3
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.