Exercise 4.2.5 (Action of $D_8$ on the set of left cosets of $H$)

Proof. The left cosets of $H$ in $D_8$ are

(a)

We label these cosets with the bijection $f$:

$k$	$1$	$2$	$3$	$4$
$f(k)$	$1H$	$\mathit{𝑟𝐻}$	$r^{2}H$	$r^{3}H$

If $g\in D_8$, we write ${\sigma }_g={\pi }_H(g)\in S_4$. Since $D_8=⟨s,r⟩$, we compute ${\sigma }_s$ and ${\sigma }_r$:

$$\begin{array}{llll}\hfill s\cdot 1H=\{s,1\}=1H& \Rightarrow {\sigma }_s(1)=1& \hfill & \\ \hfill s\cdot \mathit{𝑟𝐻}=\{\mathit{𝑠𝑟},\mathit{𝑠𝑟𝑠}\}=\{r^{3}s,r^3\}=r^{3}H& \Rightarrow {\sigma }_s(2)=4& \hfill & \\ \hfill s\cdot r^{2}H=\{s{r}^2,s{r}^{2}s\}=\{r^{2}s,r^2\}=r^{2}H& \Rightarrow {\sigma }_s(3)=3& \hfill & \\ \hfill s\cdot r^{3}H=\{s{r}^3,s{r}^{3}s\}=\{\mathit{𝑟𝑠},r\}=\mathit{𝑟𝐻}& \Rightarrow {\sigma }_s(4)=2,& \hfill & \end{array}$$

so ${\sigma }_s=(24)$.

Moreover,

$$\begin{array}{llll}\hfill r\cdot 1H=\mathit{𝑟𝐻}& \Rightarrow {\sigma }_r(1)=2& \hfill & \\ \hfill r\cdot \mathit{𝑟𝐻}=r^{2}H& \Rightarrow {\sigma }_r(2)=3& \hfill & \\ \hfill r\cdot r^{2}H=r^{3}H& \Rightarrow {\sigma }_r(3)=4& \hfill & \\ \hfill r\cdot r^{3}H=H& \Rightarrow {\sigma }_r(4)=1,& \hfill & \end{array}$$

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

Since ${\pi }_H:D_8\to S_4$ defined by ${\pi }_H(g)={\sigma }_g$ is a homomorphism,

$$\begin{array}{llll}\hfill {\sigma }_e& =()& \hfill & \\ \hfill {\sigma }_r& =(1234)& \hfill & \\ \hfill {\sigma }_{r^2}& =(13)(24)& \hfill & \\ \hfill {\sigma }_{r^3}& =(1432)& \hfill & \\ \hfill {\sigma }_s& =(24)& \hfill & \\ \hfill {\sigma }_{\mathit{𝑠𝑟}}& =(14)(23)& \hfill & \\ \hfill {\sigma }_{s{r}^2}& =(13)& \hfill & \\ \hfill {\sigma }_{s{r}^3}& =(12)(34).& \hfill & \end{array}$$

Therefore $\ker <!--\; FUNCTION\; APPLICATION\; -->({\pi }_H)=\{e\}$, thus this representation is faithful. This shows that

$$D_8\simeq ⟨(1234),(24)⟩.$$

With Gap:

     gap> r := (1,2,3,4);
     (1,2,3,4)
     gap> s := (2, 4);
     (2,4)
     gap> G := Group(r,s);
     Group([ (1,2,3,4), (2,4) ])
     gap> List(G);
     [ (), (1,3)(2,4), (1,4,3,2), (1,2,3,4), (2,4), (1,3), (1,4)(2,3), (1,2)(3,4) ]
     gap> StructureDescription(G);
     "D8"

(b)

We repeat part (a) with the bijection $g$:

$k$	$1$	$2$	$3$	$4$
$g(k)$	$1H$	$r^{2}H$	$\mathit{𝑟𝐻}$	$r^{3}H$

Then

$$\begin{array}{llll}\hfill s\cdot 1H=1H& \Rightarrow {\tau }_s(1)=1& \hfill & \\ \hfill s\cdot r^{2}H=r^{2}H& \Rightarrow {\tau }_s(2)=2& \hfill & \\ \hfill s\cdot \mathit{𝑟𝐻}=r^{3}H& \Rightarrow {\tau }_s(3)=4& \hfill & \\ \hfill s\cdot r^{3}H=\mathit{𝑟𝐻}& \Rightarrow {\tau }_s(4)=3,& \hfill & \end{array}$$

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

Moreover,

$$\begin{array}{llll}\hfill r\cdot 1H=\mathit{𝑟𝐻}& \Rightarrow {\tau }_r(1)=3& \hfill & \\ \hfill r\cdot r^{2}H=r^{3}H& \Rightarrow {\tau }_r(2)=4& \hfill & \\ \hfill r\cdot \mathit{𝑟𝐻}=r^{2}H& \Rightarrow {\tau }_r(3)=2& \hfill & \\ \hfill r\cdot r^{3}H=H& \Rightarrow {\tau }_r(4)=1,& \hfill & \end{array}$$

so ${\tau }_r=(1324)$.

