Dummit and Foote Abstract Algebra, section 4.1, exercise 9:
Assume G acts transitively on the finite set A and let H be a normal subgroup of G. Let O1, O2, ..., Or be the distinct orbits of H on A.
(a) Prove that G permutes the sets O1, O2, ..., Or in the sense that for any g∈G, gOi is an orbit (i.e. the orbits are blocks). Prove that G is transitive on the set of orbits. Deduce that all the orbits of H on A have the same cardinality.
(b) Prove that if a∈O1, then | O1 | = | H : H∩Ga | and thus r = | G : HGa |.
Proof: (a) First of all, notice that if H is a normal subgroup, then for all f∈G, k1∈H, fk1f-1 = k2 for some k2∈H, so fk1 = k2f, which is to say the two element can be rewritten in opposite order (while losing the original element in H).
Note that O1 is of the same cardinality as gO1, since gh*a = gj*a ⇒ h*a = j*a. Now, assume that gO1∩O1 ≠ O1, so that there is some m∉O1 such that m = gh*a for some h∈H. Now, if gO1∩O1 is not the null set, then gh1*a = h2*a for some h1, h2∈H, so h2-1gh1 = z∈Ga. Rewrite the two elements so that h2-1h3g = z, so h4g = z, thus g = h5z. Retrace steps: m = h5zh*a. Transpose and rewrite z and h, and collapse the two h terms on the left into one. Thus, m = h6z*a = h6*a∈O1, a contradiction. Thus, the orbits behave like blocks on A. Since G acts transitively on A, any element n of A for n = v*a, belongs to the orbit produced by vO1. So G acts transitively on the orbits in that every element's orbit can be constructed from O1 by the left multiplication of elements in G, so for any i, p∈G, iO1 = (ip-1)pO1.
(b) Clearly, HGa ≤ GO1. Let x∈GO1 and assume x∉HGa. This means that xO1 = O1, so that x*a = h*a for some h∈H. This entails h-1x = z for some z∈Ga, so that x = hz, which obviously violates x's assumption. Therefore HGa = GO1. Since wO1 = qO1 ⇔ q-1w∈GO1 ⇔ wGO1 = qGO1, we have r = | G : GO1 | = | G : HGa |.
As well, we have h*a = j*a ⇔ j-1h ∈ Ha = H∩Ga ⇔ hH∩Ga = jH∩Ga, so that | O1 | = | H : H∩Ga |.
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