There is this one problem in Herstein's book which recieved more comments than anything else.

"Show that is a finite abelian group has subgroups of orders and then it has a subgroup of order ".

An unfair way to solve this problem is to use the fundamental theorem on finite abelian groups, meaning since divide then so does their least common multiple. But then there exists a subgroup of that order.

I found a simplified proof to this using the first Sylow theorem. I will prove a stronger result. Let be a divisor of the order of a finite abelian group then it has a subgroup of that order. This is more than what is being asked but it works "easily".*

First, let be subgroups of a finite group. So that . Then , because if and then is a divisor of and since their orders are relatively prime it means . Now if is such a group so that for all then the group product is a subgroup. In particular if the group is abelian then this property must hold and so the group product is always a subgroup. The fundamental observation is that if are subgroups with orders which are relatively prime then is a subgroup (we are assuming we are working in a abelian group of course) is a subgroup of order . The proof behind this rests in the identity that , since as we explained that it means . Now finally we can complete the proof. Let be a divisor of the order of the finite abelian group. Certainly if then there is a subgroup of this order. So it is safe to assume that , in that case the we can factorize where each is a divisor of . By Sylow's theorem there is a subgroup having this order. If we let be the subgroup product of all then it has order because for . Now by above it shows that has order .

*)I use this expression because maybe Sylow's first theorem is considered heavy machinery. I do not consider it to be, but I have seen others think like that.