Exercise 4.2.3 (Left regular representation of $D_8$)

Proof. Let $r$ and $s$ be the usual generators for the dihedral group of order $8$.

(a)

To label the elements of $D_8$, we use the following bijection:

$k$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$f(k)$	$1$	$r$	$r^2$	$r^3$	$s$	$\mathit{𝑠𝑟}$	$s{r}^2$	$s{r}^3$

For all $i\in \mathbb{Z}$,

$$s(s{r}^i)=r^i\text{and}r(s{r}^i)=s{r}^{i-1},$$

then

$$\begin{array}{llll}\hfill s\cdot 1=s& \Rightarrow {\sigma }_s(1)=5,& \hfill & \\ \hfill s\cdot r=\mathit{𝑠𝑟}& \Rightarrow {\sigma }_s(2)=6,& \hfill & \\ \hfill s\cdot r^2=s{r}^2& \Rightarrow {\sigma }_s(3)=7,& \hfill & \\ \hfill s\cdot r^3=s{r}^3& \Rightarrow {\sigma }_s(4)=8,& \hfill & \\ \hfill s\cdot s=1& \Rightarrow {\sigma }_s(5)=1,& \hfill & \\ \hfill s\cdot \mathit{𝑠𝑟}=r& \Rightarrow {\sigma }_s(6)=2,& \hfill & \\ \hfill s\cdot s{r}^2=r^2& \Rightarrow {\sigma }_s(7)=3,& \hfill & \\ \hfill s\cdot s{r}^3=r^3& \Rightarrow {\sigma }_s(8)=4,& \hfill & \\ \hfill & & \hfill \end{array}$$

so ${\sigma }_s=(15)(26)(37)(48)$.

Moreover

$$\begin{array}{llll}\hfill r\cdot 1=r& \Rightarrow {\sigma }_r(1)=2,& \hfill & \\ \hfill r\cdot r=r^2& \Rightarrow {\sigma }_r(2)=3,& \hfill & \\ \hfill r\cdot r^2=r^3& \Rightarrow {\sigma }_r(3)=4,& \hfill & \\ \hfill r\cdot r^3=1& \Rightarrow {\sigma }_r(4)=1,& \hfill & \\ \hfill r\cdot s=s{r}^3& \Rightarrow {\sigma }_r(5)=8,& \hfill & \\ \hfill r\cdot \mathit{𝑠𝑟}=s& \Rightarrow {\sigma }_r(6)=5,& \hfill & \\ \hfill r\cdot s{r}^2=\mathit{𝑠𝑟}& \Rightarrow {\sigma }_r(7)=6,& \hfill & \\ \hfill r\cdot s{r}^3=s{r}^2& \Rightarrow {\sigma }_r(8)=7,& \hfill & \\ \hfill & & \hfill \end{array}$$

so ${\sigma }_r=(1234)(5876)$.

Since $D_8=⟨r,s⟩$, this gives the image of each element of $D_8$ under the left regular representation of $D_8$ in $S_8$ relative to the bijection :

$$\begin{array}{llll}\hfill {\sigma }_1& =()& \hfill & \\ \hfill {\sigma }_r& =(1234)(5876)& \hfill & \\ \hfill {\sigma }_{r^2}& =(13)(24)(57)(68)& \hfill & \\ \hfill {\sigma }_{r^3}& =(1432)(5678)& \hfill & \\ \hfill {\sigma }_s& =(15)(26)(37)(48)& \hfill & \\ \hfill {\sigma }_{\mathit{𝑠𝑟}}& =(16)(27)(38)(45)& \hfill & \\ \hfill {\sigma }_{s{r}^2}& =(17)(28)(35)(46)& \hfill & \\ \hfill {\sigma }_{s{r}^3}& =(18)(25)(36)(47).& \hfill & \end{array}$$

(b)

We relabel the elements of $D_8$ with the bijection $g$:

$k$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$g(k)$	$1$	$s$	$r$	$\mathit{𝑠𝑟}$	$r^2$	$s{r}^2$	$r^3$	$s{r}^3$

If $\tau$ is the homomorphism associate to this action,

$$\begin{array}{llll}\hfill s\cdot 1=s& \Rightarrow {\tau }_s(1)=2,& \hfill & \\ \hfill s\cdot s=1& \Rightarrow {\tau }_s(2)=1,& \hfill & \\ \hfill s\cdot r=\mathit{𝑠𝑟}& \Rightarrow {\tau }_s(3)=4,& \hfill & \\ \hfill s\cdot \mathit{𝑠𝑟}=r& \Rightarrow {\tau }_s(4)=3,& \hfill & \\ \hfill s\cdot r^2=s{r}^2& \Rightarrow {\tau }_s(5)=6,& \hfill & \\ \hfill s\cdot s{r}^2=r^2& \Rightarrow {\tau }_s(6)=5,& \hfill & \\ \hfill s\cdot r^3=s{r}^3& \Rightarrow {\tau }_s(7)=8,& \hfill & \\ \hfill s\cdot s{r}^3=r^3& \Rightarrow {\tau }_s(8)=7,& \hfill & \\ \hfill & & \hfill \end{array}$$

so ${\tau }_s=(12)(34)(56)(78)$.

Moreover

$$\begin{array}{llll}\hfill r\cdot 1=r& \Rightarrow {\tau }_r(1)=3,& \hfill & \\ \hfill r\cdot s=s{r}^3& \Rightarrow {\tau }_r(2)=8,& \hfill & \\ \hfill r\cdot r=r^2& \Rightarrow {\tau }_r(3)=5,& \hfill & \\ \hfill r\cdot \mathit{𝑠𝑟}=s& \Rightarrow {\tau }_r(4)=2,& \hfill & \\ \hfill r\cdot r^2=r^3& \Rightarrow {\tau }_r(5)=7,& \hfill & \\ \hfill r\cdot s{r}^2=\mathit{𝑠𝑟}& \Rightarrow {\tau }_r(6)=4,& \hfill & \\ \hfill r\cdot r^3=1& \Rightarrow {\tau }_r(7)=1,& \hfill & \\ \hfill r\cdot s{r}^3=s{r}^2& \Rightarrow {\tau }_r(8)=6,& \hfill & \\ \hfill & & \hfill \end{array}$$

so ${\tau }_r=(1357)(2864)$.

Since $D_8=⟨r,s⟩$, this gives the image of each element of $D_8$ under the left regular representation of $D_8$ in $S_8$ relative to the bijection $g$:

$$\begin{array}{llll}\hfill {\tau }_1& =()& \hfill & \\ \hfill {\tau }_r& =(1357)(2864)& \hfill & \\ \hfill {\tau }_{r^2}& =(15)(37)(26)(48)& \hfill & \\ \hfill {\tau }_{r^3}& =(1753)(2468)& \hfill & \\ \hfill {\tau }_s& =(12)(34)(56)(78)& \hfill & \\ \hfill {\tau }_{\mathit{𝑠𝑟}}& =(14)(27)(36)(58)& \hfill & \\ \hfill {\tau }_{s{r}^2}& =(16)(25)(38)(47)& \hfill & \\ \hfill {\tau }_{s{r}^3}& =(18)(23)(45)(67).& \hfill & \end{array}$$

Since ${\tau }_s=(12)(34)(56)(78)$ is not in the group of part (a), these two subgroups of $S_8$ are different.