Question

Prove that $\text{Aut\hspace{0.17em}}\mathbb{Z}\cong {\mathbb{Z}}_2$

Short answer

It is proved that$\text{Aut\hspace{0.17em}}\mathbb{Z}\cong {\mathbb{Z}}_2$

Step-by-step solution

 Step 1: Isomorphic Groups

Two groups G and H are isomorphic if there exists an isomorphism f from $G$onto$H$.

Prove that Aut ℤ≅ℤ2

Consider $f\in \text{Aut\hspace{0.17em}}\mathbb{Z}$

Since the group role="math" localid="1651578675195" $\mathrm{\mathbb{Z}}$ is acyclic generated by element 1 so, the function f is completely determined by image 1 under $f$that is $f\left(1\right)$.

As , $f\in \text{Aut\hspace{0.17em}}\mathbb{Z}$the function f is an isomorphism and therefore $f\left(1\right)$, is generator of $f\left(\mathbb{Z}\right)=\mathbb{Z}$.

Since, the group $\mathrm{\mathbb{Z}}$ has exactly two generators, and only two candidates for $f\left(1\right)$are $1$and $-1$. Therefore, $f\left(1\right)$or $f\left(1\right)=-1$.

Case 1: If $f\left(1\right)=1$.

For any $n\in \mathrm{\mathbb{Z}}$,

$$\begin{gathered} f\left(n\right)=f(1+1+\mathrm{...}+1) \\ =f\left(1\right)+f\left(1\right)+\mathrm{...}+f\left(1\right)\left(n\text{\hspace{0.17em}times}\right) \\ =1+1+\mathrm{...}+1\left(n\text{\hspace{0.17em}times}\right) \\ =n \end{gathered}$$

Therefore, $f\left(n\right)=n$for all $n\in \mathbb{Z}$

Hence, the function$f=i$ is the identity function.

Prove that Aut ℤ≅ℤ2

Case 2: If $f\left(1\right)=-1$

For any $n\in \mathbb{Z}$

$$\begin{gathered} f\left(n\right)=f(1+1+\mathrm{...}+1) \\ =f\left(1\right)+f\left(1\right)+\mathrm{...}+f\left(1\right)\left(n\text{\hspace{0.17em}times}\right) \\ =(-1)+(-1)+\mathrm{...}+(-1)\left(n\text{\hspace{0.17em}times}\right) \\ =-n \end{gathered}$$

Therefore, role="math" localid="1651579531506" $f\left(n\right)=-n$for all role="math" localid="1651579535904" $n\in \mathbb{Z}$

Hence, the set $\text{Aut\hspace{0.17em}}\mathbb{Z}\subseteq \{i,g\}$ where $g\left(n\right)=-n$for all $n\in \mathbb{Z}$

Prove that Aut ℤ≅ℤ2

Now, show that $\{i,g\}\subseteq \text{Aut}\mathbb{Z}$where $g\left(n\right)=-n$for all $n\in \mathrm{\mathbb{Z}}$

As the identity homomorphism is an automorphism of $\mathrm{\mathbb{Z}},$$i\in \text{Aut\hspace{0.17em}}\mathbb{Z}$

Also, the function $g:\mathbb{Z}\to \mathbb{Z}$is given by $g\left(n\right)=-n$is an automorphism of $\mathrm{\mathbb{Z}}$.

Therefore, $g\in \text{Aut}\mathbb{Z}$.

Hence, the set $\{i,g\}\subseteq \text{Aut}\mathbb{Z}$, where $g\left(n\right)=-n$for all $n\in \mathbb{Z}$.

From the above conclusions

$\text{Aut}\mathbb{Z}=\{i,g\}$

Therefore, the group $\text{Aut}\mathbb{Z}$is a group of order $2$and therefore, cyclic. Every cyclic group of order role="math" localid="1651580260383" $n$is isomorphic to ${\mathbb{Z}}_n$.

Hence, it is proved that $\text{Aut}\mathbb{Z}\cong {\mathbb{Z}}_2$