Exercise 4.5.4 (Sylow subgroups of $D_{12}$ and $S_3 imes S_3$)

Question

Exhibit all Sylow 2-subgroups and Sylow $3$-subgroups of $D_{12}$ and $S_3 	imes S_3$.

Richard Ganaye · Accepted Answer

\begin{proof} 

$\bullet$ Let $G$ be the group $D_{12}$. Then the number $n_2$ of  Sylow  2-subgroups satisfies $n_2 \mid 12$ and $n_2 \equiv 1 \pmod 2$, so $n_2 = 1$ or $n_2 = 3$.

Note that 
$$H = \langle sr, r^3 angle = \{1,sr, r^3, sr^4\}$$
is a Sylow $2$-subgroup. Moreover
\begin{align*}
s H s^{-1} &= \langle s(sr) s^{-1}, sr^3s^{-1} angle\
&= \langle sr^5, r^3 angle,\
(sr^{-1}) H (sr^{-1})^{-1} &= \langle sr^{-1} (sr) sr^{-1}, sr^{-1} r^3 (sr^{-1})angle\
&= \langle sr^3, r^3 angle.
\end{align*}
Therefore $n_2 \geq 3$, so $n_2 = 3$. The three Sylow $2$-subgroups of $D_{12}$ are
\begin{align*}
 \langle sr, r^3 angle, \qquad  \langle sr^3, r^3 angle, \qquad  \langle sr^4, r^3 angle.
\end{align*}

Since $D_{12}$ has a unique subgroup $P = \langle r^2 angle = \{1,r^2,r^4\}$ of order $3$, this subgroup is normal in $G$, so $n_3=1$.
The unique Sylow $3$-subgroup of $D_{12}$ is
\begin{align*}
 \langle  r^2 angle.
\end{align*}
(The unicity is proved in Exercise 5.)

\bigskip
$\bullet$ Let $G$ be the group $S_3 	imes S_3$, of order $36 = 2^2 \cdot 3^2$. Then $n _2 = 1, 3$ or $9$, and $n_3 = 1$ or $n_3 = 4$.

There are $9$ Sylow $2$-subgroups of $G$, given by
\begin{align*}
&\langle (1 \ 2), (1\ 2) angle,\qquad \langle (1 \ 2), (1\ 3) angle,\qquad \langle (1 \ 2), (2\ 3) angle,\
&\langle (1 \ 3), (1\ 2) angle,\qquad \langle (1 \ 3), (1\ 3) angle,\qquad \langle (1 \ 3), (2\ 3) angle,\
&\langle (2 \ 3), (1\ 2) angle,\qquad \langle (2\ 3), (1\ 3) angle,\qquad \langle (2\ 3), (2\ 3) angle.
\end{align*}
We know that $n_2 \leq 9$, so this list is complete.

$G$ has a subgroup $H$ of order $9$. This  $3$-Sylow of $S_3 	imes S_3$ is
$$H = \langle ((1\ 2\ 3), ()), ((), (1\ 2\ 3)) angle.$$

If $(\gamma, \delta) \in S_3	imes S_3$, then using $\langle (1\ 2\ 3) \unlhd S_3 angle $,
\begin{align*}
(\gamma, \delta)  H (\gamma, \delta) ^{-1} &=  \langle (\gamma (1\ 2\ 3)\gamma^{-1}, ()), ((), \delta (1\ 2\ 3) \delta^{-1}) angle\
&=\langle ((1\ 2\ 3), ()), ((), (1\ 2\ 3)) angle\
&= H.
\end{align*}
Thus $H$ is normal in $G$, and so $H$ is the unique Sylow $3$-subgroup of $S_3 	imes S_3$.
\end{proof}



Note : We confirm these results on $S_3 	imes S_3$ with GAP, where $G = S_3 	imes S_3 \simeq \langle (1\ 2\ 3), (1\ 2), (4\ 5\ 6),(4\ 5) angle$.
\begin{verbatim}
gap> a := (1,2,3);; b := (1,2);; c := (4,5,6);; d := (4,5);;
gap> G := Group([a,b,c,d]);; Order(G); 
36
gap> T := Subgroup(G,[]);;
gap> L:=IntermediateSubgroups(G,T).subgroups;;
gap> for H in L do
>     if Order(H) = 4 or Order(H) = 9 then
>         Print(Order(H)," : ",H, "
");
>     fi;
> od;
4 : Group( [ (5,6), (2,3) ] )
4 : Group( [ (4,6), (2,3) ] )
4 : Group( [ (4,5), (2,3) ] )
4 : Group( [ (5,6), (1,2) ] )
4 : Group( [ (4,6), (1,2) ] )
4 : Group( [ (4,5), (1,2) ] )
4 : Group( [ (5,6), (1,3) ] )
4 : Group( [ (4,6), (1,3) ] )
4 : Group( [ (4,5), (1,3) ] )
9 : Group( [ (4,5,6), (1,2,3) ] )
\end{verbatim}

Exercise 4.5.4 (Sylow subgroups of $D_{12}$ and $S_3 \times S_3$)

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Exercise 4.5.4 (Sylow subgroups of $D_{12}$ and $S_3 \times S_3$)

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