Since ${\lambda }_H:D_8\to S_4$ defined by ${\lambda }_H(g)={\tau }_g$ is a homomorphism,

$$\begin{array}{llll}\hfill {\tau }_e& =()& \hfill & \\ \hfill {\tau }_r& =(1324)& \hfill & \\ \hfill {\tau }_{r^2}& =(12)(34)& \hfill & \\ \hfill {\tau }_{r^3}& =(1423)& \hfill & \\ \hfill {\tau }_s& =(34)& \hfill & \\ \hfill {\tau }_{\mathit{𝑠𝑟}}& =(14)(32)& \hfill & \\ \hfill {\tau }_{s{r}^2}& =(12)& \hfill & \\ \hfill {\tau }_{s{r}^3}& =(13)(24).& \hfill & \end{array}$$

We observe that the subgroups $\{{\sigma }_g\mid g\in D_8\}$ and $\{{\tau }_g\mid g\in D_8\}$ are distinct (but these subgroups are conjugate, i.e., if $\gamma =(23)$, then ${\tau }_g=\gamma {\sigma }_g{\gamma }^{-1}$ for every $g\in D_8$)

(c)

Let $K=⟨\mathit{𝑠𝑟}⟩=\{1,\mathit{𝑠𝑟}\}$. The cosets of $K$ are $$\begin{array}{llll}\hfill 1K& =\{1,\mathit{𝑠𝑟}\},& \hfill & \\ \hfill \mathit{𝑟𝐾}& =\{r,\mathit{𝑟𝑠𝑟}\}=\{r,s\},& \hfill & \\ \hfill r^{2}K& =\{r^2,r^2\mathit{𝑠𝑟}\}=\{r^2,s{r}^3\},& \hfill & \\ \hfill r^{3}K& =\{r^3,r^3\mathit{𝑠𝑟}\}=\{r^3,s{r}^2\}.& \hfill & \end{array}$$

We label these cosets with the bijection $h$:

$k$	$1$	$2$	$3$	$4$
$h(k)$	$1K$	$\mathit{𝑟𝐾}$	$r^{2}K$	$r^{3}K$

If we write ${\pi }_K:g\mapsto {\phi }_g$ the associate representation, then

$$\begin{array}{llll}\hfill s\cdot 1K=\{s,r\}=\mathit{𝑟𝐾}& \Rightarrow {\phi }_s(1)=2& \hfill & \\ \hfill s\cdot \mathit{𝑟𝐾}=\{\mathit{𝑠𝑟},1\}=1K& \Rightarrow {\phi }_s(2)=1& \hfill & \\ \hfill s\cdot r^{2}K=\{s{r}^2,r^3\}=r^{3}K& \Rightarrow {\phi }_s(3)=4& \hfill & \\ \hfill s\cdot r^{3}K=\{s{r}^3,r^2\}=r^{2}K& \Rightarrow {\phi }_s(4)=3,& \hfill & \end{array}$$

so ${\phi }_s=(12)(34)$.

Moreover,

$$\begin{array}{llll}\hfill r\cdot 1K=\mathit{𝑟𝐾}& \Rightarrow {\phi }_r(1)=2& \hfill & \\ \hfill r\cdot \mathit{𝑟𝐾}=r^{2}K& \Rightarrow {\phi }_r(2)=3& \hfill & \\ \hfill r\cdot r^{2}K=r^{3}K& \Rightarrow {\phi }_r(3)=4& \hfill & \\ \hfill r\cdot r^{3}K=K& \Rightarrow {\phi }_r(4)=1,& \hfill & \end{array}$$

so ${\phi }_r=(1234)$.

Then the image of this representation is

$$\begin{array}{llll}\hfill \{{\phi }_g\mid g\in H\}& =⟨{\phi }_s,{\phi }_r⟩& \hfill & \\ \hfill & =⟨(12)(34),(1234)⟩& \hfill & \\ \hfill & =\{(),(24),(12)(34),(1234),(13),(13)(24),(1432),(14)(23)\}& \hfill & \\ \hfill & =\{{\sigma }_g\mid g\in H\}& \hfill & \end{array}$$

is the same subgroup of $S_4$ as in part (a).