Brian Bi
Exercise 6.10.2 Let $$s_1, s_2 \in S$$. Since $$G$$ acts transitively on $$S$$, there is some $$g$$ such that $$gs_1 = s_2$$. Let $$H_1 = \{h \mid h \in G, s_1 \in hU\}$$ and likewise $$H_2 = \{h \mid h \in G, s_2 \in hU\}$$. We are asked to show that $$|H_1| = |H_2|$$. Suppose $$h \in H_1$$. Then $$s_2 = gs_1 \in ghU$$, so $$gh \in H_2$$. The map $$h \mapsto gh$$ is one-to-one, so $$|H_2| > |H_1|$$. We can similarly show that if $$h \in H_2$$ then $$g^{-1}h \in H_1$$ so $$|H_1| > |H_2|$$. Therefore $$|H_1| = |H_2|$$.
Exercise 6.10.3 The orbit of $$U$$ is also the orbit of $$H$$, so we can simply replace $$U$$ with $$H$$ in the description. That is, we are given that $$H$$ is a subset of $$G$$ such that the sets $$gH$$ partition $$G$$ and that $$1 \in H$$. So suppose $$h \in H$$. Then the two subsets $$hH$$ and $$1H$$ must either be disjoint or else coincide. But $$hH$$ and $$1H$$ both contain the element $$h$$, implying they coincide. But $$1H$$ contains 1, so $$hH$$ must contain 1 as well; that is, $$h^{-1} \in H$$. Now suppose $$a, b \in H$$. Again, the two subsets $$aH$$ and $$1H$$ must either be disjoint or coincide, but both contain $$a$$, so they coincide. And $$aH$$ contains $$ab$$, so $$1H$$ must also contain $$ab$$, that is, $$ab \in H$$. So $$H$$ contains 1 and is closed under inverses and the operation of $$G$$; $$H$$ is a subgroup of $$G$$.