Homogeneous set

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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. $\blacksquare$

See also

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