Showing a finite abelian group is cyclic assuming something about all homomorphic images of it Let $G$ be a finite abelian group such that $|G|\ne p^n$ for any prime $p$. If every homomorphic image $\varphi (G)$ with $|\varphi (G)| < |G|$ is cyclic, then show $G$ is cyclic.
This is an old qual problem, and I'm not quite sure where to begin. Perhaps I should look at the prime factorization of $|G|$. Maybe the fact that $|\varphi (G)| | |G|$ comes into play too. Any hints about where to start would be appreciated (not looking for a full solution yet).
 A: Let $|G| = p_1^{\alpha_1}\cdots p_r^{\alpha_r}$. By the structure theorem of finite abelian groups you can write $G \simeq G_{p_1}\times\cdots\times G_{p_r}$ such that each $G_{p_i}$ is a group of order $p_i^{\alpha_i}$. Now, consider the projections $G\to G_{p_i}$ and thus each $G_{p_i}$ must be cyclic. Since the direct product of cyclic groups of coprime order is cyclic, we are done.
A: Here is an argument not using the classification of finite abelian groups, but only Lagrange's and Cauchy's theorems and the Chinese remainder theorem.
For $p$ a prime number, let $G[p^\infty]$ be the subgroup of $G$ of all elements whose order is a power of $p$. Note that $G[p^\infty]$ is a subgroup since $G$ is abelian. The order of $G[p^\infty]$ is a power of $p$ by Cauchy's theorem. Then
$$
G=\bigoplus_{p|\#G}G[p^\infty].
$$
Indeed, the sum is direct since the subgroups
$$
G[p^\infty]\quad\text{and}\quad \sum_{q> p}G[q^\infty]
$$
have relatively prime orders. In fact, the order of the second one is a divisor of a product of powers of $q$, for certain prime numbers $q> p$, by Lagrange's theorem.
Moreover, if $x\in G$ and $n$ is its order, then $\langle x\rangle $ is isomorphic to $\mathbf Z/n$. Since we do have
$$
\mathbf Z/n=\sum_{p|n}(\mathbf Z/n)[p^\infty]
$$
by the Chinese remainder theorem, one also has
$$
x\in\langle x\rangle\subseteq \sum_{p|n}\langle x\rangle[p^\infty]\subseteq\sum_{p|\#G}G[p^\infty].
$$
This proves the direct sum decomposition above.
Since $\#G\neq p^n$ with $p$ prime, the direct sum above has at least two non trivial summands. By hypothesis, each of the summands is cyclic. By the Chinese remainder theorem, $G$ is cyclic.
A: Chapter 2 of the book The Theory of Finite Groups: An Introduction, by Kurzweil and Stellmacher, contains this elementary theorem, which does not depend on the structure theorem of finite abelian groups:

If $G$ is a finite abelian group and $U$ is a cyclic subgroup of maximal order in $G$, then there exists a complement $V$ of
  $U$ in $G$, that is, $G=UV$ and $U \cap V =1$. In particular, $G \cong U \times V$.

Applying this result to our problem, we conclude that $V$ is cyclic because it is a homomorphic image of $G$. But then $G=UV$ and $U \cap V =1$ imply that $G$ is cyclic.
