Exercise 2.5.20 ($N_{QD_{16}}(\langle au \sigma angle)$and $ N_{QD_{16}}(\langle au \sigma^4 angle)$)

Question

Use the lattice of subgroups of $QD_{16}$ (cf. Exercise 11) to help find the normalizers
$$ 	ext{\bf(a)} \ N_{QD_{16}}(\langle 	au \sigma angle) \qquad  	ext{\bf(b)}\  N_{QD_{16}}(\langle 	au \sigma^4 angle).$$

Richard Ganaye · Accepted Answer

\begin{proof} We have proved in Exercise 11 that $Z(QD_{16}) = \langle \sigma^4 angle$. The lattice of subgroups of $QD_{16}$ is given by
 
%      \bigskip    
%      \begin{center}
%\begin{tikzpicture}
%
ode (1)   at (0,0) {$\langle 1 angle$};
%
%
ode (2)   at (0,2) {$\langle \sigma^4angle$};
%
ode (3)   at (-1.7,2) {$\langle	au angle$};
%
ode (4)   at (-3.4,2) {$\langle  	au\sigma^4 angle$};
%
ode (5)   at (-5.1,2) {$\langle 	au \sigma^6 angle$};
%
ode (6)   at (-6.8,2) {$\langle 	au \sigma^2 angle$};
%
%
ode (7)    at (6,4) {$\langle 	au \sigma^7 angle$};
%
ode (8)    at (3,4) {$\langle 	au \sigma angle$};
%
ode (9)    at (0,4) {$\langle \sigma^2 angle$};
%
ode (10)    at (-3,4) {$\langle \sigma^4, 	au  angle$};
%
ode (11)    at (-6,4) {$\langle \sigma^4, 	au  \sigma^2 angle$};
%
%
ode (12)    at (3,6) {$\langle \sigma^2, 	au \sigma angle$};
%
ode (13)    at (0,6) {$\langle \sigma angle$};
%
ode (14)    at (-3,6) {$\langle \sigma^2, 	au  angle$};
%
%
ode (15)    at (0,8) {$QD_{16}$};
%
%\draw (1) edge  (2) edge (3) edge (4) edge (5) edge (6);
%\draw (2) edge  (7) edge (8) edge (9) edge (10) edge (11);
%\draw (11) edge (5) edge (6);
%\draw (10) edge (3) edge (4);
%\draw (9) edge  (13);
%\draw (15) edge  (13) edge (12) edge (14);
%\draw (14) edge  (9) edge (10) edge (11);
%\draw (12) edge  (9) edge (8) edge (7);
%\end{tikzpicture}
%\end{center}

\bigskip    
      \begin{center}
\begin{tikzpicture}

ode (1)   at (0,0) {$\langle 1 angle$};

ode (2)   at (0,2) {$\langle \sigma^4angle$};

ode (3)   at (-1.7,2) {$\langle	au angle$};

ode (4)   at (-3.4,2) {$\langle  	au\sigma^4 angle$};

ode (5)   at (-5.1,2) {$\langle 	au \sigma^6 angle$};

ode (6)   at (-6.8,2) {$\langle 	au \sigma^2 angle$};

ode (7)    at (5,4) {$\langle 	au \sigma^7 angle$};

ode (8)    at (3,4) {$\langle 	au \sigma angle$};

ode (9)    at (0,4) {$\langle \sigma^2 angle$};

ode (10)    at (-3,4) {$\langle \sigma^4, 	au  angle$};

ode (11)    at (-6,4) {$\langle \sigma^4, 	au  \sigma^2 angle$};

ode (12)    at (3,6) {$\langle \sigma^2, 	au \sigma angle$};

ode (13)    at (0,6) {$\langle \sigma angle$};

ode (14)    at (-3,6) {$\langle \sigma^2, 	au  angle$};

ode (15)    at (0,8) {$QD_{16}$};

