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Difference between revisions of "Homogeneous set"

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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>.
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== Structure of a group acting on its own cosets ==
  
 
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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'''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>.
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'''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
 
''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
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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>
 
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>
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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>.
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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
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<cmath> N \subseteq \bigcap_{\alpha \in G} \alpha H \alpha^{-1} = \text{Ker}(\phi). </cmath>
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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.
  
 
== See also ==
 
== See also ==

Revision as of 23:08, 21 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.

See also

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