Difference between revisions of "Homogeneous set"
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− | Let <math>G</math> be a group acting on a set <math>S</math>. If <math>S</math> has only one orbit, then the operation of <math>G</math> on <math>S</math> is said to be ''transitive'', and the <math>G</math>-set <math>S</math> is called '''homogeneous''', or that <math>S</math> is a homogeneous set under <math>G</math>. | + | Let <math>G</math> be a group [[group action|acting]] on a set <math>S</math>. If <math>S</math> has only one orbit, then the operation of <math>G</math> on <math>S</math> is said to be ''transitive'', and the <math>G</math>-set <math>S</math> is called '''homogeneous''', or that <math>S</math> is a homogeneous set under <math>G</math>. |
If <math>G</math> operates on a set <math>S</math>, then each of the [[orbit]]s of <math>S</math> is homogenous under the induced operation of <math>G</math>. | If <math>G</math> operates on a set <math>S</math>, then each of the [[orbit]]s of <math>S</math> is homogenous under the induced operation of <math>G</math>. | ||
− | == | + | == Groups acting on their own cosets; structure of homogeneous sets == |
Let <math>G</math> be a group, <math>H</math> a subgroup of <math>G</math>, and <math>N</math> the [[normalizer]] of <math>H</math>. Then <math>G</math> operates on the left on <math>G/H</math>, the set of left [[coset]]s of <math>G</math> modulo <math>H</math>; evidently, <math>G/H</math> is a homogenous <math>G</math>-set. Furthermore, <math>N</math> operates on <math>G/H</math> from the right, by the operation <math>n: gH \mapsto gHn = gnH</math>. The operation of <math>H</math> is trivial, so <math>N/H</math> operates likewise on <math>G/H</math> from the right. Let <math>\phi : (N/H)^0 \to \mathfrak{S}_{G/H}</math> be the [[homomorphism]] of the opposite group of <math>N/H</math> into the group of permutations on <math>G/H</math> represented by this operation. | Let <math>G</math> be a group, <math>H</math> a subgroup of <math>G</math>, and <math>N</math> the [[normalizer]] of <math>H</math>. Then <math>G</math> operates on the left on <math>G/H</math>, the set of left [[coset]]s of <math>G</math> modulo <math>H</math>; evidently, <math>G/H</math> is a homogenous <math>G</math>-set. Furthermore, <math>N</math> operates on <math>G/H</math> from the right, by the operation <math>n: gH \mapsto gHn = gnH</math>. The operation of <math>H</math> is trivial, so <math>N/H</math> operates likewise on <math>G/H</math> from the right. Let <math>\phi : (N/H)^0 \to \mathfrak{S}_{G/H}</math> be the [[homomorphism]] of the opposite group of <math>N/H</math> into the group of permutations on <math>G/H</math> represented by this operation. | ||
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<cmath> N \subseteq \bigcap_{\alpha \in G} \alpha H \alpha^{-1} = \text{Ker}(\phi). </cmath> | <cmath> N \subseteq \bigcap_{\alpha \in G} \alpha H \alpha^{-1} = \text{Ker}(\phi). </cmath> | ||
Since <math>\text{Ker}(\phi)</math> is evidently a normal subgroup of <math>G</math>, it is thus the ''largest'' normal subgroup of <math>G</math> that <math>H</math> contains. | Since <math>\text{Ker}(\phi)</math> is evidently a normal subgroup of <math>G</math>, it is thus the ''largest'' normal subgroup of <math>G</math> that <math>H</math> contains. | ||
+ | |||
