TITLE: Follow-up to question: Aut(G) for G = Klein 4-group is isomorphic to $S_3$
QUESTION [9 upvotes]: This is most likely a lack of understanding of wording on my part. I was considerind the Klein 4-group as the set of four permutations: the identity permutation, and three other permutations of four elements, where each of those is made up of two transposes, (i.e., 1 $\rightarrow$ 2, 2 $\rightarrow$ 1 and 3 $\rightarrow$ 4, 4 $\rightarrow$ 3) taken over the three possible such combinations of four elements.
Here, then, is my question. I am assuming (?) that Aut(G) in this case is the set of permutations of the four elements of the Klein 4-group - or the three non-identity ones for the purpose of showing isomorphic to $S_3$. If this is the case, what does it mean to have a permutation of these three permutations that I mentioned above? As always, thanks for your help.

REPLY [4 votes]: Let $G=\{1,a_1(=a_2a_3),a_2(=a_1a_3),a_3(=a_1a_2)\}$ be Klein's $4$-group. Consider the map:
\begin{alignat}{2}
  \varphi: S_3&\longrightarrow& \operatorname{Aut}(G) \\
  \sigma&\longmapsto& \varphi_\sigma:G &\longrightarrow G \\
  &&1&\longmapsto 1 \\
  &&a_i&\longmapsto a_{\sigma(i)} \\
\end{alignat}

Good definition of $\varphi$ (codomain). Firstly, for every $\sigma\in S_3$, $\varphi_\sigma\in \operatorname{Sym}(G)$; moreover, $\forall\sigma\in S_3$:

$$\varphi_\sigma(1a_i)=\varphi_\sigma(a_i)=1\varphi_\sigma(a_i)=\varphi_\sigma(1)\varphi_\sigma(a_i)$$
and ($i=1,2,3$; $\{i,j,k\}=\{1,2,3\}$):
\begin{alignat}{1}
&\varphi_\sigma(a_i^2)=(\varphi_\sigma(a_i))^2&&\iff 1=a_{\sigma(i)}^2 &&&:\text{true}\space\forall\sigma\in S_3 \\
&\varphi_\sigma(a_ia_j)=\varphi_\sigma(a_i)\varphi_\sigma(a_j)&&\iff a_{\sigma(k)}=a_{\sigma(i)}a_{\sigma(j)} &&&:\text{true}\space\forall\sigma\in S_3 \\
\end{alignat}
Therefore, $\varphi_\sigma\in\operatorname{Aut}(G)$ for every $\sigma\in S_3$.

Injectivity of $\varphi$.

\begin{alignat}{1}
\varphi_\sigma=\varphi_\tau &\Longrightarrow a_{\sigma(i)}=a_{\tau(i)}, \forall i \\
&\Longrightarrow \sigma(i)=\tau(i), \forall i \\
&\Longrightarrow \sigma=\tau \\
\end{alignat}

Surjectivity of $\varphi$. Every automorphism of $G$, say $\lambda$, must send $1$ to $1$, $a_1$ to some element of $\{a_1,a_2,a_3\}$ (say $a_{\bar i}$), $a_2$ to some element of $\{a_1,a_2,a_3\}\setminus\{a_{\bar i}\}$ (say $a_{\bar j})$, and $a_3$ to the lonely element of $\{a_1,a_2,a_3\}\setminus\{a_{\bar i},a_{\bar j}\}$ (say $a_{\bar k}$). Thus, $\lambda=\varphi_\sigma$, where $\sigma$ is the permutation defined by $\sigma(1)=\bar i$, $\sigma(2)=\bar j$, $\sigma(3)=\bar k$.

$\varphi$ is a group homomorphism. In fact:


\begin{alignat}{1}
\varphi_{\sigma\tau}(1) = 1 =\varphi_\sigma(1)=\varphi_\sigma(\varphi_\tau(1))=(\varphi_\sigma\varphi_\tau)(1)
\end{alignat}
and
\begin{alignat}{1}
\varphi_{\sigma\tau}(a_i) &= a_{(\sigma\tau)(i)} \\
&= a_{\sigma(\tau(i))} \\
&= \varphi_\sigma(a_{\tau(i)}) \\
&= \varphi_\sigma(\varphi_\tau(a_i)) \\
&= (\varphi_\sigma\varphi_\tau)(a_i) \\
\end{alignat}
whence $\varphi_{\sigma\tau}=\varphi_\sigma\varphi_\tau$.
By this all:
$$\operatorname{Aut}(G)\stackrel{\varphi}{\cong}S_3$$