Homogeneous set

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 $G$ .

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. 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 $x\in N$ . $\blacksquare$

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Homogeneous set

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