Let denote the symmetric group on . A subgroup is said to be a transitive subgroup if the natural action on the set is transitive.
Q : Find all isomorphism classes of transitive subgroups of size n of .
Let us try to find one subgroup .., well, generated by the -cycle . Clearly it is transitive, and in fact, more is true, namely, the stabilizer of any element is the trivial permutation.
Can we think of more ?
Let us look at and look at [this is the cycle notation of permutations]. has order . Now if we can find a which somehow can make 1 go to n+1, then the subgroup generated by would be transitive.
In fact, we can find such that or rather . Just look at . Then =
The subgroup generated by is isomorphic to , so we see that is also a transitive subgroup of size 2n in
In fact, more is true !
Claim : Let be ANY subgroup of size . Then it can be embedded as a transitive subgroup of .
Proof : Let and let be the symmetric group on a set of size n, which we can assume to be G ! We have the following (usual) embedding of into got by the action of on itself by a modified left translation, namely
which sends , where is thought of as set bijections of G and itself.
- The map is a set bijection because
- is a group homomorphism – easy check. [This is why we need instead of
- is a 1-1 function (or an embedding) because the map is identity if and only if which happens if and only if
So G can be identified as a subgroup of via . Now we check that it is a transitive subgroup !
Any can be moved to by the element because .
Infact, what is the stabilizer of any under this action of G ? It is the identity ! So this action of G is in fact regular (or transtive+free = simply transitive, namely given and , there is a unique element in G which takes one to the other)