Let G be a (not necessarily finite) group with two subgroups H and K such that K ! H ! G. The purpose of this question is to establish the index formula |G : K| = |G : H|·|H : K|. Let T be a transversal to K in H and U be a transversal to H in G. (a) By considering the coset Hg or otherwise, show that if g is an element of G, then Kg = Ktu for some t ∈ T and some u ∈ U. (b) If t, t! ∈ T and u, u! ∈ U with Ktu = Kt!u!, first show that Hu = Hu! and deduce u = u!, and then show that t = t!. (c) Deduce that TU = {tu | t ∈ T, u ∈ U } is a transversal to K in G and that |G : K| = |G : H|·|H : K|. (d) Show that this formula follows immediately from Lagrange’s Theorem if G is a finite group.

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