Difference between revisions of "Homogeneous set"

Latest revision as of 17:45, 9 September 2008

Let $G$ be a group acting on a set $S$ . If $S$ has only one orbit, then the operation of $G$ on $S$ is said to be transitive, and the $G$ -set $S$ is called homogeneous, or that $S$ is a homogeneous set under $G$ .

If $G$ operates on a set $S$ , then each of the orbits of $S$ is homogenous under the induced operation of $G$ .

Groups acting on their own cosets; structure of homogeneous sets

Let $G$ be a group, $H$ a subgroup of $G$ , and $N$ the normalizer of $H$ . Then $G$ operates on the left on $G/H$ , the set of left cosets of $G$ modulo $H$ ; evidently, $G/H$ is a homogenous $G$ -set. Furthermore, $N$ operates on $G/H$ from the right, by the operation $n: gH \mapsto gHn = gnH$ . The operation of $H$ is trivial, so $N/H$ operates likewise on $G/H$ from the right. Let $\phi : (N/H)^0 \to \mathfrak{S}_{G/H}$ be the homomorphism of the opposite group of $N/H$ into the group of permutations on $G/H$ represented by this operation.

Proposition 1. The homomorphism $\phi$ induces an isomorphism from $(N/H)^0$ to the group of $G$ -automorphisms on $G/H$ .

Proof. First, we prove that the image of $\phi$ is a subset of the set of automorphisms on $G/H$ . Evidently, each element of $N/H$ is associated with a surjective endomorphism; also if $xHn = xHm,$ it follows that $Hnm^{-1} = H$ , whence $nm^{-1} \in H$ ; for $n,m \in N$ , this means $n \equiv m \pmod{H}$ . Therefore each element of $N/H$ is associated with a unique automorphism of the $G$ -set $G/H$ .

Next, we show that each automorphism $f$ of $G/H$ has an inverse image under $\phi$ . Evidently, the stabilizer of $f(H)$ is the same as the stabilizer of $H$ , which is $H$ itself. Suppose that $x$ is an element of $G$ such that $f(H) = xH$ . If $k$ is an element of the stabilizer of $xH$ , then $x^{-1}kxH \subseteq H$ , whence $x^{-1}kxH \subseteq H$ , or $k \in xHx^{-1}$ . Since every element of $xHx^{-1}$ stabilizes $xH$ , it follows that $xHx^{-1}$ is the stabilizer of $xH = f(H)$ . Therefore $xHx^{-1} = H$ , so $x\in N$ . $\blacksquare$

Let $\phi : G \to \mathfrak{S}_{G/H}$ the homomorphism corresponding to the action of $G$ on $G/H$ . An element $\alpha$ of $G$ is in the kernel of $\phi$ if and only if it stabilizes every left coset modulo $H$ ; since the stabilizers of these cosets are the conjugates of $H$ (proven in the article on stabilizers), it follows that $\text{Ker}(\phi)$ is the intersection of the conjugates of $H$ .

If $N$ is a normal subgroup of $G$ that is contained in $H$ , then for all $\alpha \in G$ , then $N = \alpha N \alpha^{-1} \subseteq \alpha H \alpha^{-1}$ . Therefore $N \subseteq \bigcap_{\alpha \in G} \alpha H \alpha^{-1} = \text{Ker}(\phi).$ Since $\text{Ker}(\phi)$ is evidently a normal subgroup of $G$ , it is thus the largest normal subgroup of $G$ that $H$ contains.

Proposition 2. Let $G$ be a group acting transitively on a set $S$ ; let $a$ be an element of $S$ , $H$ the stabilizer of $a$ , and $K$ a subgroup of $H$ . Then there exists a unique $G$ -morphism $f : G/K \to G/S$ for which $f(K) = a$ ; this mapping is surjective, and if $H=K$ , is is an isomorphism

Proof. We first note that if $f$ is a mapping satisfying this requirement, then for any $\alpha \in G$ , $f(\alpha K) = \alpha a$ ; thus $f$ is unique if it exists.

We next observe that for $\alpha, \beta \in G$ , the relation $\alpha K = \beta K$ implies $\alpha\beta^{-1} \in H$ , so $\alpha \beta^{-1}$ stabilizes $a$ and $\alpha a = \beta a$ . In other words, the equivalence relation $\alpha \equiv \beta \pmod{H}$ (with left equivalence) is compatible with the equivalence relation $\alpha a = \alpha b$ . Thus the mapping $f: \alpha H \mapsto \alpha a$ from $G/K$ to $E$ is well defined. Since $S$ is homogeneous, for each $b \in S$ , there exists $\alpha \in G$ such that $\alpha a = b$ ; then $f(\alpha K) = b$ , so $f$ is surjective.

