Suppose that $G$ is a finite abelian group that does not contain a subgroup isomorphic to $\mathbb Z_p\oplus\mathbb Z_p$ for any prime $p$. Prove that $G$ is cyclic.
Attempt: If $G$ is a finite abelian group, then let $H$ be any subgroup of $G$
It's given that $H \not\simeq \mathbb{Z}_p \oplus \mathbb{Z}_p$ which can be due to a variety of reasons like : $|H|$ may not be $p^2$ or $H$ may not contain any element of order $p$ etc
Hence, the process of finding an element $g$ such that $|g| = |G|$ seems difficult, hence, probably the best bet would be to first assume that $G$ is not cyclic.
Hence, $O(g) \neq |G|~ \forall ~ g \in G$ .
Also, $G \not\simeq Z_{|G|}$ since $G$ is not cyclic.
How do i arrive at a contradiction from here that if $G$ is not cyclic, it must contain a subgroup $H \simeq \mathbb Z_p \oplus \mathbb Z_p $.
Please note that this question occurs in Gallian before normal and factor groups are introduced.
Thank you for help.
Best Answer
Hint: decompose $G$ into the direct sum of its primary components; then examine each primary component and deduce it's cyclic because it has a minimum nonzero subgroup (that is, the intersection of its nonzero subgroups is nonzero).