|Cycle type||Number of elements||elements|
Any subgroup of must be generated by some subset of these elements, and must have order (http://planetmath.org/OrderGroup) dividing , so must be one of , or .
Think of as acting on the set of “letters” by permuting them. Then each subgroup of acts either transitively or intransitively. If is transitive, then by the orbit-stabilizer theorem, since there is only one orbit we have that the order of is a multiple of . Thus all the transitive subgroups are of orders , or . If is intransitive, then has at least two orbits on . If one orbit is of size for , then can naturally be thought of as (isomorphic to) a subgroup of . Thus all intransitive subgroups of are isomorphic to subgroups of
Looking first at subgroups of order , we note that is one such subgroup (and must be transitive, by the above analysis):
Any other subgroup of order must contain at least one element of order , and must also contain an element of order . It is easy to see that if contains two elements of order three that are not inverses, then , while if contains exactly two elements of order three which are inverses, then it contains at least one element with cycle type . But any such element together with a -cycle generates . Thus is the only subgroup of of order .
We look next at order subgroups. These subgroups are -Sylow subgroups of , so they are all conjugate and thus isomorphic. The number of them is odd and divides , so is either or . But has three conjugate subgroups of order that are all isomorphic to , the dihedral group with elements:
and so these are the only subgroups of order (which must also be transitive).
All subgroups of order must be intransitive by the above analysis since , so by the above, a subgroup of order must be isomorphic to and thus must be the image of an embedding of into . is generated by transpositions (as is for any ), so we can determine embeddings of into by looking at the image of transpositions. But the images of the three transpositions in are determined by the images of and since . So we may send and to any pair of transpositions in with a common element; there are four such pairs and thus four embeddings. These correspond to four distinct subgroups of , all conjugate, and all isomorphic to :
(The fact that transpositions in must be mapped to transpositions in rather than elements of cycle type is left to the reader).
We shall see that some subgroups of order are transitive while others are intransitive. A subgroup of order four is clearly isomorphic to either or to . The only elements of order are the -cycles, so each -cycle generates a subgroup isomorphic to , which also contains the inverse of the -cycle. Since there are six -cycles, has three cyclic subgroups of order , and each is obviously transitive:
Two of the are of cycle type . Then the third is as well, since the product of any pair of the elements of of cycle type is the third such. In this case, the group is
This group acts transitively.
One of the is of cycle type and the other two are of cycle type . In this case, the two -cycles must be disjoint, since otherwise their product is a -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 .
Finally, we have a number of subgroups of order and generated by elements of those orders; all of these are intransitive.
Of these, the only proper nontrivial normal subgroups of are and the group (see the article on normal subgroups of the symmetric groups).
|Date of creation||2013-03-22 17:40:54|
|Last modified on||2013-03-22 17:40:54|
|Last modified by||rm50 (10146)|