Let $\mathbb{Z}/n\mathbb{Z}$, $\left(\mathbb{Z}/n\mathbb{Z}\right)^{\times}$, and $\mathcal{D}_n$ be the ring of integers modulo $n$, its multiplicative group of order $\varphi(n)$, and the set of divisors of $n$, respectively. The map $\phi: \left(\mathbb{Z}/n\mathbb{Z}\right)^{\times}\times\mathbb{Z}/n\mathbb{Z}\rightarrow\mathbb{Z}/n\mathbb{Z}$, given by $\phi(g,x):=gx$ is a left group action on $\mathbb{Z}/n\mathbb{Z}$. I am having a hard time in showing that $\phi$ decomposes $\mathbb{Z}/n\mathbb{Z}$ into $|\mathcal{D}_n|$-many distinct orbits given by $\left(\mathbb{Z}/n\mathbb{Z}\right)^{\times}\cdot\bar{\delta}$, where $\delta$ is a divisor of $n$. Let me give an example. If $n=12$, then $\mathbb{Z}/12\mathbb{Z}=\{\bar{0},\dots,\bar{11}\}$, $\left(\mathbb{Z}/12\mathbb{Z}\right)^{\times}=\{\bar{1}, \bar{5},\bar{7}, \bar{11}\}$, and $\mathcal{D}_{12}=\{1,2,3,4,6,12\}$ and we have: $$\left(\mathbb{Z}/12\mathbb{Z}\right)^{\times}\cdot\bar{1}=\{\bar{1}, \bar{5},\bar{7}, \bar{11}\},$$ $$\left(\mathbb{Z}/12\mathbb{Z}\right)^{\times}\cdot\bar{2}=\{\bar{2}, \bar{10}\},$$ $$\left(\mathbb{Z}/12\mathbb{Z}\right)^{\times}\cdot\bar{3}=\{\bar{3}, \bar{9}\},$$ $$\left(\mathbb{Z}/12\mathbb{Z}\right)^{\times}\cdot\bar{4}=\{\bar{4}, \bar{8}\},$$ $$\left(\mathbb{Z}/12\mathbb{Z}\right)^{\times}\cdot\bar{6}=\{\bar{6}\},$$ $$\left(\mathbb{Z}/12\mathbb{Z}\right)^{\times}\cdot\bar{0}=\{\bar{0}\},$$ and $$\bigcup_{\delta|12}\left(\mathbb{Z}/12\mathbb{Z}\right)^{\times}\cdot\bar{\delta}=\mathbb{Z}/12\mathbb{Z}.$$ I would be very grateful for any help or insights.

  • $\begingroup$ crossposted : mathoverflow.net/questions/395022/… $\endgroup$ – verret Jun 10 at 18:36
  • $\begingroup$ "... its multiplicative group of order $n$..." The multiplicative group is of order $\varphi(n)$ (Euler's phi), not $n$. $\endgroup$ – Arturo Magidin Jun 10 at 18:54
  • $\begingroup$ @ArturoMagidin Yes, of course. Thank you. $\endgroup$ – L. Cardoso Jun 10 at 18:57

Let $k\ge l$ be two positive integers and let $p$ be a prime number, then all integers $a\in\{1,\ldots,p^l-1\}$ coprime to $p^l$ are clearly coprime to $p^k$ as well, so the homomorphism $$(\mathbb{Z}/p^k\mathbb{Z})^\times\to\left(\mathbb{Z}/p^l\mathbb{Z}\right)^\times$$ is surjective. Note that for all pairs of coprime positive integers $m_1,m_2$, we have $$\left(\mathbb{Z}/m_1m_2\mathbb{Z}\right)^\times\cong \left(\mathbb{Z}/m_1\mathbb{Z}\right)^\times\times\left(\mathbb{Z}/m_2\mathbb{Z}\right)^\times.$$ It follows that for all positive integers $n$ and $d\mid n$, the natural map $$\left(\mathbb{Z}/n\mathbb{Z}\right)^\times \to \left(\mathbb{Z}/d\mathbb{Z}\right)^\times$$ is surjective. In particular, the kernel has cardinality $\varphi(n)/\varphi(d)$.

Now let $n$ and $d\mid n$ be positive integers and let $x\in \mathbb{Z}/n\mathbb{Z}$ of order $d$. Consider the action of $\left(\mathbb{Z}/n\mathbb{Z}\right)^\times$ on $\mathbb{Z}/n\mathbb{Z}$ and in particular the stabilizer of $x$. This stabilizer may be identified with all elements $g\in \{1,\ldots, n-1\}$ coprime to $n$ such that $gx=x$, which may be rewritten as $n\mid (g-1)x$, or $d\mid g-1$. These $g$ make up precisely the kernel of the aforementioned homomorphism $\left(\mathbb{Z}/n\mathbb{Z}\right)^\times\to\left(\mathbb{Z}/d\mathbb{Z}\right)^\times$.

We conclude that the cardinality of the stabilizer is $\varphi(n)/\varphi(d)$, and with the orbit-stabilizer theorem it follows that the orbit of $x$ consists of precisely $$\frac{\varphi(n)}{\varphi(n)/\varphi(d)}=\varphi(d)$$ elements. Finally, note that automorphisms preserve order, so the orbit of $x$ can only contain elements of order $d$, and there exist precisely $\varphi(d)$ elements of order $d$ in $\mathbb{Z}/n\mathbb{Z}$. We are done.

  • $\begingroup$ The relationship between $x$ and $d$ seems unclear to me; did you mean to choose $d$ to be the smallest integer with $dx \equiv 0 (mod n)$ (which is not hard to prove must be a divisor of $n$)? $\endgroup$ – user44191 Jun 10 at 20:55
  • $\begingroup$ @user44191 Yes, $d$ is the order of $x$ in $\mathbb{Z}/n\mathbb{Z}$. Thanks for pointing that out, I've edited my post. $\endgroup$ – Mastrem Jun 10 at 21:01
  • $\begingroup$ @Mastrem First of all, thank you for your contribution. I have a question though, for the natural homomorphism from $\left(\mathbb{Z}/n\mathbb{Z}\right)^{\times}$ to $\left(\mathbb{Z}/d\mathbb{Z}\right)^{\times}$, are you assuming $n$ and $d|n$ to be coprime? $\endgroup$ – L. Cardoso Jun 11 at 17:19
  • $\begingroup$ @L.Cardoso No, I am not. If that were the case, the proof of surjectivity would be much easier, since then $(\mathbb{Z}/n\mathbb{Z})^\times \cong \left(\mathbb{Z}/d\mathbb{Z}\right)^\times\times \left(\mathbb{Z}/(n/d)\mathbb{Z}\right)^\times$ $\endgroup$ – Mastrem Jun 11 at 17:29
  • $\begingroup$ @Mastrem OK. Running the risk of a stupid question, what is the natural homomorphism you used then? $\endgroup$ – L. Cardoso Jun 11 at 17:31

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