[Math] Finite abelian group with common nontrivial subgroup $H_0$ is cyclic.

Question

let $G$ be a finite abelian group such that it contains a subgroup $H_{0}\neq (e)$ which lies in every subgroup $H\neq(e)$, prove that $G$ is cyclic. what is order of $G$?
i try to solve this problem, but i can not finish..kindly help me, or give any other solution.
I try, since $G$ is finite abelian group, so it isomorphic direct product of its sylow subgroup, $G\thickapprox S(P_{1})*S(P_{2}*)….S(P_{k})$ since $H_{0}\subset S(P_{i})$ and $S(P_{i})\cap S(P_{j})= \phi$ this imlies $k=1$ i.e. $G\thickapprox S(P_{1})$,
from here how i conclude $G$ is cyclic?

Answer 1

Actually the intersection of two distinct Sylow subgroups is $\{e\}$ rather than the empty set. A finite abelian group is a product of cyclic groups. If there is more than one group in the product, you again get a pair of subgroups with trivial intersection $\{e\}$, and therefore they couldn't both contain $H$, giving the required contradiction.