# automorphism group of a cyclic group

###### Theorem 1.

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

###### Proof.

Choose a generator $x$ for $\mathbb{Z}/n\mathbb{Z}$. If $\rho\in\operatorname{Aut}(\mathbb{Z}/n\mathbb{Z})$, 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 prime11 If they were not, say $(a,n)=d$, then $(x^{a})^{n/d}=(x^{a/d})^{n}=1$ so that $x^{a}$ would not generate.. Clearly, then, every $a$ relatively prime to $n$ induces an automorphism. We can therefore define a surjective map

 $\Phi:\operatorname{Aut}(\mathbb{Z}/n\mathbb{Z})\to(\mathbb{Z}/{n}\mathbb{Z})^{% \times}:\rho_{a}\mapsto a\pmod{n}$

$\Phi$ 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

 $\Phi(\rho_{a}\circ\rho_{b})=\Phi(\rho_{ab})=ab\pmod{n}=\Phi(\rho_{a})\Phi(\rho% _{b})$

## References

• 1 Dummit, D., Foote, R.M., Abstract Algebra, Third Edition, Wiley, 2004.
Title automorphism group of a cyclic group AutomorphismGroupOfACyclicGroup 2013-03-22 18:42:35 2013-03-22 18:42:35 rm50 (10146) rm50 (10146) 6 rm50 (10146) Theorem msc 20A05 msc 20F28