Exercise 3.1.19 (Quotient group $M/Z(M)$ where $M$ is the modular group of order $16$)

Question

Let $G$ be the modular group of order $16$ (whose lattice was computed in Exercise 11 of Section 2.5):
$$G = \langle v, u \mid v^8 = u^2 = 1,\ v u = u v^5 angle$$
and let $\overline{G} = G/\langle v^4 angle$ be the quotient of $G$ by the subgroup generated by $v^4$ (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 $\overline{u}^a \overline{v}^b$, for some integers $a$ and $b$.
\item[{\bf(c)}] Find the order of each of the elements of $\overline{G}$ exhibited in (b).
\item[{\bf(d)}]Write each of the following elements of $\overline{G}$ in the form $\overline{u}^a \overline{v}^b$, for some integers $a$ and $b$ as in (b):$\quad \overline{v u}, \quad \overline{u v^{-2} u}, \quad \overline{u^{-1} v^{-1} u v}$.
\item[{\bf(e)}] Prove that $\overline{G} \simeq Z_2	imes Z_4$.
\end{enumerate}

\bigskip
(*) The center of the modular group is not $\langle v^4 angle$, but the larger subgroup $\langle v^2 angle$ (see below). This is not a problem, because $\langle v^4 angle \leq Z(G)$, hence $\langle v^4 angle$ is normal. (Note of R.G.)

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

\begin{proof} 
We have proved in Exercise 2.5.11 that $G = \langle v, u \mid v^8 = u^2 = 1, v u = u v^3 angle$ has order $16$, and is given explicitly by
\begin{align}
G = \{1, v, v^2,\ldots, v^7, u, u v, u v^2,\ldots, u v^7\},
\end{align}
where these $16$ elements are distinct. Moreover we proved that the center of $QD_{16}$ is $Z(QD_{16}) = \langle v^2 angle$ ($v^2 \in Z(G)$ because $v^2$ and $v$ commute, and $v^2 u = uv^{10} = uv^2$). Since $\langle v^4 angle \leq Z(G)$, $\langle v^4 angle$ is a normal subgroup of $G$.

\begin{enumerate}

\item[{\bf(a)}] Since $\langle v^4 angle = \{1, v^4\}$,
$$|\overline{G}| = |G|/ |\langle v^4 angle | = 16/2 = 8.$$

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

\item[{\bf(c)}] We have proved in Exercise 2.5.14 that for all integers $k\in \Z$,
\begin{align}
u v^k &= v^{5k} u,\qquad v^k u = u v^{5k}.
\end{align}

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

For all integers $j$,
$$(u v^j)^2 = u v^j u v^j = u u v^{5j} v^j = v^{6j}.$$

Since $\overline{v}^4 = \overline{1}$, we obtain $(\overline{u v^j})^2 = \overline{v^{2j}}$, and $(\overline{u v^j})^4 = \overline{v^{4j}} = \overline{1}$. Moreover, for all integers $j$,
$$\overline{v}^{2j} = 1 \iff  4 \mid  2j \iff 2 \mid j.$$
Therefore $|\overline{u v^j}| = 2$ if $j$ is even and $|\overline{u v^j}| = 4$ if $j$ is odd.

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

\item[{\bf(d)}] By definition of $G$, $vu = uv^5$,thus we  $\overline{v}\, \overline{u} = \overline{u} \,\overline{v}^5 =  \overline{u} \,\overline{v}$. Moreover $G = \langle u, v angle$, hence 
$G$ is an abelian group. Consequently
\begin{align*}
\overline{u v^{-2} u} &= \overline{u}^2 \overline{v}^{-2} = \overline{u}^2 \overline{v}^{2},\
 \overline{u^{-1} v^{-1} u v}&= \overline{1}.
\end{align*}

\item[{\bf(e)}] Since $G = \langle u, v angle$, then $\overline{G} = \langle \overline{u}, \overline{G} angle$ (cf. Exercise 16).
Moreover,
$$ \overline{u}^2  = \overline{v}^4 = \overline{1},\qquad \overline{u}\, \overline{v} = \overline{v}\, \overline{u}.$$
Since $Z_2 	imes Z_4 = \langle a, b \mid a^2 = b^4 = 1,\ ab = ba angle$, by van Dyck's Theorem, there is a surjective homomorphism $\varphi : \Z_2 	imes Z_4 	o \overline{G}$ such that $\varphi(a) = u$ and $\varphi(b) = v$. Moreover, $|Z_2	imes Z_4| = |\overline{G}| = 8$, so $\varphi$ is an isomorphism.

Therefore
$$\overline{G} \simeq Z_2 	imes Z_4.$$
\end{enumerate}
\end{proof}

Exercise 3.1.19 (Quotient group $M/Z(M)$ where $M$ is the modular group of order $16$)

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$x$	$\bar{1}$	$\bar{v}$	$\bar{v^{2}}$	$\bar{v^{3}}$	$\bar{u}$	$\bar{uv}$	$\bar{u v^{2}}$	$\bar{u v^{3}}$

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

Exercise 3.1.19 (Quotient group $M/Z(M)$ where $M$ is the modular group of order $16$)

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