automorphism group of a cyclic group

Theorem 1.

The automorphism group of the cyclic group $\mathrm{Z}\mathrm{/}n\mathit{}\mathrm{Z}$ is ${\mathrm{(}\mathrm{Z}\mathrm{/}n\mathit{}\mathrm{Z}\mathrm{)}}^{\mathrm{\times }}$, which is of order $\varphi \mathit{}\mathrm{(}n\mathrm{)}$ (here $\varphi$ is the Euler totient function).

Proof.

Choose a generator $x$ for $\mathbb{Z}/n\mathbb{Z}$. If $\rho \in \mathrm{Aut}(\mathbb{Z}/n\mathbb{Z})$, then $\rho (x)=x^a$ for some integer $a$ (defined up to multiples of $n$); further, since $x$ generates $\mathbb{Z}/n\mathbb{Z}$, it is clear that $a$ uniquely determines $\rho$. Write ${\rho }_a$ for this automorphism. Since ${\rho }_a$ is an automorphism, $x^a$ is also a generator, and thus $a$ and $n$ are relatively prime. Clearly, then, every $a$ relatively prime to $n$ induces an automorphism. We can therefore define a surjective map

$$\mathrm{\Phi }:\mathrm{Aut}(\mathbb{Z}/n\mathbb{Z})\to {(\mathbb{Z}/n\mathbb{Z})}^{\times }:{\rho }_a\mapsto a(modn)$$

$\mathrm{\Phi }$ is also obviously injective, so all that remains is to show that it is a group homomorphism. But for every $a,b\in {(\mathbb{Z}/n\mathbb{Z})}^{\times }$, we have

$$({\rho }_a\circ {\rho }_b)(x)={\rho }_a(x^b)={(x^b)}^a=x^{ab}={\rho }_{ab}(x)$$

and thus

$$\mathrm{\Phi }({\rho }_a\circ {\rho }_b)=\mathrm{\Phi }({\rho }_{ab})=ab(modn)=\mathrm{\Phi }({\rho }_a)\mathrm{\Phi }({\rho }_b)$$

∎

Title	automorphism group of a cyclic group
Canonical name	AutomorphismGroupOfACyclicGroup
Date of creation	2013-03-22 18:42:35
Last modified on	2013-03-22 18:42:35
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	6
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 20A05
Classification	msc 20F28