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 $n = 20$:
$U_{20}(2) = C^\times_{20} = C_4\times C_2$
$U_{20}(4) \cong C_4$
$U_{20}(5) \cong U_{20}(10) \cong C_2$
Best Answer
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?