If you find any mistakes, please make a comment! Thank you.

## Characterize the automorphisms of cyclic group of order n

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.3 Exercise 2.3.26

Let $Z_n = \langle \alpha \rangle$ be a cyclic group of order $n$ and for each integer $a$, define $\sigma_a : Z_n \rightarrow Z_n$ by $\sigma_a(x) = x^a$.

(1) Prove that $\sigma_a$ is an automorphism of $Z_n$ if and only if $a$ and $n$ are relatively prime.
(2) Prove that $\sigma_a = \sigma_b$ if and only if $a \equiv b \pmod n$.
(3) Prove that every automorphism of $Z_n$ is equal to $\sigma_a$ for some $a$.
(4) Prove that $\sigma_a \circ \sigma_b = \sigma_{ab}$. Deduce that the mapping $\overline{a} \rightarrow \sigma_a$ is an isomorphism of $(\mathbb{Z}/(n))^\times$ to $Z_n$.

Solution:

(1) We saw in Exercise 2.3.25 that this mapping is surjective; since $Z_n$ is finite, it is bijective. By Exercise 1.6.22, $\varphi$ is a homomorphism, thus an automorphism.

(2) ($\Rightarrow$) Suppose $\sigma_a = \sigma_b$. Then $\sigma_a(\alpha) = \sigma_b(\alpha)$, so that $\alpha^a = \alpha^b$. Since $|\alpha| = n$, then, we have $a \equiv b \pmod n$.

($\Leftarrow$) Suppose $a \equiv b \pmod n$; then we have $a-b = kn$ for some integer $k$, so that $a = kn+b$. Now for all $x \in Z_n$ we have $$\sigma_a(x) = x^a = x^{kn+b} = x^b = \sigma_b(x);$$ hence $\sigma_a = \sigma_b$.

(3) Let $\varphi$ be an automorphism of $Z_n$. Now $\varphi(\alpha) = \alpha^k$ for some $k$, since $\alpha$ generates $Z_n$. Then $$\varphi(\alpha^i) = \varphi(\alpha)^i = \alpha^{ki} = (\alpha^i)^k = \sigma_k(\alpha^i)$$ for all integers $i$; since every element of $Z_n$ is of the form $\alpha^i$, we have that $\varphi = \sigma_k$.

(4) Define a mapping $\Phi : (\mathbb{Z}/(n))^\times \rightarrow \mathsf{Aut}\ Z_n$ by $\Phi(\overline{a}) = \sigma_a$. By part 2 above, $\Phi$ is well defined. Moreover, letting $x \in Z_n$ be arbitrary, we have $$(\sigma_a \circ \sigma_b)(x) = \sigma_a(\sigma_b(x)) = \sigma_a(x^b) = (x^b)^a = x^{ab} = \sigma_{ab}(x).$$ Thus $\sigma_a \circ \sigma_b = \sigma_{ab}$. In other words, $\Phi(\overline{a} \overline{b}) = \Phi(\overline{a}) \Phi(\overline{b})$. Thus $\Phi$ is a homomorphism.

Finally, by part 3 above, $\phi$ is surjective. Since $\mathsf{Aut}\ Z_n$ and $(\mathbb{Z}/(n))^\times$ are finite $\phi$ is bijective. Thus $\Phi$ is an isomorphism.