*Algebra*

## Section 6.10. Operations on Subsets

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 \(G\).