If $H=K$ , then $\alpha = \beta \pmod{H}$ is equivalent to the relation $\alpha a = \beta a$ . It then follows that $f$ is injective, and thus an isomorphism. $\blacksquare$

Theorem. Let $G$ be a group. Then every homogeneous $G$ -set is isomorphic to a $G$ -set of the form $G/H$ , for some subgroup $H$ of $G$ . Also, if $H,H'$ are subgroups of $G$ , then the homogeneous $G$ -sets $G/H$ , $G/H'$ are isomorphic if and only if $H$ and $H'$ are conjugate subgroups of $G$ .

Proof. Suppose $S$ is a homogeneous $G$ -set; let $H$ be the stabilizer of $S$ . Then by the previous proposition, the homogeneous $G$ -sets $S$ and $G/H$ are isomorphic.

Suppose now that $G/H$ and $G/H'$ are isomorphic left $G$ -sets; let $f: G/H \to G/H'$ be a $G$ -isomorphism. Evidently, $H$ is its own stabilizer. By transport of structure, the stabilizer of $H$ is also the stabilizer of $f(H)$ . Let $\alpha$ be an element of $G$ such that $f(H) = \alpha H'$ . But the stabilizer of $\alpha H'$ is $\alpha H' \alpha^{-1}$ , which is the image of $H$ under $\text{Int}(\alpha)$ , and which is equal to $H$ . Thus $H$ and $H'$ are conjugates.

Conversely, suppose that $H' = \alpha H \alpha^{-1}$ , for some $\alpha \in G$ . Then $H'$ is the stabilizer of $\alpha H$ , so by Proposition 2, the $G$ -sets $G/H$ and $G/H'$ are isomorphic. $\blacksquare$

@@ Line 1: / Line 1: @@
-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>.
+== 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.
-'''Proposition.'''  The homomorphism <math>\phi</math> induces an isomorphism from <math>(N/H)^0</math> to the group of <math>G</math>-[[automorphism]]s on <math>G/H</math>.
+'''Proposition 1.'''  The homomorphism <math>\phi</math> induces an isomorphism from <math>(N/H)^0</math> to the group of <math>G</math>-[[automorphism]]s on <math>G/H</math>.
 ''Proof.''  First, we prove that the image of <math>\phi</math> is a subset of the set of automorphisms on <math>G/H</math>.  Evidently, each element of <math>N/H</math> is associated with a [[surjective]] [[endomorphism]]; also if
@@ Line 12: / Line 14: @@
 Next, we show that each automorphism <math>f</math> of <math>G/H</math> has an inverse image under <math>\phi</math>.  Evidently, the [[stabilizer]] of <math>f(H)</math> is the same as the stabilizer of <math>H</math>, which is <math>H</math> itself.  Suppose that <math>x</math> is an element of <math>G</math> such that <math>f(H) = xH</math>.  If <math>k</math> is an element of the stabilizer of <math>xH</math>, then <math>x^{-1}kxH \subseteq H</math>, whence <math>x^{-1}kxH \subseteq H</math>, or <math>k \in xHx^{-1}</math>.  Since every element of <math>xHx^{-1}</math> stabilizes <math>xH</math>, it follows that <math>xHx^{-1}</math> is the stabilizer of <math>xH = f(H)</math>.  Therefore <math>xHx^{-1} = H</math>, so <math>x\in N</math>.  <math>\blacksquare</math>
+Let <math>\phi : G \to \mathfrak{S}_{G/H}</math> the homomorphism corresponding to the action of <math>G</math> on <math>G/H</math>.  An element <math>\alpha</math> of <math>G</math> is in the [[kernel]] of <math>\phi</math> if and only if it stabilizes every left coset modulo <math>H</math>; since the stabilizers of these cosets are the [[conjugate (group theory) | conjugates]] of <math>H</math> (proven in the article on [[stabilizer]]s), it follows that <math>\text{Ker}(\phi)</math> is the intersection of the conjugates of <math>H</math>.
+If <math>N</math> is a [[normal subgroup]] of <math>G</math> that is contained in <math>H</math>, then for all <math>\alpha \in G</math>, then <math>N = \alpha N \alpha^{-1} \subseteq \alpha H \alpha^{-1}</math>.  Therefore
+<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.
+'''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 ==
+* [[Homogeneous principal set]]
 * [[Orbit]]
 * [[Stabilizer]]
 [[Category:Group theory]]

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Difference between revisions of "Homogeneous set"

Latest revision as of 17:45, 9 September 2008

Groups acting on their own cosets; structure of homogeneous sets

See also