TITLE: 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
QUESTION [7 upvotes]: 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.

REPLY [11 votes]: 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.