Question

Prove that$\text{Aut}({\mathbb{Z}}_2\times {\mathbb{Z}}_2)\cong S_3$

Short answer

It is proved that$\text{Aut}({\mathbb{Z}}_2\times {\mathbb{Z}}_2)\cong S_3$

Step-by-step solution

 Step 1: Automorphism Groups

Under any automorphism identity, an element is mapped to identity element of the group.

Prove that Aut(ℤ2×ℤ2)≅S3

Consider group ${\mathbb{Z}}_2\times {\mathbb{Z}}_2=\{(0,0),(0,1),(1,0),(1,1)\}$

Label the elements of group ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$as follows:

$\begin{array}{c}e=(0,0)\\ a=(0,1)\\ b=(1,0)\\ c=(1,1)\end{array}$

It is known that under any automorphism identity, element is mapped to identity element of the group. Therefore role="math" localid="1651581879662" $\varphi \left(e\right)=e$for any $\varphi =\text{Aut}({\mathbb{Z}}_2\times {\mathbb{Z}}_2)$.

So, any automorphism ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$is nothing but the permutation of elements $a,b,c\in {\mathbb{Z}}_2\times {\mathbb{Z}}_2$.

Since, the set of permutation of a finite group with n elements is isomorphic to $S_3$.

Therefore, it is proved that $\text{Aut}({\mathbb{Z}}_2\times {\mathbb{Z}}_2)\cong S_3$