Exercise 4.2.1 ($V_4$ as permutation group)

Proof. Let $G=\{1,a,b,c\}\simeq Z_2\times Z_2$.

(a)

We label $1,a,b,c$ by the bijection $f_1$, where

$k$	$1$	$2$	$3$	$4$
$f_1(k)$	$1$	$a$	$c$	$b$

By definition of the permutation representation ${\pi }_1$, for all $i,j\in [[1,4]]$, and for all $g\in G$,

$${\pi }_1(g)(i)=j⟺g\cdot g_i=g_j⟺\mathit{𝑔𝑓}(i)=f(j).$$

Hence, if $\phi$ is the homomorphism afforded by the action of $G$ on itself, i.e., $\phi (g)(x)=\mathit{𝑔𝑥}$ for all $x\in G$, then $\phi (g)(f_1(i))=f_1(j)=f_1({\pi }_1(g)(i))$ for all $i\in [[1,4]$, so

$$f_1\circ {\pi }_1(g)=\phi (g)\circ f_1\text{ for all }g\in G,$$

thus the following diagram is commutative:

(These two actions are isomorphic.)

Then, if we write ${\sigma }_a={\pi }_1(a)$,

$$\begin{array}{lll}\hfill a\cdot 1=a\text{ and so }{\sigma }_a(1)=2,& & \hfill \\ \hfill a\cdot a=1\text{ and so }{\sigma }_a(2)=1,& & \hfill \\ \hfill a\cdot c=b\text{ and so }{\sigma }_a(3)=4,& & \hfill \\ \hfill a\cdot b=c\text{ and so }{\sigma }_a(4)=3,& & \hfill \\ \hfill & & \hfill \end{array}$$

therefore ${\sigma }_a=\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 4\hfill & \hfill 3\hfill \end{array}\right)=(12)(34)$.

Similarly

$$\begin{array}{lll}\hfill b\cdot 1=b\text{ and so }{\sigma }_b(1)=4,& & \hfill \\ \hfill b\cdot a=c\text{ and so }{\sigma }_b(2)=3,& & \hfill \\ \hfill b\cdot c=a\text{ and so }{\sigma }_b(3)=2,& & \hfill \\ \hfill b\cdot b=1\text{ and so }{\sigma }_b(4)=1,& & \hfill \\ \hfill & & \hfill \end{array}$$

therefore ${\sigma }_b=\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 4\hfill & \hfill 3\hfill & \hfill 2\hfill & \hfill 1\hfill \end{array}\right)=(14)(23)$, and

$$\begin{array}{lll}\hfill c\cdot 1=c\text{ and so }{\sigma }_b(1)=3,& & \hfill \\ \hfill c\cdot a=b\text{ and so }{\sigma }_b(2)=4,& & \hfill \\ \hfill c\cdot c=1\text{ and so }{\sigma }_b(3)=1,& & \hfill \\ \hfill c\cdot b=a\text{ and so }{\sigma }_b(4)=2,& & \hfill \\ \hfill & & \hfill \end{array}$$

therefore ${\sigma }_c=\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 3\hfill & \hfill 4\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right)=(13)(24)$,

$${\pi }_1(a)=(12)(34){\pi }_1(b)=(14)(23){\pi }_1(c)=(13)(24).$$

(b)

Now we relabel $1,a,b,c$ by the bijection $f_2$, where

$k$	$1$	$2$	$3$	$4$
$f_2(k)$	$1$	$b$	$c$	$a$

The corresponding permutation representation ${\pi }_2$ is defined by $f_2\circ {\pi }_2(g)=\phi (g)\circ f_2$ (or equivalently by ${\pi }_2(g)=f_2^{-1}\circ \phi (g)\circ f_2$) for all $g\in G$.

If we write ${\tau }_a={\pi }_2(a)$, then

$$\begin{array}{lll}\hfill a\cdot 1=a\text{ and so }{\tau }_a(1)=4,& & \hfill \\ \hfill a\cdot b=c\text{ and so }{\tau }_a(2)=3,& & \hfill \\ \hfill a\cdot c=b\text{ and so }{\tau }_a(3)=2,& & \hfill \\ \hfill a\cdot a=1\text{ and so }{\tau }_a(4)=1,& & \hfill \\ \hfill & & \hfill \end{array}$$

therefore ${\tau }_a=\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 4\hfill & \hfill 3\hfill & \hfill 2\hfill & \hfill 1\hfill \end{array}\right)=(14)(23)$,

$$\begin{array}{lll}\hfill b\cdot 1=b\text{ and so }{\tau }_b(1)=2,& & \hfill \\ \hfill b\cdot b=1\text{ and so }{\tau }_b(2)=1,& & \hfill \\ \hfill b\cdot c=a\text{ and so }{\tau }_b(3)=4,& & \hfill \\ \hfill b\cdot a=c\text{ and so }{\tau }_b(4)=3,& & \hfill \\ \hfill & & \hfill \end{array}$$

therefore ${\tau }_b=\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 4\hfill & \hfill 3\hfill \end{array}\right)=(12)(34)$.

Since ${\pi }_2:x\mapsto {\tau }_x$ is a homomorphism

$${\tau }_c={\tau }_{\mathit{𝑎𝑏}}={\tau }_a{\tau }_b=(14)(23)\cdot (12)(34)=(13)(24).$$

So

$${\pi }_2(a)=(14)(23),{\pi }_2(b)=(12)(34),{\pi }_2(c)=(13)(24).$$

So

even though ${\sigma }_a\ne {\tau }_a,{\sigma }_b\ne {\tau }_b,{\sigma }_c\ne {\tau }_c$. □