Exercise 3.1.17 (Quotient group $D_{16}/Z$)

Question

Let $G$ be the dihedral group of order $16$ (whose lattice appears in Section 2.5):
$$G = \langle r,s \mid r^8 = s^2 = 1,\ rs = sr^{-1} \rangle$$
and let $\overline{G} = G/\langle r^4 \rangle$ be the quotient of $G$ by the subgroup generated by $r^4$ (this subgroup is the center of $G$, hence is normal).
\begin{enumerate}
\item[{\bf(a)}] Show that the order of $\overline{G}$ is $8$.
\item[{\bf(b)}] Exhibit each element of $\overline{G}$ in the form $\overline{s}^a \overline{r}^b$, for some integers $a$ and $b$.
\item[{\bf(c)}] Find the order of each element of $\overline{G}$ exhibited in (b).
\item[{\bf(d)}] Write each of the following elements of $\overline{G}$ in the form $\overline{s}^a \overline{r}^b$, for some integers $a$ and $b$ as in (b):  $\quad \overline{rs}, \quad \overline{sr^{-2} s} ,\quad \overline{s^{-1} r^{-1} s r}$.
\item[{\bf(e)}] Prove that $\overline{H} = \langle \overline{s}, \overline{r}^2 \rangle$ is a normal subgroup of $\overline{G}$ and $\overline{H}$ is isomorphic to the Klein $4$-group. Describe the isomorphic type of the complete preimage of $\overline{H}$ in $G$.
\item[{\bf(f)}] Find the centre of $\overline{G}$ and describe the isomorphism type of $\overline{G}/Z(\overline{G})$.
\end{enumerate}

Richard Ganaye · Accepted Answer

\begin{proof} We know that
\begin{align*}
G &= \langle r,s \mid r^8 = s^2 = 1,\ rs = sr^{-1} angle\
\end{align*}
is explicitly 
\begin{align}
G&= \{1,r,r^2,\ldots,r^7, s, sr, sr^2,\ldots, sr^7\},
\end{align}
where these $16$ elements are distinct.
\begin{enumerate}
\item[{\bf(a)}] By Lagrange's Theorem, if $N \leq G$, the number of cosets $xN$, where  $x \in G$ is $|G|/|N|$. If $N \unlhd G$, then $|G/N| = |G|/|N|$.

For $G  = D_{16}$ and $N = \langle r^4 angle = \{1, r^4\}$, this gives
$$|\overline{G}| = 16/2 = 8.$$

\item[{\bf(b)}] 
By (1),  $G = \{1,r,r^2,\ldots,r^7, s, sr, sr^2,\ldots, sr^7\}$,  then 
$$G/\langle r^4 angle =\{\overline{1}, \overline{r}, \ldots, \overline{r}^7, \overline{s}, \overline{s}\overline{r},  \overline{s}\overline{r}^2, \ldots \overline{s}\overline{r}^7\}$$
where $\overline{r}^4 = \overline{1}$. By removing duplicates ($\overline{r}^i = \overline{r}^{4+i}$), this gives
\begin{align*}
\overline{G} = G/\langle r^4 angle &=\{\overline{1}, \overline{r}, \overline{r}^2, \overline{r}^3, \overline{s}, \overline{s}\overline{r},  \overline{s}\overline{r}^2,  \overline{s}\overline{r}^3\}.
\end{align*}
Since $|\overline{G}| = 8$ by part (a), these $8$ elements are distinct.

\item[{\bf(c)}] Since $(sr^j)^2 = 1$ for all $j$, we obtain $(\overline{sr^j})^2 = \overline{1}$, where $\overline{sr^j} 
e 1$ by part (b). Therefore $|\overline{sr^j}| = 2$.

Since $\overline{r}^4 = \overline{1}$, and $\overline{r}^2 
e \overline{1}$ by part (b), we obtain $|\overline{r}| = 4$. Therefore $|\overline{r}^i | = \frac{4}{4 \wedge i}$.

Explicitly,
\begin{center}
\begin{tabular}{|c|cccccccc|}
\hline
 $x$  & $\overline{1}$ & $\overline{r}$ & $\overline{r^2}$  & $\overline{r^3}$ & $\overline{s}$ & $\overline{sr}$ & $\overline{sr^2}$ & $\overline{sr^3}$ \
\hline
 $|x|$&$1$  &$4$  & $2$  & $4$ & $2$ &$2$  & $2$ & $$2$$ \
\hline
\end{tabular}
\end{center}

\item[{\bf(d)}] Since $rs = sr^{-1}$, we obtain $\overline{r}\, \overline{s} = \overline{s} \overline{r}^{-1}  = \overline{s} \overline{r^3}$.
By induction, we obtain $\overline{r^j s} = \overline{s r^{-j}}$ for all integers $j$, where $ \overline{r}^{-j} =  \overline{r}^{3j}$ . Thus
\begin{align*}
\overline{sr^{-2} s} &= \overline{s s r^{2}} = \overline{r^2},\
 \overline{s^{-1} r^{-1} s r} &= \overline{s^{-1} s r r} = \overline{r^2}.
 \end{align*}

\item[{\bf(e)}] Since $\overline{G} = \langle \overline{s}, \overline{r} angle$ and $\overline{H} = \langle \overline{s},  \overline{r^2} angle$, it is sufficient to verify 
\begin{align*}
\overline{s}\,  \overline{s}\,  \overline{s}^{-1} &= \overline{s} \in \overline{H},\
\overline{s}\,  \overline{r}^2 \overline{s}^{-1}&= \overline{r}^{-2} = \overline{r}^2 \in \overline{H},\
\overline{r} \, \overline{s}\,  \overline{r}^{-1} &= \overline{s}\,  \overline{r}^{-2} =  \overline{s} \, \overline{r}^{2}  \in \overline{H},\
\overline{r} \, \overline{r}^2 \overline{r}^{-1} &= \overline{r}^2  \in \overline{H}.
\end{align*}
Therefore $$\overline{H} \unlhd \overline{G}.$$

