Suppose G is finite abelian and for every divisor d of G the

Suppose G is finite, abelian, and for every divisor d of |G|, there is a unique subgroup H of G with |H|=d. Prove that G is cyclic. [ Note: This is trivial if you use the Fundamental Theorem on finite abelian groups. If you use it, you will get less than half the credit.]

Solution

Let G be a finite abelian group

Let 0(G) =n

If d\ there is a subgroup H of G with order d.

Let H consist of a1, a2....ad where d is finite

As this is a subgroup there is identity element

let ai be the identity element

Since d divides n,

the order of any element in a would be less than d

Let ai be an element such that ai^d =e

Then it follows that ai generates H and hence H is cyclic

If possible let G be not cyclic

that is there exists an element a\' such that a\' not equal ai^k for some integer n

Let k = m mod d

Then a^m where m <d does not belong to H which is a contradiction.

Hence G is also cyclic.