Exercise 4.4.17 ($\mathrm{Aut}(Z_n)$ for $n = 2,3,4,5,6$)

Proof. I rewrite the proof of Proposition 16, in a more detailed way.

Here $Z_n=⟨x⟩$, where $|x|=n$.

If $a\in \mathbb{Z}$ satisfies $a\wedge n=1$, we define

Then

• For all $y,z\in Z_n$, since $Z_n$ is abelian,

  $${\psi }_a(y){\psi }_a(z)=y^a{z}^a={(\mathit{𝑦𝑧})}^a={\psi }_a(\mathit{𝑦𝑧}),$$

  so ${\psi }_a$ is a homomorphism.

• ${\psi }_a$ is injective: since $a\wedge n=1$, there are integers $b$ and $c$ such that $\mathit{𝑎𝑏}-\mathit{𝑐𝑛}=1$. Suppose that ${\psi }_a(y)={\psi }_a(z)$. Then $y^a=z^a$, thus $y^{\mathit{𝑎𝑏}}=z^{\mathit{𝑎𝑏}}$. Since $y^n=z^n=1$,

  $$y=y{(y^n)}^c=y^{1+\mathit{𝑐𝑛}}=y^{\mathit{𝑎𝑏}}={(y^a)}^b={(z^a)}^b=z^{\mathit{𝑎𝑏}}=z^{1+\mathit{𝑐𝑛}}=z{(z^n)}^c=z.$$

  This shows that ${\psi }_a$ is injective.

• Since $Z_n$ is a finite set, and ${\psi }_a=Z_n\to Z_n$ is injective, ${\psi }_a$ is also surjective, so

Consider now the map

(Then $\psi ={\mathrm{\Psi }}^{-1}$, for the map $\mathrm{\Psi }$ defined in the text.)

• $\psi$ is well defined: If $\bar{a}=\bar{b}$, then $b=a+\mathit{𝑘𝑛}$ for some integer $k$, thus for all $y\in Z_n$, ${\psi }_b(y)=y^b=y^{a+\mathit{𝑘𝑛}}=y^a{(y^n)}^k=y^a={\psi }_a(y)$, so ${\psi }_a={\psi }_b$.
• $\psi$ is injective: For all $\bar{a},\bar{b}\in \mathbb{Z}∕\mathit{𝑛\mathbb{Z}}$, if $\psi (\bar{a})=\psi (\bar{b})$, then ${\psi }_a(y)={\psi }_b(y)$ for all $y\in Z_n$. In particular $x^a={\psi }_a(x)={\psi }_b(x)=x^b$, where $|x|=n$, thus $a\equiv b(\mathit{𝑚𝑜𝑑}n)$, so $\bar{a}=\bar{b}$. This shows that $\psi$ is injective.
• $\psi$ is surjective: Let $\sigma$ be any element of $\mathrm{Aut}(Z_n)$. Then $\sigma (x)=x^a$ for some $a\in \mathbb{Z}$. Moreover $n=|x|=|\sigma (x)|=|x^a|=\frac{n}{a\wedge n}$, thus $a\wedge n=1$. Moreover, for all integers $k$, ${\psi }_a(x^k)={(x^k)}^a=x^{\mathit{𝑘𝑎}}={(x^a)}^k=\sigma {(x)}^k=\sigma (x^k)$. Since every element $y$ of $Z_n$ is of the form $y=x^k$ for some integer $k$, this shows that $\sigma ={\psi }_a=\psi (\bar{a})$, where $\bar{a}\in {(\mathbb{Z}∕\mathit{𝑛\mathbb{Z}})}^{\times }$, so $\psi$ is surjective.

• $\psi$ is a homomorphism: If $\bar{a},\bar{b}\in {(\mathbb{Z}∕\mathit{𝑛\mathbb{Z}})}^{\times }$, then for all $y\in Z_n$,

  $$\begin{array}{llll}\hfill [\psi (\bar{a})\psi (\bar{b}](y)=& ={\psi }_a({\psi }_b(y))& \hfill & \\ \hfill & ={(y^b)}^a& \hfill & \\ \hfill & =y^{\mathit{𝑎𝑏}}& \hfill & \\ \hfill & ={\psi }_{\mathit{𝑎𝑏}}(y)& \hfill & \\ \hfill & =[\psi (\overline{\mathit{𝑎𝑏}})](y).& \hfill & \end{array}$$

  Therefore $\psi (\bar{a})\psi (\bar{b}=\psi (\overline{\mathit{𝑎𝑏}})$, so $\psi$ is a homomorphism.

