# 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” $\Omega=\{1,2,3,4\}$ 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 $\lvert\Omega\rvert=4$. Thus all the transitive subgroups are of orders $4,8$, or $12$. If $G$ is intransitive, then $G$ has at least two orbits on $\Omega$. If one orbit is of size $k$ for $1\leq k<4$, 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

 $\displaystyle S_{2}\times S_{2}\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2% \mathbb{Z}\cong V_{4}$ $\displaystyle S_{1}\times S_{3}\cong S_{3}$

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:

 $\displaystyle\{e,(1324),\ (1423),\ (12)(34),\ (14)(23),\ (13)(24),\ (12),\ (34)\}$ $\displaystyle\{e,(1234),\ (1432),\ (13)(24),\ (12)(34),\ (14)(23),\ (13),\ (24)\}$ $\displaystyle\{e,(1342),\ (1243),\ (14)(23),\ (13)(24),\ (12)(34),\ (14),\ (23)\}$

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 $S_{3}$:

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

(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:

 $\displaystyle\{e,\ (1234),\ (13)(24),\ (1432)\}$ $\displaystyle\{e,\ (1243),\ (14)(23),\ (1342)\}$ $\displaystyle\{e,\ (1324),\ (12)(34),\ (1423)\}$

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):

 $\xymatrix@R1pc@C1pc{&&S_{4}\ar@{-}[llddd]\ar@{-}[d]\ar@{-}[rrdd]&&&&(24)\\ &&A_{4}\ar@{-}[ldddd]\ar@{-}[rddd]&&&&(12)\\ &&&&D_{8}\ar@{-}[ldd]\ar@{-}[dd]\ar@{-}[rdd]&&(8)\\ S_{3}\ar@{-}[rdd]\ar@{-}[rrddd]&&&&&&(6)\\ &&&\langle(12)(34),(13)(24)\rangle\ar@{-}[rdd]&\langle(12),(34)\rangle\ar@{-}[% dd]\ar@{-}[lldd]&\langle(1234)\rangle\ar@{-}[ldd]&(4)\\ &\langle(123)\rangle\ar@{-}[rdd]&&&&&(3)\\ &&\langle(12)\rangle\ar@{-}[d]&&\langle(12)(34)\rangle\ar@{-}[lld]&&(2)\\ &&\{e\}&&&&(1)}$
Title subgroups of $S_{4}$ SubgroupsOfS4 2013-03-22 17:40:54 2013-03-22 17:40:54 rm50 (10146) rm50 (10146) 13 rm50 (10146) Topic msc 20B30 msc 20B35