\bigskip
First $\overline{H} \supseteq \{\overline{1}, \overline{r}^2, \overline{s}, \overline{s}\, \overline{r}^2\}$. Moreover, since $H = \langle s, r^2 angle$ has order $8$ (see the lattice of $D_{16}$ page 70), then  $|\overline{H}| = |H|/ |\langle r^4 angle| = 4$. Hence
$$\overline{H} =\{\overline{1}, \overline{r}^2, \overline{s}, \overline{s}\, \overline{r}^2\}.$$
Since $|\overline{H}| = 4$, and $\overline{H}$ does not contain any element of order $4$, $\overline{H}$ is not cyclic, therefore
$$\overline{H} \simeq Z_2 	imes Z_2$$
is isomorphic to the Klein $4$-group $V_4$.

\bigskip
If $H =  \langle s, r^2 angle$, then $\pi(H) =  \langle \overline{s},  \overline{r^2} angle = \overline{H}$, where $\pi$ is the natural projection $g \mapsto g \cdot  \langle r^4 angle$. Then  $\pi^{-1}(\overline{H}) = H$.

So the complete preimage of $\overline{H} $ is $H =  \langle s, r^2 angle$. I repeat (copy and paste) the arguments given in Exercise 2.5.15:

Note that $\langle s, r^2 angle \supseteq \{1, r^2, r^4,r^6, s, sr^2, sr^4, sr^6\}$. The given lattice of subgroups of $D_{16}$ shows that $\langle s, r^2 angle$ is a maximal subgroup of order $8$, so
$$ \langle s, r^2 angle = \{1, r^2, r^4,r^6, s, sr^2, sr^4, sr^6\}.$$
Moreover $sr^2 = sr r = r^7 s r = r^{7} r^{7} s = r^{14} s = r^6 s$, thus
$$s^2 =1,\qquad (r^2)^4 = 1, \qquad s (r^2) =(r^2)^3 s.$$
Since $D_8 = \langle 	au, \sigma \mid 	au^2 = \sigma^4 = 1,\ 	au \sigma = \sigma^3 	au angle$, the van Dyck's Theorem shows that there is a surjective homomorphism $\varphi : D_8 	o \langle s, r^2 angle $ such that $\varphi(	au) = s$ and $\varphi(\sigma) = r^2$. Moreover $|D_8| = |\langle s, r^2 angle | = 8$, thus $\varphi$ is an isomorphism.
$$H = \langle s, r^2 angle  \simeq D_8.$$

\item[{\bf(f)}] First $e \in Z(\overline{G})$, and $\overline{r}^2 \in Z(\overline{G})$, since $\overline{r^2}$ and $\overline{r}$ commute,  and
$$\overline{r}^2 \overline{s} =   \overline{s}\,  \overline{r}^{-2} = \overline{s} \, \overline{r}^{2}.$$

Conversely, suppose that $ z = \overline{s}^i\,  \overline{r}^{j} \in Z$ ($0\leq i < 2,\ 0 \leq j < 3$). Then
\begin{itemize}
\item If $i = 1$, then in particular $ (\overline{s}\,  \overline{r}^{j}) \overline{r}= \overline{r}   (\overline{s}\,  \overline{r}^{j})$, so $\overline{s} \, \overline{r}^{j+1} = \overline{s}\,  \overline{r}^{j-1}$, which gives $\overline{r}^2 = \overline{1}$: this is impossible, because the order of $\overline{r}$ is $4$.

\item So $i = 0$, and $ \overline{r}^{j}\overline{s} = \overline{s} \,  \overline{r}^{j})$, thus $\overline{s}\,  \overline{r}^{-j} = \overline{s} \,  \overline{r}^{j}$, therefore $\overline{r}^{-j}  = \overline{r}^{j}$, so $\overline{r}^{2j} =\overline{1}$, where $|\overline{r}| = 4$. Then $4 \mid 2j$, so $2 \mid j$, where $0 \leq j < 4$. This gives $j = 0$ or $j = 2$, so $z \in \{\overline{1}, \overline{r}^2\}$.
\end{itemize}
In conclusion, 
$$Z(\overline{G}) = \{\overline{1}, \overline{r}^2\} = \langle \overline{r}^2 angle.$$
\end{enumerate}

Then $|\overline{G}/Z(\overline{G})| = 4$. Put $ho = \overline{r} $ and $\sigma = \overline{s}$ in $\overline{G}$, and $Z = Z(\overline{G})$. Then 
$$\overline{G}/Z(\overline{G}) = \{ Z , ho Z,  \sigma Z,  \sigma ho Z\}$$
(where $(ho Z)^2  = (\sigma Z)^2 = (\sigmaho Z)^2 = Z$) has no element of order $4$, so
$$\overline{G}/Z(\overline{G}) \simeq Z_2 	imes Z_2.$$

(Alternatively, we may prove that $\overline{G} \simeq D_8$.)
\end{proof}


$x$	$\bar{1}$	$\bar{r}$	$\bar{r^{2}}$	$\bar{r^{3}}$	$\bar{s}$	$\bar{sr}$	$\bar{s r^{2}}$	$\bar{s r^{3}}$

$\| x \|$	$1$	$4$	$2$	$4$	$2$	$2$	$2$	2

Exercise 3.1.17 (Quotient group $D_{16}/Z$)

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Exercise 3.1.17 (Quotient group $D_{16}/Z$)

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