In conclusion, $\psi$ is an isomorphism. This proves

Therefore the order of $\mathrm{Aut}(Z_n)$ is

Now we write the solution of the exercise.

• If $n=2$,

  then $\phi (n)=1$, thus $\mathrm{Aut}(G)=\{{\mathrm{id}}_G\}$.

• If $n=3$,

  then $G=⟨x⟩=\{1,x,x^2\}$, where $x^3=1$ and $\phi (n)=2$, thus

  $$\mathrm{Aut}(G)=\{{\psi }_1,{\psi }_2\},$$

  where ${\psi }_1(y)=y^1=y$ for all $y\in G$, so ${\psi }_1={\mathrm{id}}_G$.

  For all $y\in Z_n$, ${\psi }_2(y)=y^2$, so

  $${\psi }_2(1)=1,{\psi }_2(x)=x^2,{\psi }_2(x^2)=x^4=x.$$

  So ${\psi }_2$ exchanges $x$ and $x^2$. if we label $1,x,x^2$ with $0,1,2$, this shows that

  $$\mathrm{Aut}(Z_3)\simeq ⟨(12)⟩\simeq Z_2.$$

• If $n=4$,

  then $G=\{1,x,x^2,x^3\}$, where $x^4=1$ and $\phi (n)=2$, thus $\mathrm{Aut}(G)=\{{\psi }_1,{\psi }_3\}$, where ${\psi }_1={\mathrm{id}}_G$, and ${\psi }_3(y)=y^3$ for all $y\in G$, so

  $${\psi }_3(1)=1,{\psi }_3(x)=x^3,{\psi }_3(x^2)=x^2,{\psi }_3(x^3)=x.$$

  If we label $1,x,x^2,x^3$ with $0,1,2,3$, then

  $$\mathrm{Aut}(G)\simeq ⟨(13)⟩\simeq Z_2.$$

• If $n=5$,

  then $G=⟨x⟩=\{1,x,x^2,x^3,x^4\}$, where $x^5=1$ and $\phi (n)=4$, thus

  $$\mathrm{Aut}(G)=\{{\psi }_1,{\psi }_2,{\psi }_3,{\psi }_4\},$$

  where ${\psi }_k(y)=y^k$ for all $y\in G$, so

  $$\begin{array}{llll}\hfill & {\psi }_1(1)=1,{\psi }_1(x)=x,{\psi }_1(x^2)=x^2,{\psi }_1(x^3)=x^3,{\psi }_1(x^4)=x^4,& \hfill & \\ \hfill & {\psi }_2(1)=1,{\psi }_2(x)=x^2,{\psi }_2(x^2)=x^4,{\psi }_2(x^3)=x,{\psi }_2(x^4)=x^3,& \hfill & \\ \hfill & {\psi }_3(1)=1,{\psi }_3(x)=x^3,{\psi }_3(x^2)=x,{\psi }_3(x^3)=x^4,{\psi }_3(x^4)=x^2,& \hfill & \\ \hfill & {\psi }_4(1)=1,{\psi }_4(x)=x^4,{\psi }_4(x^2)=x^3,{\psi }_4(x^3)=x^2,{\psi }_4(x^4)=x,& \hfill & \\ \hfill & & \hfill \end{array}$$

  With the labelling $1,2,3,4$ for $x,x^2,x^3,x^4$, we obtain

  $$\mathrm{Aut}(G)\simeq \{(),(1243),(1342),(14)(23)\}=⟨(1243)⟩\simeq Z_4.$$

• If $n=6$,

  then $G=⟨x⟩=\{1,x,x^2,x^3,x^4,x^5\}$, where $x^6=1$ and $\phi (n)=2$, thus

  $$\mathrm{Aut}(G)=\{{\psi }_1,{\psi }_5\},$$

  where ${\psi }_k(y)=y^k$ for all $y\in G$, so

  $$\begin{array}{llll}\hfill & {\psi }_1(1)=1,{\psi }_1(x)=x,{\psi }_1(x^2)=x^2,{\psi }_1(x^3)=x^3,{\psi }_1(x^4)=x^4,{\psi }_1(x^5)=x^5& \hfill & \\ \hfill & {\psi }_5(1)=1,{\psi }_5(x)=x^5,{\psi }_5(x^2)=x^4,{\psi }_5(x^3)=x^3,{\psi }_5(x^4)=x^2,{\psi }_5(x^5)=x& \hfill & \\ \hfill & & \hfill \end{array}$$

  With the labelling $1,2,3,4,5$ for $x,x^2,x^3,x^4,x^5$, we obtain

  $$\mathrm{Aut}(G)\simeq ⟨(15)(24)⟩\simeq Z_2.$$