subgroups of $S_4$

The symmetric group on $4$ letters, $S_4$, has $24$ elements. Listed by cycle type, they are:

Cycle type	Number of elements	elements
$1,1,1,1$	$1$	$()$
$2,1,1$	$6$	$(12),(13),(14),(23),(24),(34)$
$3,1$	$8$	$(123),(132),(124),(142),(134),(143),(234),(243)$
$2,2$	$3$	$(12)(34),(13)(24),(14)(23)$
$4$	$6$	$(1234),(1243),(1324),(1342),(1423),(1432)$

Any subgroup of $S_4$ must be generated by some subset of these elements, and must have order (http://planetmath.org/OrderGroup) dividing $24$, so must be one of $1,2,3,4,6,8$, or $12$.

Think of $S_4$ as acting on the set of “letters” $\mathrm{\Omega }=\{1,2,3,4\}$ by permuting them. Then each subgroup $G$ of $S_4$ acts either transitively or intransitively. If $G$ is transitive, then by the orbit-stabilizer theorem, since there is only one orbit we have that the order of $G$ is a multiple of $|\mathrm{\Omega }|=4$. Thus all the transitive subgroups are of orders $4,8$, or $12$. If $G$ is intransitive, then $G$ has at least two orbits on $\mathrm{\Omega }$. If one orbit is of size $k$ for , then $G$ can naturally be thought of as (isomorphic to) a subgroup of $S_k\times S_{n-k}$. Thus all intransitive subgroups of $S_4$ are isomorphic to subgroups of

$$S_2\times S_2\cong \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\cong V_4$$

$$S_1\times S_3\cong S_3$$

Looking first at subgroups of order $12$, we note that $A_4$ is one such subgroup (and must be transitive, by the above analysis):

$$A_4=\{e,(12)(34),(13)(24),(14)(23),(123),(132),(124),(142),(134),(143),(234),(243)\}$$

Any other subgroup $G$ of order $12$ must contain at least one element of order $3$, and must also contain an element of order $2$. It is easy to see that if $G$ contains two elements of order three that are not inverses, then $G=A_4$, while if $G$ contains exactly two elements of order three which are inverses, then it contains at least one element with cycle type $2,2$. But any such element together with a $3$-cycle generates $A_4$. Thus $A_4$ is the only subgroup of $S_4$ of order $12$.

We look next at order $8$ subgroups. These subgroups are $2$-Sylow subgroups of $S_4$, so they are all conjugate and thus isomorphic. The number of them is odd and divides $24/8=3$, so is either $1$ or $3$. But $S_4$ has three conjugate subgroups of order $8$ that are all isomorphic to $D_8$, the dihedral group with $8$ elements:

$$\mathrm{\{}e,(1324),(1423),(12)(34),(14)(23),(13)(24),(12),(34)\}$$

$$\mathrm{\{}e,(1234),(1432),(13)(24),(12)(34),(14)(23),(13),(24)\}$$

$$\mathrm{\{}e,(1342),(1243),(14)(23),(13)(24),(12)(34),(14),(23)\}$$

and so these are the only subgroups of order $8$ (which must also be transitive).

All subgroups of order $6$ must be intransitive by the above analysis since $4\nmid 6$, so by the above, a subgroup of order $6$ must be isomorphic to $S_3$ and thus must be the image of an embedding of $S_3$ into $S_4$. $S_3$ is generated by transpositions (as is $S_n$ for any $n$), so we can determine embeddings of $S_3$ into $S_4$ by looking at the image of transpositions. But the images of the three transpositions in $S_3$ are determined by the images of $(12)$ and $(13)$ since $(23)=(12)(13)(12)$. So we may send $(12)$ and $(13)$ to any pair of transpositions in $S_4$ with a common element; there are four such pairs and thus four embeddings. These correspond to four distinct subgroups of $S_4$, all conjugate, and all isomorphic to $S_3$:

$$\mathrm{\{}e,(12),(13),(23),(123),(132)\}$$

$$\mathrm{\{}e,(13),(14),(34),(134),(143)\}$$

$$\mathrm{\{}e,(23),(24),(34),(234),(243)\}$$

$$\mathrm{\{}e,(12),(14),(24),(124),(142)\}$$

(The fact that transpositions in $S_3$ must be mapped to transpositions in $S_4$ rather than elements of cycle type $2,2$ is left to the reader).

