$G$ 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.

First, decompose $G=C_1\times\cdots\times C_n$, via the Fundamental theorem of finitely generated abelian groups, where the $C_n$ are cyclic. Let's show that the orders of the $C_i$ are pairwise coprime. Let's start, say, with $C_1$ and $C_2$.
Suppose $d$ divides both $|C_1|$ and $|C_2|$. Then $C_1$ has an element $\alpha$ of order $d$, and $C_2$ also has an element $\beta$ of order $d$. Then the elements of the form $(\alpha,\beta^i)$ and $(\alpha^i,\beta)$, $i=0,\ldots,d-1$ have order $d$. Thus we've found $2d-1$ elements of order $d$, so the hypothesis implies $2d-1\leq d$ and thus $d=1$.
Thus, $|C_1|$ and $|C_2|$ are coprime, and therefore $D_1:=C_1\times C_2$ is cyclic (Product of two cyclic groups is cyclic iff their orders are co-prime).
But now we have $G=D_1\times C_3\times\cdots C_n$, a product of $n-1$ cyclic groups. Proceeding by induction on the number of terms, we conclude that $G$ is cyclic.
