Exercise 4.5.6 (Sylow $3$-subgroups of $A_4$, of $S_4$)

Question

Exhibit all Sylow $3$-subgroups of $A_4$ and all Sylow $3$-subgroups of $S_4$.

Richard Ganaye · Accepted Answer

\begin{proof} 
The lattice of subgroups of $A_4$ given p. 110 shows that the is exactly 4 subgroups of $A_4$ of order $3$, which are Sylow $3$-subgroups of $A_4$. They are
\begin{align}
\langle (1\ 2 \ 3) angle, \qquad \langle (1\ 2 \ 4) angle, \qquad \langle (1\ 3 \ 4) angle, \qquad \langle (2 \ 3\ 4) angle.
\end{align}

Since $|S_4| = 4! = 2^3\cdot 3$, they are also Sylow $3$-subgroups of $S_4$. If $n_3$ is the number of Sylow $3$-subgroups of $S_4$, then by Sylow's Theorem
$$n_3 \equiv 1 \pmod 3\qquad 	ext{and} \qquad n_3 \mid 24,$$
thus $n_3 = 1$ or $n_3 = 4$. This shows that that the preceding list is complete. The Sylow $3$-subgroups of $S_4$ are given by (1).
\end{proof}

Note: The group $G$ of rotations of the cube is isomorphic to $S_4$. This shows that the four Sylow $3$-subgroups of $G$ are given by the rotations of angle $k \cdot2 \pi/3$ around the four big diagonals.

\bigskip
Note 2: With GAP:
\begin{verbatim}
gap> S4:=SymmetricGroup(4);;
gap> T := Subgroup(S4,[]);;
gap> L:=IntermediateSubgroups(S4,T).subgroups;;
gap> for H in L do
>     if Order(H) = 3 then
>        Print(H,"
");
>     fi;
> od;
Group( [ (2,4,3) ] )
Group( [ (1,3,2) ] )
Group( [ (1,3,4) ] )
Group( [ (1,4,2) ] )
\end{verbatim}

Exercise 4.5.6 (Sylow $3$-subgroups of $A_4$, of $S_4$)

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Exercise 4.5.6 (Sylow $3$-subgroups of $A_4$, of $S_4$)

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