0 votes 0 votes Let G be abelian, H and K subgroups of G with orders n, m. Then G has subgroup of order lcm(n,m) I have gone through prrof here:- https://math.stackexchange.com/questions/465742/let-g-be-abelian-h-and-k-subgroups-of-orders-n-m-then-g-has-subgrou But this uses assertion :- if G is abelian and n divides |G| then G has a subgroup of order n.How is this assertion true. there are some existing examples where n divides G but there is no subgroup with that order.Please clear with explanation. Mathematical Logic discrete-mathematics group-theory abelian-group + – rahul sharma 5 asked Jun 13, 2017 rahul sharma 5 660 views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply junaid ahmad commented Jun 15, 2017 reply Follow Share You are correct whenever order of group divisible by order of subgroup it does not ensure that subgroup is group.IT IS NEGATIVITY TEST LIKE PUMPING LEMMA.we need further investigation to prove subgroup is a group,you can use closure property to ensure it is a subgroup. 0 votes 0 votes rahul sharma 5 commented Jun 15, 2017 reply Follow Share thanks.Can you provide me answer to this question? 0 votes 0 votes rahul sharma 5 commented Jun 19, 2017 reply Follow Share Can anyone provide me a proof of this question? 0 votes 0 votes Please log in or register to add a comment.