Exercise 3.2.14 ( $S_4$ does not have a normal subgroup of order $8$ or a normal subgroup of order $3$)

Warning: I use the results on Sylow Theorems (Section 4.5).

Proof. Since $| S_{4} | : 4! = 24 = 2^{3} \cdot 3$ , a subgroup of order $8$ is a $2$ -Sylow subgroup of $S_{4}$ . The number $n_{2}$ of $2$ -Sylow satisfies by Theorem 18 p.139

n_{2} \equiv 1 (𝑚𝑜𝑑 2), n_{2} ∣ 3,

so $n_{2} = 1$ or $n_{2} = 3$ .

By Exercise 13, we know a subgroup of order $8$ , isomorphic to $D_{8}$ , given by

\begin{array}{l} H = ⟨ (1 2 3 4), (1 2) (3 4) ⟩ = {(), (1 2 3 4), (1 4) (2 3), (1 3) (2 4), (1 3), (2 4), (1 4 3 2), (1 2) (3 4)} . \end{array}

But a different numbering of the vertices of the square gives another subgroup $K \neq H$ , by exchanging $3$ and $4$ , given by

\begin{array}{l} K = ⟨ (1 2 4 3), (1 2) (3 4) ⟩ = {(), (1 2 4 3), (1 3) (2 4), (1 4) (2 3), (1 4), (2 3), (1 3 4 2), (1 2) (3 4)} . \end{array}

Therefore $n_{2} > 1$ , so $n_{2} = 3$ . This shows that there are only three subgroups of order $8$ .

(The third subgroup of order $8$ is

\begin{array}{l} L = ⟨ (1 3 2 4), (1 3) (2 4) ⟩ = {(), (3, 4), (1, 2), (1, 2) (3, 4), (1, 3) (2, 4), (1, 3, 2, 4), (1, 4, 2, 3), (1, 4) (2, 3)} .) \end{array}

By Theorem 18 and Corollary 20 of Section 4.5 (or direct computation), these three subgroups are conjugate and are not normal in $S_{4}$ .

Consider now a subgroup $A$ of order $3$ . Since $3$ is prime, $A$ is cyclic. The only elements of order $3$ in $S_{4}$ are $3$ -cycles, therefore

A = ⟨ (a b c) ⟩,

where $a, b, c$ are three distinct elements among ${1, 2, 3, 4}$ so there are exactly $4$ subgroups of order $3$ :

A_{1} = ⟨ (1 2 3) ⟩, A_{2} = ⟨ (1 2 4) ⟩, A_{3} = ⟨ (1 3 4) ⟩, A_{4} = ⟨ (2 3 4) ⟩ .

These four subgroups are distinct, and conjugate: if $σ = (\begin{matrix} a & b & c & d \\ a^{'} & b^{'} & c^{'} & d^{'} \end{matrix})$ , then

σ (a b c) σ^{- 1} = (σ (a) σ (b) σ (c)) = (a^{'} b^{'} c^{'}) .

For instance $(3 4) (1 2 3) {(3 4)}^{- 1} = (1 2 4)$ , therefore $A_{1}$ and $A_{2}$ are conjugate subgroups. This shows that $A_{1}, A_{2}, A_{3}, A_{4}$ are not normal subgroups.

In conclusion, $S_{4}$ does not have a normal subgroup of order $8$ or a normal subgroup of order $3$ . □

With GAP:

gap> S4:=SymmetricGroup(4);;
gap> T := Subgroup(S4,[]);;
gap> L:=IntermediateSubgroups(S4,T).subgroups;;
gap> for G in L do
>     Print(Order(G)," : ",G,"\t", IsNormal(S4,G),"\n");
> od;
2 : Group( [ (1,3)(2,4) ] ) false
2 : Group( [ (1,4)(2,3) ] ) false
2 : Group( [ (1,2)(3,4) ] ) false
2 : Group( [ (3,4) ] ) false
2 : Group( [ (2,4) ] ) false
2 : Group( [ (2,3) ] ) false
2 : Group( [ (1,4) ] ) false
2 : Group( [ (1,3) ] ) false
2 : Group( [ (1,2) ] ) false
3 : Group( [ (2,4,3) ] ) false
3 : Group( [ (1,3,2) ] ) false
3 : Group( [ (1,3,4) ] ) false
3 : Group( [ (1,4,2) ] ) false
4 : Group( [ (1,4)(2,3), (1,3)(2,4) ] ) true
4 : Group( [ (3,4), (1,2)(3,4) ] ) false
4 : Group( [ (2,4), (1,3)(2,4) ] ) false
4 : Group( [ (2,3), (1,4)(2,3) ] ) false
4 : Group( [ (1,3,2,4), (1,2)(3,4) ] ) false
4 : Group( [ (1,2,3,4), (1,3)(2,4) ] ) false
4 : Group( [ (1,2,4,3), (1,4)(2,3) ] ) false
6 : Group( [ (3,4), (2,4,3) ] ) false
6 : Group( [ (1,3), (1,3,2) ] ) false
6 : Group( [ (1,3), (1,3,4) ] ) false
6 : Group( [ (1,4), (1,4,2) ] ) false
8 : Group( [ (1,4)(2,3), (1,3)(2,4), (3,4) ] ) false
8 : Group( [ (1,4)(2,3), (1,2)(3,4), (2,4) ] ) false
8 : Group( [ (1,3)(2,4), (1,2)(3,4), (2,3) ] ) false
12 : Group( [ (1,4)(2,3), (1,3)(2,4), (2,4,3) ] ) true

Chapter

Problem name

Text (in LaTeX)

Warning: I use the results on Sylow Theorems (Section 4.5).
\begin{proof} 
Since $|S_4| : 4! = 24 = 2^3\cdot 3$, a subgroup of order $8$ is a $2$-Sylow subgroup of $S_4$. The number $n_2$ of $2$-Sylow satisfies by Theorem 18 p.139
$$n_2 \equiv 1 \pmod 2, \qquad n_2 \mid 3,$$
so $n_2 = 1$ or $n_2 = 3$.