\draw (1) edge  (2) edge (3) edge (4) edge (5) edge (6);
\draw (2) edge  (7) edge (8) edge (9) edge (10) edge (11);
\draw (11) edge (5) edge (6);
\draw (10) edge (3) edge (4);
\draw (9) edge  (13);
\draw (15) edge  (13) edge (12) edge (14);
\draw (14) edge  (9) edge (10) edge (11);
\draw (12) edge  (9) edge (8) edge (7);
\end{tikzpicture}
\end{center}

\begin{enumerate}
\item[{\bf(a)}]
    Note that $	au \sigma \in N_{QD_{16}}(\langle 	au \sigma angle)$. Moreover
  \begin{align*}
  \sigma^2 (	au \sigma) \sigma^{-2} &= \sigma^2 	au \sigma^{-1}\
  &=	au \sigma^6 \sigma^{-1}\
  &= 	au \sigma^5\
  &= (	au \sigma)\sigma^4 \in \langle 	au \sigma angle,
  \end{align*}
because $\sigma^4 \in  \langle 	au \sigma angle$ and $	au \sigma \in  \langle 	au \sigma angle.$ This shows that
$$\langle \sigma^2, 	au \sigma \leq N_{QD_{16}}(\langle 	au \sigma angle).$$
The preceding lattice shows that $N_{QD_{16}}(\langle 	au \sigma angle) = \langle \sigma^2, 	au \sigma angle $ or $N_{QD_{16}}(\langle 	au \sigma angle) = QD_{16}$.

But 
\begin{align*}
\sigma (	au  \sigma) \sigma^{-1} &= \sigma 	au\
&= 	au \sigma^3 
ot \in \langle 	au \sigma angle = \{1, 	au \sigma, \sigma^4, 	au \sigma^3 \},
\end{align*}
  so $N_{QD_{16}}(\langle 	au \sigma angle) 
e QD_{16}$. This proves
  $$N_{QD_{16}}(\langle 	au \sigma angle) = \langle \sigma^2, 	au \sigma angle .$$
  
\item[{\bf(b)}]
Since $\sigma^4$ is in the center of $QD_{16}$, then $\sigma^4 \in N_{QD_{16}}(\langle 	au \sigma^4 angle)$. Moreover
\begin{align*}
	au (	au \sigma^4) 	au^{-1} &= \sigma^4 	au = 	au \sigma^4\
\end{align*}
so
$$\langle \sigma^4, 	au angle \leq N_{QD_{16}}(\langle 	au \sigma^4 angle).$$
But
\begin{align*}
\sigma^2 (	au \sigma^4) \sigma^{-2} &=\sigma^2 	au \sigma^2\
&= 	au \sigma^6 \sigma^2\
&= 	au 
ot \in \langle 	au \sigma^4 angle = \{1, 	au \sigma^4\}.
\end{align*}
This shows that $\langle \sigma^2, 	au 
ot \subseteq N_{QD_{16}}(\langle 	au \sigma^4 angle)$. The preceding lattice shows that
$$N_{QD_{16}}(\langle 	au \sigma^4 angle) = \langle \sigma^4, 	au angle.$$
\end{enumerate}
\end{proof}

With Sagemath, we add to the instructions given in the note of Exercise 18 the following instructions
\begin{verbatim}
sage: H = G.subgroup([s * t])
....: K = G.normalizer(H)
....: w = ''
....: for h in K:
....:     w = w + wp(h) + ' '
....: print(w)
....: 
1 ts^3 ts s^4 s^6 ts^7 ts^5 s^2 
sage: G.normalizer(H)== G.subgroup([s * t, s^2])
True
sage: L = G.subgroup([s^4 * t])
....: M = G.normalizer(L)
....: w = ''
....: for h in M:
....:     w = w + wp(h) + ' '
....: print(w)
....: 
1 t s^4 ts^4 
sage: G.normalizer(L) == G.subgroup([s^4, t])
True
\end{verbatim}
This confirms the results of (a) and (b).

Exercise 2.5.20 ($N_{QD_{16}}(\langle \tau \sigma \rangle)$and $ N_{QD_{16}}(\langle \tau \sigma^4 \rangle)$)

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Exercise 2.5.20 ($N_{QD_{16}}(\langle \tau \sigma \rangle)$and $ N_{QD_{16}}(\langle \tau \sigma^4 \rangle)$)

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