+ | '''Proposition 2.''' Let <math>G</math> be a group acting transitively on a set <math>S</math>; let <math>a</math> be an element of <math>S</math>, <math>H</math> the stabilizer of <math>a</math>, and <math>K</math> a subgroup of <math>H</math>. Then there exists a unique <math>G</math>-morphism <math>f : G/K \to G/S</math> for which <math>f(K) = a</math>; this mapping is [[surjective]], and if <math>H=K</math>, is is an [[isomorphism]] | ||
+ | |||
+ | ''Proof.'' We first note that if <math>f</math> is a mapping satisfying this requirement, then for any <math>\alpha \in G</math>, <math>f(\alpha K) = \alpha a</math>; thus <math>f</math> is unique if it exists. | ||
+ | |||
+ | We next observe that for <math>\alpha, \beta \in G</math>, the relation <math>\alpha K = \beta K</math> implies <math>\alpha\beta^{-1} \in H</math>, so <math>\alpha \beta^{-1}</math> stabilizes <math>a</math> and <math>\alpha a = \beta a</math>. In other words, the [[equivalence relation]] <math>\alpha \equiv \beta \pmod{H}</math> (with left equivalence) is compatible with the equivalence relation <math>\alpha a = \alpha b</math>. Thus the mapping <math>f: \alpha H \mapsto \alpha a</math> from <math>G/K</math> to <math>E</math> is well defined. Since <math>S</math> is homogeneous, for each <math>b \in S</math>, there exists <math>\alpha \in G</math> such that <math>\alpha a = b</math>; then <math>f(\alpha K) = b</math>, so <math>f</math> is surjective. | ||
+ | |||
+ | If <math>H=K</math>, then <math>\alpha = \beta \pmod{H}</math> is ''equivalent'' to the relation <math>\alpha a = \beta a</math>. It then follows that <math>f</math> is [[injective]], and thus an isomorphism. <math>\blacksquare</math> | ||
+ | |||
+ | '''Theorem.''' Let <math>G</math> be a group. Then every homogeneous <math>G</math>-set is isomorphic to a <math>G</math>-set of the form <math>G/H</math>, for some subgroup <math>H</math> of <math>G</math>. Also, if <math>H,H'</math> are subgroups of <math>G</math>, then the homogeneous <math>G</math>-sets <math>G/H</math>, <math>G/H'</math> are isomorphic if and only if <math>H</math> and <math>H'</math> are conjugate subgroups of <math>G</math>. | ||
+ | |||
+ | ''Proof.'' Suppose <math>S</math> is a homogeneous <math>G</math>-set; let <math>H</math> be the stabilizer of <math>S</math>. Then by the previous proposition, the homogeneous <math>G</math>-sets <math>S</math> and <math>G/H</math> are isomorphic. | ||
+ | |||
+ | Suppose now that <math>G/H</math> and <math>G/H'</math> are isomorphic left <math>G</math>-sets; let <math>f: G/H \to G/H'</math> be a <math>G</math>-isomorphism. Evidently, <math>H</math> is its own stabilizer. By transport of structure, the stabilizer of <math>H</math> is also the stabilizer of <math>f(H)</math>. Let <math>\alpha</math> be an element of <math>G</math> such that <math>f(H) = \alpha H'</math>. But the stabilizer of <math>\alpha H'</math> is <math>\alpha H' \alpha^{-1}</math>, which is the image of <math>H</math> under <math>\text{Int}(\alpha)</math>, and which is equal to <math>H</math>. Thus <math>H</math> and <math>H'</math> are conjugates. | ||
+ | |||
+ | Conversely, suppose that <math>H' = \alpha H \alpha^{-1}</math>, for some <math>\alpha \in G</math>. Then <math>H'</math> is the stabilizer of <math>\alpha H</math>, so by Proposition 2, the <math>G</math>-sets <math>G/H</math> and <math>G/H'</math> are isomorphic. <math>\blacksquare</math> | ||
== See also == | == See also == | ||
+ | * [[Homogeneous principal set]] | ||
* [[Orbit]] | * [[Orbit]] | ||
* [[Stabilizer]] | * [[Stabilizer]] | ||
[[Category:Group theory]] | [[Category:Group theory]] |
Latest revision as of 17:45, 9 September 2008
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.