Exercise 2.5.18 (Centralizers of elements of $QD_{16}$)

Question

Use the lattice to help find the centralizer of every element in $QD_{16}$ (cf. Exercise 11).

Richard Ganaye · Accepted Answer

\begin{proof} 
By Exercise 11, we know the lattice of subgroups of $QD_{16}$:

%      \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}

\bigskip
We have proved in Exercise 11 that the center of $G = QD_{16}$ is $\langle \sigma^4 angle$, so every centralizer of some element $x \in QD_{16}$ contains $\langle \sigma^4 angle$.

For instance, the centralizer of $	au \sigma^3$ contains $	au \sigma^3$ and $\sigma^4$, thus contains $(	au \sigma^3)(\sigma^4) = 	au \sigma^7$, therefore
$$\langle 	au \sigma^7angle  \leq C_G(	au \sigma^3).$$
But $\sigma^2$ don't commute with $	au \sigma^7$: $ (	au \sigma^7) \sigma^2 = 	au \sigma$ and $\sigma^2 (	au \sigma^7)  = 	au \sigma^6 \sigma^7 = 	au \sigma^5$. Therefore $C_G(	au \sigma^3)$ doesn't contain $\langle \sigma^2, 	au \sigmaangle$. Hence the preceding lattice shows that 
$$\langle 	au \sigma^7 angle= C_G(	au \sigma^3).$$

More generally we obtain

\bigskip
\begin{center}
\begin{tabular}{|c|cccccccc|}
\hline
 $x$  & $1$ & $\sigma$ & $\sigma^2$  & $\sigma^3$ & $\sigma^4$ & $\sigma^5$ & $\sigma^6$ & $\sigma^7$ \
\hline
 $C_G(x)$&$QD_{16}$  &$\langle  \sigma angle$  & $\langle  \sigmaangle$  & $\langle \sigma angle$ & $QD_{16}$ &$\langle  \sigma angle$  & $\langle  \sigma angle$  & $\langle \sigmaangle$ \
\hline
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}{|c|cccccccc|}
\hline
 $x$  & $	au$ & $	au\sigma$ & $	au\sigma^2$  & $	au\sigma^3$ & $	au\sigma^4$ & $	au\sigma^5$ & $	au\sigma^6$ & $	au\sigma^7$ \
\hline
 $C_G(x)$&$\langle  \sigma^4, 	au angle$  &$\langle 	au \sigma angle$  & $\langle  \sigma^4,	au \sigma^2angle$  & $\langle  	au \sigma^7angle$ & $\langle \sigma^4, 	au  angle$ &$\langle  	au\sigma angle$  & $\langle   \sigma^4, 	au \sigma^2angle$  & $\langle  	au \sigma^7angle$ \
\hline
\end{tabular}

\end{center}
\end{proof}
With Sagemath (I am more comfortable with Sagemath than with GAP):
\begin{verbatim}
sage: G = SemidihedralGroup(4)
....: t, s = G.gens(); t,s
....: 
((2,4)(3,7)(6,8), (1,2,3,4,5,6,7,8))
sage: s^8 ,t^2,  t * s == s^3 * t
((), (), True)
sage: def wp(x):
....:      """ for the quasidihedral group QD_16 """
....:      u, v = -1,-1
....:      for i in range(8):
....:          for j in range(2):
....:              if  s^i * t^j  == x:
....:                  u, v = i, j
....:      if u == 0:
....:          f = ""
....:      elif u == 1:
....:          f = "s"
....:      else:
....:          f = "s^{}".format(u)
....:      if v == 0:
....:          g = ""
....:      elif v == 1:
....:          g = "t"
....:      else:
....:          g = "t^{}".format(v)
....:      if u==0 and v == 0:
....:          f = '1'
....:      return( g + f)
....:      
....: print(wp(s * t))
....: 
ts
sage: for g in G:
....:    print(wp(g), '=>')
....:    rep = ''
....:    for h in G.centralizer(g).list():
....:        rep = rep + wp(h) + ' '
....:    print(rep)
....:    print
....:    
('1', '=>')
1 t s ts^3 ts s^2 s^3 ts^6 ts^4 ts^2 s^4 s^6 ts^5 ts^7 s^5 s^7

('t', '=>')
1 t s^4 ts^4

('s', '=>')
1 s s^4 s^2 s^5 s^3 s^6 s^7

('ts^3', '=>')
1 ts^3 s^4 ts^7

('ts', '=>')
1 ts s^4 ts^5

('s^2', '=>')
1 s s^2 s^4 s^3 s^5 s^6 s^7

('s^3', '=>')
1 s s^3 s^4 s^2 s^5 s^7 s^6

('ts^6', '=>')
1 ts^6 s^4 ts^2

('ts^4', '=>')
1 t s^4 ts^4

('ts^2', '=>')
1 s^4 ts^2 ts^6

('s^4', '=>')
1 t s ts^3 ts s^2 s^3 ts^6 ts^4 ts^2 s^4 s^6 ts^5 ts^7 s^5 s^7

('s^6', '=>')
1 s s^4 s^6 s^2 s^5 s^7 s^3

('ts^5', '=>')
1 s^4 ts^5 ts

('ts^7', '=>')
1 s^4 ts^7 ts^3

('s^5', '=>')
1 s s^4 s^5 s^2 s^6 s^3 s^7

('s^7', '=>')
1 s s^4 s^7 s^2 s^5 s^6 s^3 
\end{verbatim}
This confirms the preceding results.


$x$	$1$	$σ$	$σ^{2}$	$σ^{3}$	$σ^{4}$	$σ^{5}$	$σ^{6}$	$σ^{7}$

$C_{G} (x)$	$Q D_{16}$	$⟨ σ ⟩$	$⟨ σ ⟩$	$⟨ σ ⟩$	$Q D_{16}$	$⟨ σ ⟩$	$⟨ σ ⟩$	$⟨ σ ⟩$


$x$	$τ$	$τσ$	$τ σ^{2}$	$τ σ^{3}$	$τ σ^{4}$	$τ σ^{5}$	$τ σ^{6}$	$τ σ^{7}$

$C_{G} (x)$	$⟨ σ^{4}, τ ⟩$	$⟨ τσ ⟩$	$⟨ σ^{4}, τ σ^{2} ⟩$	$⟨ τ σ^{7} ⟩$	$⟨ σ^{4}, τ ⟩$	$⟨ τσ ⟩$	$⟨ σ^{4}, τ σ^{2} ⟩$	$⟨ τ σ^{7} ⟩$

Exercise 2.5.18 (Centralizers of elements of $QD_{16}$)

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Exercise 2.5.18 (Centralizers of elements of $QD_{16}$)

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