We shall see that some subgroups of order $4$ are transitive while others are intransitive. A subgroup of order four is clearly isomorphic to either $\mathbb{Z}/4\mathbb{Z}$ or to $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$. The only elements of order $4$ are the $4$-cycles, so each $4$-cycle generates a subgroup isomorphic to $\mathbb{Z}/4\mathbb{Z}$, which also contains the inverse of the $4$-cycle. Since there are six $4$-cycles, $S_4$ has three cyclic subgroups of order $4$, and each is obviously transitive:

$$\mathrm{\{}e,(1234),(13)(24),(1432)\}$$

$$\mathrm{\{}e,(1243),(14)(23),(1342)\}$$

$$\mathrm{\{}e,(1324),(12)(34),(1423)\}$$

A subgroup isomorphic to $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ has, in addition to the identity, three elements ${\sigma }_1,{\sigma }_2,{\sigma }_3$ of order $2$, and thus of cycle types $2,1,1$ or $2,2$. There are several possibilities:

• All three of the ${\sigma }_i$ are of cycle type $2,1,1$. Then the product of any two of those is a $3$-cycle or of cycle type $2,2$, which is a contradiction.

• Two of the ${\sigma }_i$ are of cycle type $2,2$. Then the third is as well, since the product of any pair of the elements of $S_4$ of cycle type $2,2$ is the third such. In this case, the group is

  $$\{e,(12)(34),(13)(24),(14)(23)\}$$

  This group acts transitively.

• One of the ${\sigma }_i$ is of cycle type $2,2$ and the other two are of cycle type $2,1,1$. In this case, the two $2$-cycles must be disjoint, since otherwise their product is a $3$-cycle, so the group looks like

  $$\{e,(12),(34),(12)(34)\}$$

  or one of its conjugates (of which there are two). These groups are intransitive, each having two orbits of size $2$.

Finally, we have a number of subgroups of order $2$ and $3$ generated by elements of those orders; all of these are intransitive.

Summing up, $S_4$ has the following subgroups up to isomorphism and conjugation:

Order	Conjugates	Group
$12$	$1$	$A_4$ (transitive)
$8$	$3$	$\{e,(1324),(1423),(12)(34),(14)(23),(13)(24),(12),(34)\}\cong D_8$ (transitive)
$6$	$4$	$\{e,(12),(13),(23),(123),(132)\}\cong S_3$ (intransitive)
$4$	$3$	$\{e,(1234),(13)(24),(1432)\}\cong \mathbb{Z}/4\mathbb{Z}$ (transitive)
$4$	$1$	$\{e,(12)(34),(13)(24),(14)(23)\}\cong V_4$ (transitive)
$4$	$3$	$\{e,(12),(34),(12)(34)\}\cong V_4$ (intransitive)
$3$	$4$	$\{e,(123),(132)\}$ (intransitive)
$2$	$6$	$\{e,(12)\}$ (intransitive)
$2$	$3$	$\{e,(12)(34)\}$ (intransitive)
$1$	$1$	$\{e\}$ (intransitive)

Of these, the only proper nontrivial normal subgroups of $S_4$ are $A_4$ and the group $\{e,(12)(34),(13)(24),(14)(23)\}\cong V_4$ (see the article on normal subgroups of the symmetric groups).

The subgroup lattice of $S_4$ is thus (listing only one group in each conjugacy class, and taking liberties identifying isomorphic images as subgroups):

Title	subgroups of $S_4$
Canonical name	SubgroupsOfS4
Date of creation	2013-03-22 17:40:54
Last modified on	2013-03-22 17:40:54
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	13
Author	rm50 (10146)
Entry type	Topic
Classification	msc 20B30
Classification	msc 20B35