**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.1 Exercise 3.1.18**

Let $G = QD_{16}$ be the quasidihedral group presented by $$\langle \sigma, \tau \ |\ \sigma^8 = \tau^2 = 1, \sigma\tau = \tau\sigma^3 \rangle.$$ Let $H = \langle \sigma^4 \rangle = \{ 1, \sigma^4 \}$.

(1) Show that the order of $G/H$ is 8.

(2) Exhibit each element of $G/H$ as $\overline{\tau}^a \overline{\sigma}^b$ for some integers $a$ and $b$.

(3) Find the order of each of the elements exhibited in the previous point.

(4) Write each of the following elements as $\overline{\tau}^a \overline{\sigma}^b$: $\overline{\sigma\tau}$, $\overline{\tau\sigma^{-2}\tau}$, $\overline{\tau^{-1}\sigma^{-1}\tau\sigma}$.

(5) Prove that $G/H \cong D_8$.

Solution:

(1)&(2) We can compute the elements of $G/H$ directly: $$G/H = \{ \overline{1} = \{ 1, \sigma^4 \}, \overline{\sigma} = \{ \sigma, \sigma^5 \}, \overline{\sigma}^2 = \{ \sigma^2, \sigma^6 \}, \overline{\sigma}^3 = \{ \sigma^3, \sigma^7 \}, \overline{\tau} = \{ \tau, \tau \sigma^4 \}, \overline{\tau}\,\overline{\sigma} = \{ \tau \sigma, \tau \sigma^5 \}, \overline{\tau}\,\overline{\sigma}^2 = \{ \tau \sigma^2, \sigma^6 \}, \overline{\tau}\, \overline{\sigma}^3 = \{ \tau \sigma^3, \sigma^7 \} \}.$$ (3)

---$x$--- | -------------Reasoning------------- | Order |

$\overline{1}$ | 1 | |

$\overline{\sigma}$ | 4 | |

$\overline{\sigma}^2$ | $(\overline{\sigma}^2)^2 = \overline{1}$ | 2 |

$\overline{\sigma}^3$ | $ |x^{-1}| = |x|$ | 4 |

$\overline{\tau}$ | 2 | |

$\overline{\tau}\,\overline{\sigma}$ | $ (\overline{\tau}\,\overline{\sigma})^2 = \overline{\sigma^4} = \overline{1}$ | 2 |

$\overline{\tau}\,\overline{\sigma}^2$ | $(\overline{\tau}\,\overline{\sigma}^2)^2 = \overline{\tau}^2 \overline{\sigma}^8 = \overline{1}$ | 2 |

$\overline{\tau}\,\overline{\sigma}^3$ | $(\overline{\tau}\,\overline{\sigma}^3)^2 = \overline{\tau}^2 \overline{\sigma}^{12} = \overline{1}$ | 2 |

(4) We have $$\overline{\sigma\tau} = \overline{\tau\sigma^3} = \overline{\tau}\,\overline{\sigma}^3, $$ $$\overline{\tau\sigma^{-2}\tau} = \overline{\tau\sigma^2\tau} = \overline{\tau\sigma\tau\sigma^3} = \overline{\tau^2\sigma^6} = \overline{\sigma^2} = \overline{\sigma}^2,$$ and $$\overline{\tau^{-1}\sigma^{-1}\tau\sigma} = \overline{\tau \sigma^3 \tau \sigma} = \overline{\tau^2 \sigma^{10}} = \overline{\sigma^2} = \overline{\sigma}^2.$$ (5) Define a mapping $\psi : \{ r, s \} \rightarrow G/H$ by $r \mapsto \overline{\sigma}$ and $s \mapsto \overline{\tau}$. Since $|\overline{\sigma}| = 4$, $|\overline{\tau}| = 2$, $$\overline{\sigma}\,\overline{\tau} = \overline{\sigma\tau} = \overline{\tau\sigma^3} = \overline{\tau}\, \overline{\sigma}^{-1},$$ and $\overline{\tau} \notin \langle \overline{\sigma} \rangle$, by a previous exercise $\psi$ extends to an injective homomorphism $\varphi : D_8 \rightarrow G/H$. Since $|G/H| = 8$, $\varphi$ is surjective. Thus $G/H \cong D_8$.