Exercise 2.5.15 (Isomorphism type of each of the three maximal subgroups of $D_{16}$)

Proof. We examine the three subgroups of $D_{16}$ of order $8$ given in Example (6).

We know that

where the elements $1,r,\dots ,r^7,s,\mathit{sr},\dots ,s{r}^7$ are distinct.

• Subgroup $⟨r⟩$.

  Since $|r|=8$, we obtain

  $$⟨r⟩\simeq Z_8.$$

• Subgroup $⟨s,r^2⟩$.

  Note that $⟨s,r^2⟩\supseteq \{1,r^2,r^4,r^6,s,s{r}^2,s{r}^4,s{r}^6\}$. The given lattice of subgroups of $D_{16}$ shows that $⟨s,r^2⟩$ is a maximal subgroup of order $8$, so

  $$⟨s,r^2⟩=\{1,r^2,r^4,r^6,s,s{r}^2,s{r}^4,s{r}^6\}.$$

  Moreover $s{r}^2=\mathit{srr}=r^7\mathit{sr}=r^7{r}^{7}s=r^{14}s=r^{6}s$, thus

  $$s^2=1,{(r^2)}^4=1,s(r^2)={(r^2)}^{3}s.$$

  Since $D_8=⟨\tau ,\sigma \mid {\tau }^2={\sigma }^4=1,\mathit{\tau \sigma }={\sigma }^3\tau ⟩$, the van Dyck’s Theorem shows that there is a surjective homomorphism $\phi :D_8\to ⟨s,r^2⟩$ such that $\phi (\tau )=s$ and $\phi (\sigma )=r^2$. Moreover $|D_8|=|⟨s,r^2⟩|=8$, thus $\phi$ is an isomorphism.

  $$⟨s,r^2⟩\simeq D_8.$$

• Subgroup $⟨\mathit{sr},r^2⟩$.

  As previously,

  $$⟨\mathit{sr},r^2⟩=\{1,r^2,r^4,r^6,\mathit{sr},s{r}^3,s{r}^5,s{r}^7\}.$$

  Since $\mathit{sr}=r^{7}s$, by multiplying on the right and on the left by $s$, we obtain also $\mathit{rs}=s{r}^7$. Therefore ${(\mathit{sr})}^2=\mathit{srsr}=\mathit{ss}{r}^{7}r=s^2{r}^8=1$. Moreover,

  $$(\mathit{sr})r^2=r^{7}s{r}^2=r^7{r}^7\mathit{sr}=r^{14}\mathit{sr}=r^6\mathit{sr}={(r^2)}^3(\mathit{sr}).$$

  Since $D_8=⟨\tau ,\sigma \mid {\tau }^2={\sigma }^4=1,\mathit{\tau \sigma }={\sigma }^3\tau ⟩$, the relations ${(\mathit{sr})}^2={(r^2)}^4=1$ and $(\mathit{sr})r^2={(r^2)}^3(\mathit{sr})$ show by the van Dyck’s Theorem that there is a surjective homomorphism $\psi :D_8\to ⟨\mathit{sr},r^2⟩$ such that $\psi (\tau )=\mathit{sr}$ and $\psi (\sigma )=r^2$. Moreover $|D_8|=|⟨\mathit{sr},r^2⟩|=8$, thus $\psi$ is an isomorphism.

  $$⟨\mathit{sr},r^2⟩\simeq D_8.$$

In conclusion,

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