# Classification of subgroups of symmetric group S4

Mathematics ·This article tries to identify the subgroups of symmetric group S4 using theorems from undergraduate algebra courses.

# Basic Fact

Below we will use the cycle notation to denote subgroup elements.

\(S_4\) has \(4!\) elements. Categorize them by cycle patterns, and we get,

\(\begin{array}
{c|c|c}
Cycle \ Pattern & No. \ of \ elements & Order \\
\hline
(1,): id. & 1 \ elem & (ord(e)=1) \\
(2,1,1):(1,2),(1,3)... & 6 & (ord(\alpha)=2) \\
(3,1): (1,2,3),(1,3,2)... & 8 & (ord(\beta)=3) \\
(2,2): (1,2)(3,4)... & 3 & (ord(\gamma)=2) \\
(4): (1,2,3,4)... & 6 & (ord(\delta)=4) \\
\end{array}\)

# Classification

From Lagrange’s Theorem, nontrivial proper subgroups of \(S_4\) should be of order p \(\mid\) 24 for some integer \(p\), i.e., \(p = 2,3,4,6,8\) or 12.

### 1.Subgroups of order 2

\(H \cong \mathbb{Z_2}\), which is equivalent to \(H = <\sigma>\) for some \(\sigma\), where ord(\(\sigma\))=2. Thus \(S_4\) has 9 subgroups:\(\lbrace <\alpha> \rbrace \cup \lbrace <\gamma> \rbrace\).

### 2.Subgroups of order 3

\(H \cong \mathbb{Z_3}\), which is equivalent to \(H = <\sigma>\) for some \(\sigma\), where ord(\(\sigma\))=3. Thus \(S_4\) has 8 subgroups:\(\lbrace <\beta> \rbrace\).

### 3.Subgroups of order 4

The class equation (or just simple reasoning) tells us that any subgroup \(H\) of order 4 is abelian, and thus the fundamental theorem of finite abelian groups tells us \(H \cong \mathbb{Z_4}\) or \(H \cong \mathbb{Z_2} \times \mathbb{Z_2}\). Next we try to find all subgroups by enumerating all cases.

#### Case 1: \(H \cong \mathbb{Z_4}\)

From cycle patterns, we know that there are 3 cyclic subgroups. (Note that not 6 subgroups: \(<1,2,3,4> = <1,4,3,2>\) and so on.)

#### Case 2: \(H \cong \mathbb{Z_2} \times \mathbb{Z_2}\)

Denote \(H= \lbrace id,a,b,c \rbrace\), then \(ord(a)=ord(b)=ord(c)=2\), and \(a*b=c,a*c=b,b*c=a\).

Consider the conjugacy class \(\lbrace \alpha \rbrace\) and \(\lbrace \gamma \rbrace\):

###### 1. \(H\backslash \lbrace e \rbrace \subset \lbrace \alpha \rbrace\):

None of such subgroup exists.(The set is not closed.)

###### 2. \(H\backslash \lbrace e \rbrace \subset \lbrace \gamma \rbrace\):

\(H^{(2)} = \lbrace e \rbrace \cup \lbrace \gamma \rbrace\), which is a nontrivial proper abelian (thus normal) subgroup of \(S_4\).

###### 3. \(H\backslash \lbrace e \rbrace \not\subset \lbrace \alpha \rbrace\) and \(H\backslash \lbrace e \rbrace \not\subset \lbrace \gamma \rbrace\):

There are three subgroups, where one element is from {\(\gamma\)} and the other two nontrivial elements are from {\(\alpha\)} correponding to the two adjacent transposition from the element. For example, \(\lbrace id,(1,2)(3,4),(1,2),(3,4) \rbrace\).

### 4.Subgroups of order 6

\(2=p \mid 2=3-1=q-1\), thus we cannot claim using corollary to Sylow’s Theorem that the subgroup is abelian. However, as no element of \(S_4\) is of order 6, thus \(S_4\) has no abelian subgroup of order 6 by FTFAP.

Using brute force approach or semidirect product \(H\) of order 2 and a normal subgroup \(N\) of order 3, we may claim that \(H=S_3\) up to isomorphism. Thus \(S_4\) has four subgroups of order 6.

### 5.Subgroups of order 8

Subgroups of order 8 are 2-Sylow subgroups of \(S_4\). Sylow’s third theorem tells us there are 1 or 3 2-Sylow subgroups. Case \(r=1\) can be ruled out, otherwise \(H\) is a normal subgroup in \(S_4\), but there is no such union(group) of conjugacy classes whose cardinality is 8. Thus \(r=3\).

H is not normal in \(S_4\), thus H is not abelian. Lemma to Sylow’s First Theorem gives us that center of H, Z, satisfies \(Z \cong \mathbb{Z_2}\) and \(G/Z \cong \mathbb{Z_2} \times \mathbb{Z_2}\). Refer back to subgroups of order 4(Section 3.2), we may simply use composition of \(H^{(2)}\) and \(\alpha\) to get the three subgroups, e.g., \(\lbrace H^{(2)}, (1,2) H^{(2)}\rbrace\).

### 6.Subgroups of order 12

\([G:H]=2\), thus \(H^{(1)}\) is a normal subgroup in \(S_4\). Check the cardinality of conjugacy classes, we can get, there exists only one subgroup of order 12, i.e., \(\lbrace e \rbrace \cup \lbrace \beta \rbrace \cup \lbrace \gamma \rbrace\).

In fact, \(H^{(1)} \cong A_4\) the alternating subgroup: elements of \(H^{(1)}\) are even permutations.

# Discussion

### 1. \(S_4\) is a solvable group.

Generate the commutator subgroup sequence, and we can get, \(S_4 = H^{(0)} \supset H^{(1)} \subset H^{(2)} \supset H^{(3)} = \lbrace e \rbrace\).