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 orbitstabilizer 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 $$, then $G$ can naturally be thought of as (isomorphic^{} to) a subgroup of ${S}_{k}\times {S}_{nk}$. Thus all intransitive subgroups of ${S}_{4}$ are isomorphic to subgroups of
$${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):
$$\text{xymatrix}\mathrm{@}R1pc\mathrm{@}C1pc\mathrm{\&}\mathrm{\&}{S}_{4}\text{ar}\mathrm{@}[llddd]\text{ar}\mathrm{@}[d]\text{ar}\mathrm{@}[rrdd]\mathrm{\&}\mathrm{\&}\mathrm{\&}\mathrm{\&}(24)\mathrm{\&}\mathrm{\&}{A}_{4}\text{ar}\mathrm{@}[ldddd]\text{ar}\mathrm{@}[rddd]\mathrm{\&}\mathrm{\&}\mathrm{\&}\mathrm{\&}(12)\mathrm{\&}\mathrm{\&}\mathrm{\&}\mathrm{\&}{D}_{8}\text{ar}\mathrm{@}[ldd]\text{ar}\mathrm{@}[dd]\text{ar}\mathrm{@}[rdd]\mathrm{\&}\mathrm{\&}(8){S}_{3}\text{ar}\mathrm{@}[rdd]\text{ar}\mathrm{@}[rrddd]\mathrm{\&}\mathrm{\&}\mathrm{\&}\mathrm{\&}\mathrm{\&}\mathrm{\&}(6)\mathrm{\&}\mathrm{\&}\mathrm{\&}\u27e8(12)(34),(13)(24)\u27e9\text{ar}\mathrm{@}[rdd]\mathrm{\&}\u27e8(12),(34)\u27e9\text{ar}\mathrm{@}[dd]\text{ar}\mathrm{@}[lldd]\mathrm{\&}\u27e8(1234)\u27e9\text{ar}\mathrm{@}[ldd]\mathrm{\&}(4)\mathrm{\&}\u27e8(123)\u27e9\text{ar}\mathrm{@}[rdd]\mathrm{\&}\mathrm{\&}\mathrm{\&}\mathrm{\&}\mathrm{\&}(3)\mathrm{\&}\mathrm{\&}\u27e8(12)\u27e9\text{ar}\mathrm{@}[d]\mathrm{\&}\mathrm{\&}\u27e8(12)(34)\u27e9\text{ar}\mathrm{@}[lld]\mathrm{\&}\mathrm{\&}(2)\mathrm{\&}\mathrm{\&}\{e\}\mathrm{\&}\mathrm{\&}\mathrm{\&}\mathrm{\&}(1)$$ 
Title  subgroups of ${S}_{4}$ 

Canonical name  SubgroupsOfS4 
Date of creation  20130322 17:40:54 
Last modified on  20130322 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 