Question

Prove that $\mathbf{N}\mathbf{=}\left\{r_0,r_1,r_2,r_3\right\}$is a normal subgroup of ${\mathbf{D}}_{\mathbf{4}}$ by listing all its right and left cosets.

Short answer

It is proved that$\mathrm{N}=\left\{{\mathrm{r}}_0,{\mathrm{r}}_1,{\mathrm{r}}_2,{\mathrm{r}}_3\right\}$ is a normal subgroup of ${\mathrm{D}}_4$.

Step-by-step solution

Find the left cosets of N

Form the multiplication table of${\mathrm{D}}_4$as:

.	${\mathrm{r}}_0$	r1	r2	r3	d	h	t	v
r0	r0	r1	r2	r3	d	h	t	v
r1	r1	r2	r3	r0	h	t	v	d
r2	r2	r3	r0	r1	t	v	d	h
r3	r3	r0	r1	r2	v	d	h	t
d	d	v	t	h	r0	r3	r2	r1
h	h	d	v	t	r1	r0	r3	r2
t	t	h	d	v	r2	r1	r0	r3
v	v	t	h	d	r3	r2	r1	r0

Therefore, the left cosets of $\mathrm{N}$are$\mathrm{N},\mathrm{dN},\mathrm{hN},\mathrm{tN}$ and $\mathrm{vN}$. This implies that:

$\mathrm{N}=\left\{{\mathrm{r}}_0,{\mathrm{r}}_1,{\mathrm{r}}_2,{\mathrm{r}}_3\right\}$

$\mathrm{dN}=\left\{\mathrm{d},\mathrm{v},\mathrm{t},\mathrm{h}\right\}$

$\mathrm{hN}=\left\{\mathrm{h},\mathrm{d},\mathrm{v},\mathrm{t}\right\}$

$\mathrm{tN}=\left\{\mathrm{t},\mathrm{h},\mathrm{d},\mathrm{v}\right\}$

$\mathrm{vN}=\left\{\mathrm{v},\mathrm{t},\mathrm{h},\mathrm{d}\right\}$

Find the right cosets of N

It is observed that $\mathrm{dN}=\mathrm{hN}=\mathrm{tN}=\mathrm{vN}=\left\{\mathrm{v},\mathrm{t},\mathrm{h},\mathrm{d}\right\}$, and the right cosets are:

$\mathrm{N}=\left\{{\mathrm{r}}_0,{\mathrm{r}}_1,{\mathrm{r}}_2,{\mathrm{r}}_3\right\}$

$\mathrm{Nd}=\left\{\mathrm{d},\mathrm{v},\mathrm{t},\mathrm{h}\right\}$

$\mathrm{Nh}=\left\{\mathrm{h},\mathrm{t},\mathrm{v},\mathrm{d}\right\}$

$\mathrm{Nt}=\left\{\mathrm{t},\mathrm{v},\mathrm{d},\mathrm{h}\right\}$

$\mathrm{Nv}=\left\{\mathrm{v},\mathrm{d},\mathrm{h},\mathrm{t}\right\}$

This implies that $\mathrm{Nd}=\mathrm{Nh}=\mathrm{Nt}=\mathrm{Nv}=\left\{\mathrm{v},\mathrm{t},\mathrm{h},\mathrm{d}\right\}$. Therefore, $\mathrm{dN}=\mathrm{hN}=\mathrm{tN}=\mathrm{vN}=\mathrm{Nd}=\mathrm{Nh}=\mathrm{Nt}=\mathrm{Nv}$. This implies that$\mathrm{N}$ is the normal subgroup.