----- > [!proposition] Proposition. ([[automorphism group of cyclic group]]) > Let $G \cong C_{n}$. The [[automorphism|automorphism group]] $\text{Aut}(G)$ is [[group isomorphism|isomorphic to]] $(\mathbb{Z} / n\mathbb{Z})^{\times}$. In particular, $|\text{Aut}(G)|=\phi(n)$, where $\phi$ denotes the [[Euler phi function]]. ^1d0443 > [!proof]- Proof. ([[automorphism group of cyclic group]]) > $G$ is [[cyclic group|cyclic]]; $\text{Aut}(G)$ is characterized by the number of ways a given [[generating set of a group|generator]] can be sent to another [[generating set of a group|generator]]. See [[automorphism|automorphism examples]]. For the [[cyclic group]] of [[order of a group|order]] $n$, an element is a [[generating set of a group|generator]] if and only if it has [[order of an element in a group|order]] $n$; therefore, there are precisely as many ways to do this as there are elements of order $n$ in $G$. By [[number of elements of each order in a cyclic group]], there are $\phi(n)$ elements of order $n$ in $G$; therefore, $|\text{Aut}(G)|=\phi(n)$. \ By construction, $|(\mathbb{Z} / n\mathbb{Z})^{\times}|=\phi(n)$. Thus $|\text{Aut}(G)|=|(\mathbb{Z}/n \mathbb{Z})^{\times}|$. \ There are $\phi(n)$ [[automorphism]]s of $G$, call them $\{ f_{i} \}_{i=1}^{\phi(n)}$. Each is determined by where it sends the [[generating set of a group|generator]] $a$ of $G$. Let $\varphi_{n}$ denote the ordered list of numbers less than $n$ [[relatively prime integers|coprime to]] $n$, indexed by $i$ as $\varphi_{n}^{(i)}$. For example, if $n=8$, then $\varphi _{n}=\{1, 3,5,7\}$ and $\varphi_{n}^{(2)}=3$. Now, we can label a given $f_{i}$ as the [[automorphism]] that maps $a \mapsto \varphi_{n}^{(i)}$. \ We claim that the map $\begin{align} \psi: \text{Aut}(G) \to & (\mathbb{Z} / n \mathbb{Z})^{\times} \\ \psi(f_{i}) := & \varphi_{n}^{(i)} \end{align}$ is the desired [[automorphism]]. $\psi$ respects the [[binary operation|group operation]] because, observing that $f_{i} \circ f_{j} (a)=a^{\varphi_{n}^{(i)}\varphi_{n}^{(j)}}$, we have $\begin{align} \psi(f_{i} \circ f_{j}) = \varphi_{n}^{(i)} \varphi_{n}^{(j)}=\psi(f_{i})\psi(f_{j}) \end{align}.$ Also observe that, by construction, the only element sent to $1$ by $\psi$ is $f_{1}$ (the identity [[automorphism]]). Therefore $\psi$ has a [[group homomorphism is injective iff kernel is trivial|trivial kernel and is thus injective]]. We already showed the two [[group]]s had the same size; hence $\psi$ is an [[group isomorphism|isomorphism]]. ^b05c31 ----- #### ---- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file[](group%20homomorphism%20is%20injective%20iff%20kernel%20is%20trivial%20iff%20is%20a%20monomorphism.md)er Reading" FROM outgoing([[]]) FLATTEN file.tags GROUP BY file.tags as Tag