# Classification of subgroups of symmetric group S4

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, $$S_4 = H^{(0)} \supset H^{(1)} \subset H^{(2)} \supset H^{(3)} = \lbrace e \rbrace$$.