If an Abelian group $G$ has order $n$ and at most one subgroup of order $d$ for all $d$ dividing $n$ then $G$ is cyclic If an Abelian group $G$ has order $n$ and at most one subgroup of order $d$ for all $d$ dividing $n$ then $G$ is cyclic.
I am trying to use the structure theorem for finitely generated abelian groups.
So I write $n=p_1^{\alpha_1}\ldots p_n^{\alpha_n}$.
I am hoping to show each of the alpha's must =1 then I will have that $G$ is isomorphic to $\prod_i^n \mathbb{Z}/p_i \mathbb{Z}$, which is cyclic. 
 A: Let $\phi$ be Euler's totient function. 
The assumption implies that $G$ has at most $\phi(d)$ elements of order $d$ whenever $d$ divides $n$. 
But then it has at least $\phi(d)$ elements of order $d$ whenever $d$ divides $n$. 
So it has elements of order $n$.
EDIT. As Thomas suggests, I should add that the following fact has been implicitly used: We have 
$$
\sum_{d|n}\ \phi(d) = n
$$
by considering the case when $G$ is cyclic.
A: I would prove this along the lines of J Rotman - An introduction to the theory of Groups, Graduate Texts in Mathematics, Springer-Verlag, 4th Edition. Pg:28
Observe that the converse of what you'd like to prove is quite a trivial issue.
For what you'd like to prove, J R has the following Recipe:
Lemma: Let $\phi$ be the Euler's totient function that counts the number of numbers less than $n$ and coprime to it, $$ \sum_{d|n} \phi (d) = n$$
Proof:
The proof of this fact follows from the following: Let G be a cyclic group of order $n$. I'll assume you can prove that there are $\phi(n)$ generators for this $G$ and in general $\phi(k)$ generators for a cyclic group of order $k$.
$$G=\bigcup gen(C)$$ where $gen(C)$ stands for generators of cyclic subgroups $C$ of G.(So $C$ ranges over all cyclic subgroups of $G$)  And this union is a disjoint union, from the definition of a generator.
Note that the disjoint union holds for all $G$ and there is nothing sacrosanct about $G$ being cyclic, except from here: I claim now that, as the subgroups are unique for each divisor, the disjoint union translates into a sum that is exactly the lemma we seek to prove.
Your Question:
As pointed out before, I'll make use of the disjoint union. 
$$ n=|G| = \sum |gen(C)| \leq \sum_{d|n}{\phi(d)}$$ 
So, now I use my lemma to conclude that, I have cyclic subgroups for each divisor of $n$ and in particular for $n$. So, that must be $G$. So, G must be cyclic. 
I would like to emphasise, that this classifies all finite cyclic groups.
A: Just the most basic group theory and a simple counting argument will suffice. By Lagrange's theorem $G$ only has elements of orders $d$ dividing $n$. If such an element exists, then the cyclic group it generates has a number $\phi(d)$ of elements of order $d$ that only depends on $d$ (this is Euler's totient function but we need to know nothing about it). By the hypothesis $G$ cannot contain other elements of order $d$, so it has either $\phi(d)$ such elements or none; set $\chi_G(d)=1$ if it does have such elements and $\chi_G(d)=0$ if it doesn't. Then counting element of $G$ by their order one has
$$
 n=\#G=\sum_{d\mid n}\chi_G(d)\phi(d).
$$
But a cyclic group of order $n$ has exactly one cyclic subgroup of order $d$ for every $d$ that divides $n$. This means that
$$
 \sum_{d\mid n}\phi(d)=n.
$$
Thus the formula for $G$ can only be satisfied if all values $\chi_G(d)$ are equal to $1$. In particular $\chi_G(n)=1$: the group $G$ contains elements of its own order $n$ and is therefore cyclic.
A: It may be easier if you use the form of the structure theorem that says that $G$ is a product of cyclic groups where the order of each divides the order of the next. 
A: By Cayley Theorem $G$ has an element $x$ of order $p_i$. It follows from your requirement that the subgroup $\langle x\rangle$ generated by $\langle x\rangle$ contains all the elements of order $p_i$.
Exercise for you: Prove that $p_i$ cannot divide the order of the factor group $G/\langle x\rangle$ (Why?)
and you are done...
P.S. If you covered this topic already, looking to the $p$-Sylow subgroups also helps, but I think this is more complicated. :)
