Exercise 2.2.4 (Center and centralizers in $S_3, D_8$ and $Q_8$)

Question

For each of $S_3,D_8$ and $Q_8$ compute the centralizers of each element and find the center of each group. Does Lagrange's Theorem (Exercise 19 in Section 1.7) simplify your work?

Richard Ganaye · Accepted Answer

\begin{proof} 
We use the Cayley's tables of $S_3,D_8$ and $Q_8$ given in Exercise 1.5.2.
\begin{itemize}
\item $G = S_3.$

For instance $\{e, 	au\} \subseteq C_G(	au)$. By Lagrange's Theorem, $|C_G(	au) |= 2$ or $6$. But $\sigma 
ot \in C_G(	au)$, because $\sigma 	au 
e 	au \sigma = \sigma^2 	au$. Hence  $\{e, 	au\} = C_G(	au)$.

Similarly $\{e, \sigma, \sigma^2\} \subseteq C_G(\sigma)$, so $|C_G(\sigma) |= 3$ or $6$. Since $	au 
ot \in C_G(\sigma)$, $\{a, \sigma, \sigma^2\} = C_G(\sigma)$.
\begin{center}
\begin{tabular}{|c|c|}
\hline
 $a$ &  $C_G(a)$   \
\hline
 $e$ &   $S_3$  \
 $\sigma$& $\{e ,\sigma, \sigma^2\}$     \
 $\sigma^2$&  $\{e ,\sigma, \sigma^2\}$   \
 $	au$ &  $\{e,	au\}$   \
 $\sigma 	au$&   $\{e,\sigma 	au\}$    \
 $\sigma^2 	au$&  $\{e,\sigma^2 	au\}$   \
\hline
\end{tabular}
\end{center}
The center $Z$ is the intersection of the centralizers of the elements of $S_3$, thus
$$Z = \bigcap_{a \in G} C_{G}(a) = \{e\}.$$

\item $G = D_8$.

For instance, $\{e, r, r^2,r^3\} \subseteq C_G(r^2s)$, and $(r^2 s) s = r^2 = s(r^2 s)$, so $|C_G(r^2s)|>4$. By Lagrange's Theorem, $|C_G(r^2s)| = 6$, so $C_G(r^2s = D_8$.
\begin{center}
\begin{tabular}{|c|c|}
\hline
 $a$ &  $C_G(a)$   \
\hline
 $e$ &   $D_8$  \
 $r$&     $\{e, r, r^2,r^3\}$\ 
 $r^2$&   $D_8$    \
 $r^3$&   $\{e, r, r^2,r^3\}$    \
 $s$ &  $\{e,s , r^2, r^2s\}$   \
 $rs$&   $\{e,rs , r^2, r^3s\}$    \
 $r^2 s$&  $\{e,s , r^2, r^2s\}$   \
  $r^3 s$&  $\{e,rs ,r^2,r^3s\}$   \
\hline
\end{tabular}
\end{center}
Hence $Z = \bigcap_{a \in G} C_{G}(a) = \{e, r^2\}.$

\bigskip
Note: This shows that $H =  \{e, r^2, s, r^2s \}  = \{e, r^2, s, sr^2 \} = C_G(s) = C_G(r^2s)$ and $K = \{e,r^2, sr, sr^3\} = \{e , r^2, r^3s, rs\} = C_G(rs) = C_G(r^3s)$ are subgroups of $D_8$.
This proves anew the results of Exercise 2.1.3.

\item $G = Q_8$.

By definition of $Q_8$, we know that $1$ and $-1$ are in the center of $Q_8$, thus $C_G(-i) = C_G(i)$. Since $i j = k 
e -k = ji$, $ j$ does not commute with $i$. Using Lagrange's Theorem, this gives
\begin{center}
\begin{tabular}{|c|c|}
\hline
 $a$ &  $C_G(a)$   \
\hline
 $1$ &   $Q_8$  \
 $-1$&     $Q_8$\ 
 $i$&   $\{1, -1 ,i ,- i\}$    \
 $-i$&   $\{1, -1 ,i ,- i\}$    \
 $j$ &  $\{1, -1 ,j ,- j\}$   \
 $-j$&   $\{1, -1 ,j ,- j\}$    \
 $k$&  $\{1, -1 ,k ,- k\}$  \
  $-k$&  $\{1, -1 ,k ,- k\}$   \
\hline
\end{tabular}
\end{center}
Hence $Z = \{1,-1\}$.
\end{itemize}
\end{proof}

With Sagemath:
\begin{verbatim}
sage: G = PermutationGroup([[(1,2),(3,4)], [(1,2,3,4)]])
sage: list(G)
[(), (1,2)(3,4), (1,2,3,4), (1,3), (1,3)(2,4), (2,4), (1,4,3,2), (1,4)(2,3)]
sage: Z = G.center(); list(Z)
[(), (1,3)(2,4)]
sage: s, r = G([(1,2),(3,4)]), G([(1,2,3,4)])
sage: G.centralizer(r)
Subgroup of (Permutation Group with generators [(1,2)(3,4), (1,2,3,4)]) 
    generated by [(1,2,3,4)]
sage: list(G.centralizer(r))
[(), (1,2,3,4), (1,3)(2,4), (1,4,3,2)]
\end{verbatim}

For $Q_8$, use
\begin{verbatim}
sage: G = PermutationGroup([[(1,2,4,7),(3,6,8,5)], [(1,3,4,8),(2,5,7,6)]])
\end{verbatim}


$a$	$C_{G} (a)$

$e$	$S_{3}$
$σ$	${e, σ, σ^{2}}$
$σ^{2}$	${e, σ, σ^{2}}$
$τ$	${e, τ}$
$στ$	${e, στ}$
$σ^{2} τ$	${e, σ^{2} τ}$


$a$	$C_{G} (a)$

$e$	$D_{8}$
$r$	${e, r, r^{2}, r^{3}}$
$r^{2}$	$D_{8}$
$r^{3}$	${e, r, r^{2}, r^{3}}$
$s$	${e, s, r^{2}, r^{2} s}$
$rs$	${e, rs, r^{2}, r^{3} s}$
$r^{2} s$	${e, s, r^{2}, r^{2} s}$
$r^{3} s$	${e, rs, r^{2}, r^{3} s}$


$a$	$C_{G} (a)$

$1$	$Q_{8}$
$- 1$	$Q_{8}$
$i$	${1, - 1, i, - i}$
$- i$	${1, - 1, i, - i}$
$j$	${1, - 1, j, - j}$
$- j$	${1, - 1, j, - j}$
$k$	${1, - 1, k, - k}$
$- k$	${1, - 1, k, - k}$

Exercise 2.2.4 (Center and centralizers in $S_3, D_8$ and $Q_8$)

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Exercise 2.2.4 (Center and centralizers in $S_3, D_8$ and $Q_8$)

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