Let be a group acting on a set . If has only one orbit, then the operation of on is said to be transitive, and the -set is called homogeneous, or that is a homogeneous set under .
If operates on a set , then each of the orbits of is homogenous under the induced operation of .
Groups acting on their own cosets; structure of homogeneous sets
Let be a group, a subgroup of , and the normalizer of . Then operates on the left on , the set of left cosets of modulo ; evidently, is a homogenous -set. Furthermore, operates on from the right, by the operation . The operation of is trivial, so operates likewise on from the right. Let be the homomorphism of the opposite group of into the group of permutations on represented by this operation.
Proposition 1. The homomorphism induces an isomorphism from to the group of -automorphisms on .
Proof. First, we prove that the image of is a subset of the set of automorphisms on . Evidently, each element of is associated with a surjective endomorphism; also if it follows that , whence ; for , this means . Therefore each element of is associated with a unique automorphism of the -set .
Next, we show that each automorphism of has an inverse image under . Evidently, the stabilizer of is the same as the stabilizer of , which is itself. Suppose that is an element of such that . If is an element of the stabilizer of , then , whence , or . Since every element of stabilizes , it follows that is the stabilizer of . Therefore , so .
Let the homomorphism corresponding to the action of on . An element of is in the kernel of if and only if it stabilizes every left coset modulo ; since the stabilizers of these cosets are the conjugates of (proven in the article on stabilizers), it follows that is the intersection of the conjugates of .
If is a normal subgroup of that is contained in , then for all , then . Therefore Since is evidently a normal subgroup of , it is thus the largest normal subgroup of that contains.
Proposition 2. Let be a group acting transitively on a set ; let be an element of , the stabilizer of , and a subgroup of . Then there exists a unique -morphism for which ; this mapping is surjective, and if , is is an isomorphism
Proof. We first note that if is a mapping satisfying this requirement, then for any , ; thus is unique if it exists.
We next observe that for , the relation implies , so stabilizes and . In other words, the equivalence relation (with left equivalence) is compatible with the equivalence relation . Thus the mapping from to is well defined. Since is homogeneous, for each , there exists such that ; then , so is surjective.
If , then is equivalent to the relation . It then follows that is injective, and thus an isomorphism.
Theorem. Let be a group. Then every homogeneous -set is isomorphic to a -set of the form , for some subgroup of . Also, if are subgroups of , then the homogeneous -sets , are isomorphic if and only if and are conjugate subgroups of .
Proof. Suppose is a homogeneous -set; let be the stabilizer of . Then by the previous proposition, the homogeneous -sets and are isomorphic.
Suppose now that and are isomorphic left -sets; let be a -isomorphism. Evidently, is its own stabilizer. By transport of structure, the stabilizer of is also the stabilizer of . Let be an element of such that . But the stabilizer of is , which is the image of under , and which is equal to . Thus and are conjugates.
Conversely, suppose that , for some . Then is the stabilizer of , so by Proposition 2, the -sets and are isomorphic.