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The following question is driving me bananas.
I am given a split extension
$0 \to \mathbb{Z}/n\mathbb{Z} \to (\mathbb{Z}/m\mathbb{Z})\ltimes (\mathbb{Z}/n\mathbb{Z})\to \mathbb{Z}/m\mathbb{Z}\to 0,$
with $m,n>2$ natural numbers, and an element $x\in \mathbb{Z}/n\mathbb{Z}$ satisfying $x^m=1$ but $x^j\neq 1$ for $j < m$. Multiplication by $x$ corresponds to the conjugation action of $\mathbb{Z}/m\mathbb{Z}$. I know also that $ 1 - x $ is a unit. I inject $\mathbb{Z}/n\mathbb{Z}$ into $\mathbb{Z}/n^2\mathbb{Z}$ by mapping the generator $1\in \mathbb{Z}/n\mathbb{Z}$ to $1\in \mathbb{Z}/n^2\mathbb{Z}$, then $2\in \mathbb{Z}/n\mathbb{Z}$ to $2\in \mathbb{Z}/n^2\mathbb{Z}$, etc. Call this mapping ι.

What is the order of $(\iota x)^m-1$ in $\mathbb{Z}/n^2\mathbb{Z}$?

Based on naive calculations and on an optimistic temprament, I'm tempted to guess that it is exactly $n$. A certain topological quantity which I'm working with is naturally an element of the ideal generated by $\frac{(\iota x)^m-1}{n}\bmod n$ in $\mathbb{Z}/n\mathbb{Z}$ (I think), and it would be so much more elegant if I knew this ideal to be all of $\mathbb{Z}/n\mathbb{Z}$.

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    $\begingroup$ If I understand the notation and question correctly, $x=2, n=3511, m=ord_n(x)=1755=ord_{n^2}(x)$ is a counter example to your guess. See wikipedia for more on Wiefrich primes and generalizations. $\endgroup$
    – Dror Speiser
    Commented Oct 24, 2010 at 22:08
  • $\begingroup$ I inject $\mathbb{Z}/n\mathbb{Z}$ into $\mathbb{Z}/n^2\mathbb{Z}$ by mapping the generator $1\in \mathbb{Z}/n\mathbb{Z}$ to $1\in \mathbb{Z}/n^2\mathbb{Z}$, then $2\in \mathbb{Z}/n\mathbb{Z}$ to $2\in \mathbb{Z}/n^2\mathbb{Z}$, etc. Call this mapping ι. Is this an embedding of groups ? $\endgroup$
    – Chandan Singh Dalawat
    Commented Oct 25, 2010 at 5:13
  • $\begingroup$ @Chandan Sinh Dalawat: $\{0,1,\ldots,n-1\}$ isn't a subgroup of $\mathbb{Z}/n^2\mathbb{Z}$ under addition. This is just an injection. @Dror Thanks! That sucks though... $\endgroup$
    – Daniel Moskovich
    Commented Oct 25, 2010 at 12:05

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Dror Speiser provides a counterexample in a comment:

$x=2$, $n=3511$, $m= \mathrm{ord}_n(x)=1755= \mathrm{ord}_{n^2}(x)$. See wikipedia for more on Wiefrich primes and generalizations.
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  • $\begingroup$ @DrorSpeiser's comment. $\endgroup$
    – LSpice
    Commented May 24, 2021 at 14:20

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