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)].

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Comments

  • Gobi Ree  On December 5, 2011 at 2:30 am

    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|.

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