6.4.3 Lagrange's Theorem Theorem 6.19: Let H be a subgroup of the group G. Then ghgeg and Hglgeg have the same cardinal number ◆ Proof:LetS={Hgg∈G}andT=gHg∈G q:S→T,φ(Ha)=alHl 1)p is an everywhere function. for ha=hb a-H?=b-lH a≠blif|a]n[b]= 2)(p is one-to-one For Ha, Hb, suppose that Ha*Hb, and cp(Ha=(p(Hb) oNto6.4.3 Lagrange's Theorem  Theorem 6.19: Let H be a subgroup of the group G. Then {gH|gG} and {Hg|gG} have the same cardinal number  Proof：Let S={Hg|gG} and T={gH|gG}  : S→T, (Ha)=a-1H。 (1)  is an everywhere function. for Ha=Hb, a -1H?=b-1H [a][b] iff [a]∩[b]= (2)  is one-to-one。 For Ha,Hb,suppose that HaHb，and (Ha)=(Hb) (3)Onto