Group theory

28.7 SUBDIVIDING A GROUP• Two cosets X 1 H and X 2 H are identical if, and only if, X2 −1X1 belongs to H. IfX2 −1X1 belongs to H then X 1 = X 2 Y i for some i, andX 1 H = X 2 Y i H = X 2 H,since by the permutation law Y i H = H. Thus the two cosets are identical.Conversely, suppose X 1 H = X 2 H. Then X2 −1X1H = H. But one element ofH (on the left of the equation) is I; thus X2 −1X1 must also be an element of H(on the right). This proves the stated result.• Every element of G is in some left coset XH. This follows trivially since Hcontains I, and so the element X i is in the coset X i H.The final step in establishing Lagrange’s theorem is, as previously, to note thateach coset contains h elements, that the cosets are disjoint and that every one ofthe g elements in G appears in one and only one distinct coset. It follows thatg = kh for some integer k.As noted earlier, Lagrange’s theorem justifies our statement that any group oforder p, where p is prime, must be cyclic and cannot have any proper subgroups:since any subgroup must have an order that divides p, this can only be 1 or p,corresponding to the two trivial subgroups I and the whole group.It may be helpful to see an example worked through explicitly, and we againuse the same six-element group.Find the left cosets of the proper subgroup H of the group G that has table 28.8 as itsmultiplication table.The subgroup consists of the set of elements H = {I, A, B}. We note in passing that it hasorder 3, which, as required by Lagrange’s theorem, is a divisor of 6, the order of G. As inall cases, H itself provides the first (left) coset, formally the cosetIH = {II, IA, IB} = {I, A, B}.We continue by choosing an element not already selected, C say, and formCH = {CI, CA, CB} = {C, D, E}.These two cosets of H exhaust G, and are therefore the only cosets, the index of H in Gbeing equal to 2.This completes the example, but it is useful to demonstrate that it would not havemattered if we had taken D, say, instead of I to form a first cosetDH = {DI, DA, DB} = {D, E, C},and then, from previously unselected elements, picked B, say:The same two cosets would have resulted. BH = {BI, BA, BB} = {B, I, A}.It will be noticed that the cosets are the same groupings of the elementsof G which we earlier noted as being the choice of adjacent column and rowheadings that give the multiplication table its ‘neatest’ appearance. Furthermore,1067