Difference between revisions of "Homogeneous set"

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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.
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'''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]]
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''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.
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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.
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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>
  
 
== See also ==
 
== See also ==

Revision as of 15:22, 22 May 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 $G$.

Structure of a group acting on its own cosets

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. $\blacksquare$

See also