By Exercise 13, we know a subgroup of order $8$, isomorphic to $D_8$, given by
\begin{align*}
H = \langle(1\ 2 \ 3 \ 4), (1\ 2)(3 \ 4)\rangle =  \{(), (1\ 2\ 3\ 4), (1\ 4)(2\ 3),(1\ 3)(2\ 4), (1\ 3), (2\ 4), (1\ 4\ 3\ 2), (1\ 2)(3\ 4)\}.
\end{align*}
But a different numbering of the vertices of the square gives another subgroup $K \ne H$, by exchanging $3$ and $4$, given by
\begin{align*}
K =  \langle(1\ 2 \ 4 \ 3), (1\ 2)(3 \ 4)\rangle = \{(), (1\ 2\ 4\ 3), (1\ 3)(2\ 4),(1\ 4)(2\ 3), (1\ 4), (2\ 3), (1\ 3\ 4\ 2), (1\ 2)(3\ 4)\}.
\end{align*}
Therefore $n_2>1$, so $n_2 = 3$. This shows that there are only three subgroups of order $8$.

(The third subgroup of order $8$ is
\begin{align*}
L = \langle(1\ 3 \ 2 \ 4), (1\ 3)(2 \ 4)\rangle =  \{(), (3,4), (1,2), (1,2)(3,4), (1,3)(2,4), (1,3,2,4), (1,4,2,3), (1,4)(2,3)\}.)
\end{align*}

By Theorem 18 and Corollary 20 of Section 4.5 (or direct computation), these three subgroups are conjugate and are not normal in $S_4$.

\bigskip
Consider now a subgroup $A$ of order $3$. Since $3$ is prime, $A$ is cyclic. The only elements of order $3$ in $S_4$ are $3$-cycles, therefore
$$A = \langle (a\ b \ c) \rangle,$$
where $a, b,c$ are three distinct elements among $\{1,2,3,4\}$ so there are exactly $4$ subgroups of order $3$:
$$A_1 = \langle (1\ 2 \ 3) \rangle,\qquad A_2 = \langle (1\ 2 \ 4)\rangle, \qquad A_3 = \langle (1\ 3 \ 4) \rangle ,\qquad A_4 = \langle (2\ 3 \ 4) \rangle.$$
These four subgroups are distinct, and conjugate: if $\sigma = \begin{pmatrix} a & b & c & d\\ a' &b' & c' & d'\end{pmatrix}$, then
$$\sigma (a\ b \ c) \sigma^{-1} =(\sigma(a)\ \sigma(b)\ \sigma(c)) =  (a'\ b' \ c').$$
For instance $(3\ 4) (1\ 2 \ 3)(3\ 4)^{-1} = (1\ 2 \ 4)$, therefore $A_1$ and $A_2$ are conjugate subgroups. This shows that $A_1,A_2,A_3,A_4$ are not normal subgroups.

\bigskip
In conclusion, $S_4$ does not have a normal subgroup of order $8$ or a normal subgroup of order $3$.
\end{proof}
With GAP:
\begin{verbatim}
gap> S4:=SymmetricGroup(4);;
gap> T := Subgroup(S4,[]);;
gap> L:=IntermediateSubgroups(S4,T).subgroups;;
gap> for G in L do
>     Print(Order(G)," : ",G,"\t", IsNormal(S4,G),"\n");
> od;
2 : Group( [ (1,3)(2,4) ] )	false
2 : Group( [ (1,4)(2,3) ] )	false
2 : Group( [ (1,2)(3,4) ] )	false
2 : Group( [ (3,4) ] )	false
2 : Group( [ (2,4) ] )	false
2 : Group( [ (2,3) ] )	false
2 : Group( [ (1,4) ] )	false
2 : Group( [ (1,3) ] )	false
2 : Group( [ (1,2) ] )	false
3 : Group( [ (2,4,3) ] )	false
3 : Group( [ (1,3,2) ] )	false
3 : Group( [ (1,3,4) ] )	false
3 : Group( [ (1,4,2) ] )	false
4 : Group( [ (1,4)(2,3), (1,3)(2,4) ] )	true
4 : Group( [ (3,4), (1,2)(3,4) ] )	false
4 : Group( [ (2,4), (1,3)(2,4) ] )	false
4 : Group( [ (2,3), (1,4)(2,3) ] )	false
4 : Group( [ (1,3,2,4), (1,2)(3,4) ] )	false
4 : Group( [ (1,2,3,4), (1,3)(2,4) ] )	false
4 : Group( [ (1,2,4,3), (1,4)(2,3) ] )	false
6 : Group( [ (3,4), (2,4,3) ] )	false
6 : Group( [ (1,3), (1,3,2) ] )	false
6 : Group( [ (1,3), (1,3,4) ] )	false
6 : Group( [ (1,4), (1,4,2) ] )	false
8 : Group( [ (1,4)(2,3), (1,3)(2,4), (3,4) ] )	false
8 : Group( [ (1,4)(2,3), (1,2)(3,4), (2,4) ] )	false
8 : Group( [ (1,3)(2,4), (1,2)(3,4), (2,3) ] )	false
12 : Group( [ (1,4)(2,3), (1,3)(2,4), (2,4,3) ] )	true
\end{verbatim}
\end{document}

Exercise 3.2.14 ( $S_4$ does not have a normal subgroup of order $8$ or a normal subgroup of order $3$)

Add answer