Exercise 3.1.18 (Quotient group $QD_{16}/Z(QD_{16})$)

Proof. We have proved in Exercise 2.5.15 that $G=⟨\sigma ,\tau \mid {\sigma }^8={\tau }^2=1,\mathit{\sigma \tau }=\tau {\sigma }^3⟩$ has order $16$, and is given explicitly by

where these $16$ elements are distinct. Moreover we proved that the center of $Q{D}_{16}$ is $Z(Q{D}_{16})=⟨{\sigma }^4⟩$.

(a)

Since $|G|=16$ and $|⟨{\sigma }^4⟩|=|\{1,{\sigma }^4\}|=2$, we obtain $$|\bar{G}|=|G∕⟨{\sigma }^4⟩|=16∕2=8.$$

(b)

By (1), we obtain $$G∕⟨{\sigma }^4⟩=\{\bar{1},\bar{\sigma },\dots ,{\bar{\sigma }}^7,\bar{\tau },\bar{\tau }\bar{\sigma },\bar{\tau }{\bar{\sigma }}^2,\dots \bar{\tau }{\bar{\sigma }}^7\}$$

where ${\bar{\sigma }}^4=\bar{1}$. By removing duplicates ( ${\bar{\sigma }}^i={\bar{\sigma }}^{4+i}$), this gives

$$\begin{array}{llll}\hfill \bar{G}=G∕⟨{\sigma }^4⟩& =\{\bar{1},\bar{\sigma },{\bar{\sigma }}^2,{\bar{\sigma }}^3,\bar{\tau },\bar{\tau }\bar{\sigma },\bar{\tau }{\bar{\sigma }}^2,\bar{\tau }{\bar{\sigma }}^3\}.& \hfill & \end{array}$$

Since $|\bar{G}|=8$ by part (a), these $8$ elements are distinct.

(c)

We have proved in Exercise 2.5.11 that for all integers $k\in \mathbb{Z}$, $$\begin{array}{lll}\hfill \tau {\sigma }^k& ={\sigma }^{3k}\tau ,{\sigma }^k\tau =\tau {\sigma }^{3k}.& \hfill \text{(2)}\end{array}$$

Therefore, for all integers $j$,

$${(\tau {\sigma }^j)}^2=\tau {\sigma }^j\tau {\sigma }^j=\mathit{\tau \tau }{\sigma }^{3j}{\sigma }^j={\sigma }^{4j}.$$

Since ${\bar{\sigma }}^4=\bar{1}$, we obtain ${(\overline{\tau {\sigma }^j})}^2=\bar{1}$, where $\overline{\tau {\sigma }^j}\ne 1$ by part (b). Therefore $|\overline{\tau {\sigma }^j}|=2$.

Moreover ${\bar{\sigma }}^2\ne \bar{1}$ by part (b), so we obtain $|\bar{\sigma }|=4$. Therefore $|{\bar{\sigma }}^i|=\frac{4}{4\wedge i}$.

Explicitly,

$x$	$\bar{1}$	$\bar{\sigma }$	$\overline{{\sigma }^2}$	$\overline{{\sigma }^3}$	$\bar{\tau }$	$\overline{\mathit{\tau \sigma }}$	$\overline{\tau {\sigma }^2}$	$\overline{\tau {\sigma }^3}$
$|x|$	$1$	$4$	$2$	$4$	$2$	$2$	$2$	$$2 $$

(d)

By definition of $G$, $\mathit{\sigma \tau }=\tau {\sigma }^3$, thus $\bar{\sigma }\bar{\tau }=\bar{\tau }{\bar{\sigma }}^3=\bar{\tau }{\bar{\sigma }}^{-1}$.

Using (1), we obtain

$$\begin{array}{llll}\hfill \overline{\tau {\sigma }^{-2}\tau }& =\overline{\mathit{\tau \tau }{\sigma }^{-6}}={\bar{\sigma }}^2& \hfill & \\ \hfill \overline{{\tau }^{-1}{\sigma }^{-1}\mathit{\tau \sigma }}& =\overline{{\tau }^{-1}\tau {\sigma }^{-3}\sigma }={\bar{\sigma }}^{-2}={\bar{\sigma }}^2.& \hfill & \end{array}$$

(e)

Since $G=⟨\sigma ,\tau ⟩$, then $\bar{G}=⟨\bar{\sigma },\bar{\tau }⟩$ (cf. Exercise 16).

Moreover,

$${\bar{\sigma }}^4={\bar{\tau }}^2=\bar{1},\bar{\sigma }\bar{\tau }=\bar{\tau }{\bar{\sigma }}^{-1}.$$

Since $D_8=⟨r,s\mid r^4=s^2=1,\mathit{rs}=s{r}^{-1}⟩$, by van Dyck’s Theorem, there is a surjective homomorphism $\phi :D_8\to \bar{G}$ such that $\phi (r)=\bar{\sigma }$ and $\phi (s)=\tau$. Moreover, $|D_8|=|\bar{G}|=8$, so $\phi$ is an isomorphism. Therefore

$$\bar{G}\simeq D_8.$$

Since the center of $D_8$ is $Z(D_8)=⟨r^2⟩$ (cf. Ex 2.2.7), the isomorphism $\phi$ maps $r$ on $\bar{\sigma }$, and $Z(D_8)$ on $Z(\bar{G})$, hence

$$Z(\bar{G})=⟨{\bar{\sigma }}^2⟩=\{\bar{1},{\bar{\sigma }}^2\}.$$