Homogeneous set
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.