## Transitive group actions induce transitive actions on the orbits of the action of a subgroup

Suppose $G \leq S_A$ acts transitively on $A$ and let $H \leq G$ be normal. Let $O_1, O_2, \ldots, O_r$ be the distinct orbits of $H$ on $A$.

1. Prove that $G$ permutes the sets $O_k$ in the sense that for each $g \in G$ and each $i \in \{1,2,\ldots,r\}$ there exists a $j$ such that $g \cdot O_i = O_j$. Prove that $G$ is transitive on $\{ O_1,O_2,\ldots,O_r \}$. Deduce that all orbits of $H$ on $A$ have the same cardinality.
2. Prove that if $a \in O_1$ then $|O_1| = [H : H \cap \mathsf{stab}_G(a)]$ and that $r = [G : H \mathsf{stab}_G(a)]$. [Hint: Note that $H \cap \mathsf{stab}_G(a) = \mathsf{stab}_H(a)$ and use the Second Isomorphism Theorem.]

1. Let $g \in G$ and $a \in A$. Now $H \cdot a$ is an arbitrary orbit of $H$ on $A$. Moreover, $g \cdot a = b$ for some $b \in A$. Then $g \cdot (H \cdot a) = gH \cdot a = Hg \cdot a = H \cdot (g \cdot a) = H \cdot b$, since $H$ is normal. Thus $G$ permutes the orbits $\{ H \cdot a \ |\ a \in A \}$.

Now let $a,b \in A$ be arbitrary; since the action of $G$ on $A$ is transitive, there exists $g \in G$ such that $g \cdot a = b$. Then $g \cdot (H \cdot a) = gH \cdot a = Hg \cdot a = H \cdot (g \cdot a) = H \cdot b$; thus the action of $G$ on the orbits $\{ H \cdot a \ |\ a \in A \}$ is transitive.

Let $a,b \in A$ be arbitrary, and let $g \in G$ such that $g \cdot a = b$. Because $H$ is normal, for every $h \in H$ there exists a unique $k \in H$ such that $gh = kg$. We have a mapping $\varphi_g : H \cdot a \rightarrow H \cdot b$ given by $\varphi_g(h \cdot a) = k \cdot b$. Because $gH = Hg$, this mapping $\varphi_g$ is bijective, and $|H \cdot a| = |H \cdot b|$. Because the action of $G$ is transitive on the orbits, all orbits have the same cardinality.

2. Let $a \in A$. We can restrict the action of $G$ on the orbits of $H$ to $H$; with this action we have $|H \cdot a| = [H : \mathsf{stab}_H(a)]$. Clearly $\mathsf{stab}_H(a) = H \cap \mathsf{stab}_G(a)$; thus $|H \cdot a| = [H : H \cap \mathsf{stab}_G(a)]$. Considering now the (transitive) action of $G$ on the orbits of $H$, we have $r = |\{ H \cdot a \ |\ a \in A \}| = [G : \mathsf{stab}_G(H \cdot a)]$.

We claim that $\mathsf{stab}_G(H \cdot a) = H \mathsf{stab}_G(a)$. Proof of claim: Suppose $g \in \mathsf{stab}_G(H \cdot a)$. Now $Hg \cdot a = gH \cdot a = g \cdot (H \cdot a) = H \cdot a$, so that in particular $g \cdot a = h \cdot a$ for some $h \in H$. Then $h^{-1}g \cdot a = a$, so that $h^{-1}g \in \mathsf{stab}_G(a)$. Thus $g \in H \mathsf{stab}_G(a)$. Now suppose $g \in H \mathsf{stab}_G(a)$. Then $g = hx$ for some $h \in H$ and $x \in \mathsf{stab}_G(a)$. Now $g \cdot (H \cdot a) = gH \cdot a = Hg \cdot a = Hhx \cdot a = H \cdot a$, so that $g \in \mathsf{stab}_G(H \cdot a)$. Thus $r = [G : H \mathsf{stab}_G(a)]$.

An easy way to show $r=|G:HG_a|$ in 2: Since all orbits of $H$ on $A$ have the same cardinality, we see that $r=|A|/|O_1|=|A|/|H:H \cap G_a|=|A|/|HG_a:G_a|$ (using the second isomorphism theorem). Also since $G$ is transitive, $|A|=|Ga|=|G:G_a|=|G:HG_a||HG_a:G_a|$ (using that subgroup index is multiplicative). Substituting this yields that $r=|G:HG_a|$.