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 .**

**Answer** :

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)

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