Exercise 3.1.33 (Normal subgroups of $D_8$ and quotient groups)

Proof.

The lattice of subgroups of $D_8$ is given by

We use Exercise 28 to reduce the number of verifications.

• As for every group $⟨1⟩⊴D_8$ and $D_8⊴D_8$, where $D_8∕⟨1⟩\simeq D_8$ and $D_8∕D_8\simeq \{1\}$.

• $N=⟨r^2⟩$. Then $N=\{1,r^2\}$.

  We know that $Z(D_8)=⟨r^2⟩$, therefore $⟨r^2⟩⊴D_8$.

  Since ${\bar{r}}^2=\bar{1}$ in $D_8∕⟨r^2⟩$, we obtain by removing duplicates

  $$\begin{array}{llll}\hfill D_8∕⟨r^2⟩& =\{\bar{1},\bar{r},{\bar{r}}^2,{\bar{r}}^3,\bar{s},\bar{s}\bar{r},\bar{s}\overline{r^2},\bar{s}\overline{r^3}\}& \hfill & \\ \hfill & =\{\bar{1},\bar{r},\bar{s},\bar{s}\bar{r}\}.& \hfill & \end{array}$$

  Since $|D_8∕⟨r^2⟩|=8∕2=4$, these four elements are distinct, and for every $\bar{x}\in D_8∕⟨r^2⟩$, ${\bar{x}}^2=\bar{1}$. Therefore

  $$D_8∕⟨r^2⟩\simeq Z_2\times Z_2.$$

• $N=⟨r⟩$. Then $N=\{1,r,r^2,r^3\}$.

  Every subgroup of $G$ with order $|G|∕2$ is normal, so $N$ is normal (alternatively $\mathit{rr}{r}^{-1}=r\in N$ and $\mathit{sr}{s}^{-1}=r^3\in N$).

  $$⟨r⟩⊴D_8.$$

  Moreover $|D_8∕⟨r⟩|=2$, hence

  $$D_8∕⟨r⟩\simeq Z_2.$$

• Similarly, $⟨r^2,\mathit{sr}⟩⊴D_8$ and $⟨r^2,s⟩⊴D_8$, and

  $$D_8∕⟨r^2,\mathit{sr}⟩\simeq Z_2,D_8∕⟨r^2,s⟩\simeq Z_2.$$

• $N=⟨s{r}^j⟩,j\in \{0,1,2,3\}$. Then $N=\{1,s{r}^j\}$.

  Moreover

  $$\begin{array}{llll}\hfill r(s{r}^j)r^{-1}& =s{r}^3{r}^j{r}^{-1}=s{r}^{j+2}.& \hfill & \end{array}$$

  Assume for the sake of contradiction that $s{r}^{j+2}\in ⟨s{r}^j⟩=\{1,s{r}^j\}$.

  Then $s{r}^{j+2}=1$ or $s{r}^{j+2}=s{r}^j$. In the first case $s\in ⟨r⟩$, and in the second case $r^2=1$. Both cases are impossible, thus $r(s{r}^j)r^{-1}\notin ⟨s{r}^j⟩$. This shows that

  $⟨s⟩,⟨\mathit{sr}⟩,⟨s{r}^2⟩,⟨s{r}^3⟩$ are not normal subgroups of $D_8$.

In conclusion, the only normal subgroups of $D_8$ are

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