I'm curious how we can construct a morphism $\text{Aut}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}) \rightarrow S_3$ and show then that it is an isomorphism.
I know that $(12),(23)$ generate $S_3$ and that $([a]_2,[b]_2)$ with $a,b \in \mathbb{Z}/2\mathbb{Z}$ generate $\mathbb{Z}/2\mathbb{Z}$. But, I don't really see how we can construct it.
If you could give me some ''hints'' I would appreciate it.
Thank you in advance
Automorphisms are bijections (of a group into itself) which preserve the order of the elements. Now $\Bbb Z_2\times \Bbb Z_2\cong V_4=\{e,a,b,ab\}$, where all the elements have order $2$. Therefore, an automorphism of $V_4$ maps:
Therefore, the elements of $\operatorname{Aut}(V_4)$ actually permutes $3$ objects, just like the elements of $S_3$ do (by definition), and you can map any of the former to the "corresponding" element of $S_3$. Formally, define a bijection $f\colon \{a,b,ab\} \to \{1,2,3\}$. For example, set: $f(a)=2, f(b)=1, f(ab)=3$; then, take $\alpha\in \operatorname{Aut}(V_4)$ defined e.g. by (see the bullet points bove): $\alpha(a)=ab, \alpha(b)=b, \alpha(ab)=a$. Now, let the bijection $f$ to come into play:
\begin{alignat}{4} &\alpha(a)=ab &&\iff f(\alpha(a))=f(ab)=3 &&&\iff f\alpha(f^{-1}(2))=3 &&&&\iff (f\alpha f^{-1})(2)=3 \\ &\alpha(b)=b &&\iff f(\alpha(b))=f(b)=1 &&&\iff f\alpha(f^{-1}(1))=1 &&&&\iff (f\alpha f^{-1})(1)=1 \\ &\alpha(ab)=a &&\iff f(\alpha(ab))=f(a)=2 &&&\iff f\alpha(f^{-1}(3))=2 &&&&\iff (f\alpha f^{-1})(3)=2 \\ \end{alignat}
Therefore, within this framework, the sought morphism maps $\alpha\mapsto (23)$, and similarly for the others elements of $\operatorname{Aut}(V_4)$.