subgroups of S4

The symmetric groupMathworldPlanetmathPlanetmath on 4 letters, S4, 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 subgroupMathworldPlanetmathPlanetmath of S4 must be generated by some subset of these elements, and must have order ( dividing 24, so must be one of 1,2,3,4,6,8, or 12.

Think of S4 as acting on the set of “letters” Ω={1,2,3,4} by permuting them. Then each subgroup G of S4 acts either transitively or intransitively. If G is transitiveMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath, then by the orbit-stabilizer theorem, since there is only one orbit we have that the order of G is a multiple of |Ω|=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 Ω. If one orbit is of size k for 1k<4, then G can naturally be thought of as (isomorphicPlanetmathPlanetmathPlanetmathPlanetmath to) a subgroup of Sk×Sn-k. Thus all intransitive subgroups of S4 are isomorphic to subgroups of


Looking first at subgroups of order 12, we note that A4 is one such subgroup (and must be transitive, by the above analysis):


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 inversesMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath, then G=A4, 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 A4. Thus A4 is the only subgroup of S4 of order 12.

We look next at order 8 subgroups. These subgroups are 2-Sylow subgroups of S4, 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 S4 has three conjugate subgroups of order 8 that are all isomorphic to D8, the dihedral groupMathworldPlanetmath with 8 elements:


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 46, so by the above, a subgroup of order 6 must be isomorphic to S3 and thus must be the image of an embeddingPlanetmathPlanetmathPlanetmath of S3 into S4. S3 is generated by transpositionsMathworldPlanetmath (as is Sn for any n), so we can determine embeddings of S3 into S4 by looking at the image of transpositions. But the images of the three transpositions in S3 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 S4 with a common element; there are four such pairs and thus four embeddings. These correspond to four distinct subgroups of S4, all conjugate, and all isomorphic to S3:


(The fact that transpositions in S3 must be mapped to transpositions in S4 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 /4 or to /2×/2. The only elements of order 4 are the 4-cycles, so each 4-cycle generates a subgroup isomorphic to /4, which also contains the inverse of the 4-cycle. Since there are six 4-cycles, S4 has three cyclic subgroups of order 4, and each is obviously transitive:


A subgroup isomorphic to /2×/2 has, in addition to the identityPlanetmathPlanetmath, three elements σ1,σ2,σ3 of order 2, and thus of cycle types 2,1,1 or 2,2. There are several possibilities:

  • All three of the σi are of cycle type 2,1,1. Then the productMathworldPlanetmathPlanetmath of any two of those is a 3-cycle or of cycle type 2,2, which is a contradictionMathworldPlanetmathPlanetmath.

  • Two of the σi are of cycle type 2,2. Then the third is as well, since the product of any pair of the elements of S4 of cycle type 2,2 is the third such. In this case, the group is


    This group acts transitively.

  • One of the σi 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


    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, S4 has the following subgroups up to isomorphismMathworldPlanetmathPlanetmathPlanetmath and conjugationMathworldPlanetmath:

Order Conjugates Group
12 1 A4 (transitive)
8 3 {e,(1324),(1423),(12)(34),(14)(23),(13)(24),(12),(34)}D8 (transitive)
6 4 {e,(12),(13),(23),(123),(132)}S3 (intransitive)
4 3 {e,(1234),(13)(24),(1432)}/4 (transitive)
4 1 {e,(12)(34),(13)(24),(14)(23)}V4 (transitive)
4 3 {e,(12),(34),(12)(34)}V4 (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 subgroupsMathworldPlanetmath of S4 are A4 and the group {e,(12)(34),(13)(24),(14)(23)}V4 (see the article on normal subgroups of the symmetric groups).

The subgroup lattice of S4 is thus (listing only one group in each conjugacy classMathworldPlanetmath, and taking liberties identifying isomorphic images as subgroups):

Title subgroups of S4
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