$. It follows that all you need to do is to find some $\zeta_n\in(\mathbb{Z}/p^n)^\times$ such that $\zeta_n \equiv_p \zeta$ and $\zeta_n^{p-1}-1\equiv_{p^n}0$. In other words, you want to lift the solution from modulo $p$ to modulo $p^n$ which is exactly what Hensel's lemma is for.


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