Let G be a finite group and H be a subgroup of G. For a in G, define aH = {a*h | h in H}. (a) Show that |aH| = |H|. (b) Show that for every pair of elements a, b in G, either aH = bH or aH and bH are disjoint. (c) Use the above to argue that the order of H must divide the order of G.

GATE 1999 · Discrete Mathematics · Group Theory · medium

Answer: |aH| = |H|; distinct cosets are disjoint and partition G; therefore |G| = [G:H]*|H| and the order of H divides the order of G (Lagrange's Theorem).

1. Part (a): cosets have size |H|: L_a is onto aH by definition of aH; it is one-to-one because a*h1 = a*h2 implies h1 = h2 after left-multiplying by a^(-1) (which exists in the group). A bijection between finite sets equates their sizes, so |aH| = |H|
2. Part (b): two cosets are equal or disjoint: suppose x is in both aH and bH, so x = a*h1 = b*h2 with h1,h2 in H. Then a = b*h2*h1^(-1), and since h2*h1^(-1) is in H, every a*h = b*(h2*h1^(-1)*h) lies in bH, giving aH ⊆ bH; by symmetry bH ⊆ aH, so aH = bH. Hence any two cosets either coincide or have empty intersection
3. Part (c): conclude |H| divides |G|: every element a lies in its own coset aH, so the distinct cosets cover G; by part (b) they are pairwise disjoint and by part (a) each has |H| elements. If there are [G:H] distinct cosets then |G| = [G:H] * |H|, an exact multiple of |H|, so |H| divides |G|