Follow-up to question: Aut(G) for G = Klein 4-group is isomorphic to $S_3$

Question

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.

Answer 1

Every group is a set (together with a binary operation on the set). Every automorphism of a group $G$ is a bijective function from the underlying set of $G$ to itself (which in addition respects the operation of $G$). So every automorphism of a group $G$ can be viewed as a subgroup of the group of all permutations on the underlying set of $G$; in fact, since an automorphism must send the identity of $G$ to itself, you can even view every automorphism of $G$ as a permutation of the set $G-\{e\}$.

Here you are getting a bit confused because you are viewing your group $G$ as a subgroup of $S_4$ (nothing wrong with that), and then you are trying to understand $\mathrm{Aut}(G)$ (which can be viewed as a subgroup of $S_{G-{e}} \cong S_3$) as acting on the set $\{1,2,3,4\}$ as well. While some automorphisms can be defined in terms of an action on $\{1,2,3,4\}$, not every automorphism can.

If you view $G$ as the set $\{\mathrm{id}, (1\;2), (3\;4), (1\;2)(3\;4)\}$, then letting $$\begin{align*} x &= (1\;2),\\ y &= (3\;4),\\ z &= (1\;2)(3\;4)\\ e &= \mathrm{id} \end{align*}$$ then the automorphisms of $G$ will always map $e$ to itself, and so you can view the automorphisms as being elements of $S_{\{x,y,z\}}$, the permutation group of $\{x,y,z\}$. The elements of the automorphism group are then (written in 2-line format): $$\begin{align*} \mathrm{id}_{\{x,y,z\}} &= \left(\begin{array}{ccc} x & y & z\\ x & y & z \end{array}\right)\\ f_1 &= \left(\begin{array}{ccc} x & y & z\\ y & z & x \end{array}\right)\\ f_2 &=\left(\begin{array}{ccc} x & y & z\\ z & x & y \end{array}\right)\\ f_3 &= \left(\begin{array}{ccc} x & y & z\\ y & x & z \end{array}\right)\\ f_4 &= \left(\begin{array}{ccc} x & y & z\\ z & y & x \end{array}\right)\\ f_5 &= \left(\begin{array}{ccc} x & y & z\\ x & z & y \end{array}\right) \end{align*}$$ and you can verify that each of them is an automorphism; since every automorphism corresponds to a unique element of $S_{\{x,y,z\}}$, and every element of this group is an automorphism, then this is the automorphism group.

However, you are trying to view your group $G$ as a subgroup of $S_4$. Can the automorphisms of $S_4$ be "induced" by some permutation of $\{1,2,3,4\}$? Equivalently:

If we view the Klein $4$-group $G$ as the subgroup of $S_4$ generated by $(1\;2)$ and $(3\;4)$, is every automorphism of $G$ induced by conjugation by an element of $S_4$?

Well, it doesn't. The reason it doesn't is that $x$, $y$, and $z$ don't all have the same cycle structure: conjugation by elements of $S_4$ (or more precisely, by elements of the normalizer of $G$ in $S_4$) will necessarily map $z$ to itself, because $z$ is the only element with its cycle structure (product of two disjoint transpositions) in $G$. So the only automorphisms that can be viewed as coming from "acting on ${1,2,3,4}$" are the identity and $f_3$

Added: However: it is possible to view the Klein $4$-group as a different subgroup of $S_4$: identify $x\mapsto (1\;2)(3\;4)$; $y\mapsto (1\;3)(2\;4)$, $z\mapsto (1\;4)(2\;3)$. It is not hard to verify that this is a subgroup of order $4$, and since it has three elements of order $2$, it is isomorphic to the Klein $4$-group. Now you can indeed realize every automorphism of $G$ via conjugation by an element of $S_4$ (in several different ways). Here's one way: the automorphism $\mathrm{id}$ can be realized as conjugation by the identity. The automorphism $f_1$ is given by conjugation by $(2\;3\;4)$; $f_2$ is given by conjugation by $(2\;4\;3)$; the automorphism $f_3$ is given by conjugation by $(2\;3)$; $f_4$ is given by conjugation by $(2\;4)$; and $f_5$ is given by conjugation by $(3\;4)$. You can now easily see that the automorphism group is indeed isomorphic to $S_3$, since it corresponds precisely to the subgroup of $S_4$ that fixes $1$ (you can take any of the other four one-point stabilizers as well).

Answer 2

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:

$$\operatorname{Aut}(G)\stackrel{\varphi}{\cong}S_3$$

Answer 3

It is also interesting to note that $Aut(V_4)$ is isomorphic to $GL_2(F_2)$, i.e. the general linear group of dimension $2$ over $\Bbb Z/2\Bbb Z$. The one-to-one correspondence between them is fairly straight forward.