$G$ is an abelian group of order $n$ with the property that $G$ has at most $d$ elements of order $d$, for any $d$ dividing $n$. Then $G$ is cyclic. $G$ is an abelian group of order $n$ with the property that $G$ has at most $d$ elements of order $d$, for any $d$ dividing $n$. Then $G$ is cyclic.
I am not getting any clue how to start. Please Help.

Induction by n. Suppose the property is true for all $k \lt n$. 
We'll prove it for $n$. If $n$ is prime then $G$ is cyclic. 
Suppose $n=pq$ where $p,q$ are coprime. 
Now let $H_1, H_2$ subgroups of $G$ of orders $p,q$ (we know from Cauchy theorem there are such subgroups). Then, by induction hypothesis, both subgroups are cyclic. 
Let $x \in H_1, y \in H_2$ of order $p, q$. Then z=xy will be of order $pq=n$ so $G$ is cyclic.
Now suppose $n=p^k, k \ge 2$ where $p$ is prime. 
Then $G$ is a p-group ( group in which every element has order $p^k$ for some $k$).
The prove for $G$ cyclic in this case can be found here. Notice that the property "G has at most d elements of order d" implies $G$ has at most $p$ elements of order $p$ hence there is a unique subgroup of order $p$.
