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Recall that the automorphisms of $C_n$ are in 1-1 correspondence with the positive integers not exceeding $n$ which are relatively prime to $n$. For the automorphism $\psi_r$ determined by the relative prime $r$, refer to $r$ as the index of $\psi_r$. Recall also that for any $m$ dividing $n$ there's exactly one isomorphic copy of $C_m$ in $C_n$ generated by e.g. $\frac{n}{m}$. One can show easily that the elements of $\textrm{Aut}(C_n)$ fixing $C_m$ element-wise are precisely those automorphisms with index $r$ having the additional property that

$r\equiv 1\,\textrm{mod }m$.

To examine the structure of such automorphisms, then, define

$U_n(m) := \{q\in C^\times_n \,\,|\,\, q\equiv 1\,\textrm{mod }m\}$

forming a subgroup of the multiplicative group $C^\times_n$ of integers modulo $n$. In what circumstances is it cyclic?

For example, with $n = 20$:

$U_{20}(2) = C^\times_{20} = C_4\times C_2$

$U_{20}(4) \cong C_4$

$U_{20}(5) \cong U_{20}(10) \cong C_2$

Justin Greenough
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    Here is an interesting related question: https://math.stackexchange.com/q/756247/368269 – Dennis Muhonen Aug 12 '19 at 19:52
  • In general I'm interested in the structure of $U_n(m)$, not merely in the case where it's cyclic. – Justin Greenough Aug 12 '19 at 20:00
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    Why is this question put on hold? I don't understand. – Justin Greenough Aug 12 '19 at 20:07
  • see https://math.stackexchange.com/questions/314846/for-what-n-is-u-n-cyclic?rq=1 – Hao S Aug 12 '19 at 20:35
  • @Justin , might be because this could be an undergrad assignment ? Without context, we don't know the difference. –  Aug 12 '19 at 21:26
  • @HaoS Yes that is the posting about the structure of the group of units as a finite abelian group but my question is somewhat different. – Justin Greenough Aug 12 '19 at 21:30
  • @RoddyMacPhee I understand, though this is in fact not an undergrad assignment. Your comment made me chuckle. – Justin Greenough Aug 12 '19 at 21:31
  • You don't have enough reputation to edit do you ? if so please do. If not provide the context in a comment, and someone might be able to be nice enough to add it to the question. –  Aug 12 '19 at 21:33
  • Why is my question on hold? Just no idea. – Justin Greenough Aug 12 '19 at 22:09
  • @JustinGreenough did you read the post and the relation to chinese remainder theorem? I think via such a method you get that it's only cyclic if your subgroup is {1}x {1}x ... x {1} x Z_p that is it's just one cartesian product which only works if n/m is 2 times a prime power right? or am I missing something? – Hao S Aug 12 '19 at 22:10
  • @HaoS why not answer the question? – Justin Greenough Aug 12 '19 at 22:14
  • because on hold means they can't except comments. –  Aug 12 '19 at 23:47
  • Well how can we remove the 'on hold' status? Admins here seem quick to judge and slow to forgive. – Justin Greenough Aug 13 '19 at 00:43
  • Twice in the last 24 hours, you have deleted your question, Justin, and then undeleted it. What kind of games are you playing? – Gerry Myerson Aug 13 '19 at 22:54
  • Well there are two copies of this question and I was unsure about which I should delete. Is there a problem? Have I violated your rather strict protocol here? I would be happier if you were to spend your time trying to help me by answering my question than belaboring this or that triviality. – Justin Greenough Aug 14 '19 at 22:08
  • I've been here for ten years, Justin, and you've been here for ten minutes. Please don't tell me what's a triviality, or what to do with my time. – Gerry Myerson Aug 15 '19 at 00:07
  • Also, the only question you ask, Justin, is, "In what circumstances is it cyclic?" but then in the comments you write, "In general I'm interested in the structure of $U_n(m)$, not merely in the case where it's cyclic." And then you accept an "answer" which does nothing but find the order of $U_n(m)$. So it's not the least bit clear what you really want to know. – Gerry Myerson Aug 15 '19 at 00:14

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Define $\gamma:U_n\rightarrow U_n$ by $x\mapsto x_m$ where $x\equiv x_m \,\,\textrm{mod } m$. Then $\gamma$ is clearly a homomorphism of groups with image yielding a complete residue system modulo $m$ since $m|n$ means that everything relatively prime to $n$ is also relatively prime to $m$. Note that $U_n(m) = \textrm{Ker}(\gamma)$, hence \begin{equation*} U_n/U_n(m)\simeq U_m. \end{equation*} Thus one can say that the order of $U_n(m)$ is $\varphi(n)/\varphi(m)$. Is this right?

Justin Greenough
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