Mathematical Reflections 2011, Issue 6 - Problem U212

Problem: 
Let $G$ be a finite abelian group such that $G$ contains a subgroup $K \neq \{e\}$ with the
property that $K \subset H$ for each subgroup $H$ of $G$ such that $H \neq \{e\}$. Prove that $G$ is a
cyclic group.

Proposed by Daniel Lopez Aguayo.

Solution:

We use the following  

Lemma.
A finite abelian group is not cyclic if and only if it contains a subgroup isomorphic to $\mathbb{Z}_p \times \mathbb{Z}_p$ for some prime $p$.

Proof.
Let $G$ be a finite abelian group. If $G$ contains a subgroup isomorphic to $\mathbb{Z}_p \times \mathbb{Z}_p$ (which is not cyclic) then $G$ can't be cyclic since every subgroup of a cyclic group is cyclic. Conversely, if $G$ is a finite abelian group which is not cyclic then (by Fundamental Theorem of Finitely Generated Abelian Groups) $G$ contains a subgroup isomorphic to $\mathbb{Z}_{p^a} \times \mathbb{Z}_{p^b}$, because if all the components in the direct product correspond to distinct primes, then $G$ would be cyclic. So, the subgroup $(p^{a-1}) \times (p^{b-1})$ of $\mathbb{Z}_{p^a} \times \mathbb{Z}_{p^b}$ is isomorphic to $\mathbb{Z}_p \times \mathbb{Z}_p$.

Now, suppose that $G$ is not a cyclic group. Then $G$ contains a subgroup isomorphic to $\mathbb{Z}_p \times \mathbb{Z}_p$ and this subgroup, by Cauchy's Theorem, contains a subgroup $H$ which is isomorphic to $\mathbb{Z}_p$. If $K \neq \{e\}$ is a proper subgroup of $H$, then, by Lagrange's Theorem, $|K|$ divides $|H|$, i.e. $|K|$ divides $p$. So $K=\{e\}$ or $K=H$, contradiction.