Intersecting co-cyclic subgroups Let $G$ be a finite abelian group of type $(m_1,\dotsc,m_r)$; that is, essentially, $G=\mathbb Z_{m_1}\oplus\dotsb\oplus\mathbb Z_{m_r}$ with integer $1<m_1\mid m_2\mid\dotsb\mid m_r$. Dropping for each $i\in[1,r]$ the $i$th direct summand, we get a system of $r$ co-cyclic subgroups of $G$ the intersection of which is the zero subgroup. Can one find fewer than $r$ co-cyclic subgroups of $G$ with zero intersection? What is the smallest $k=k(m_1,\dotsc,m_r)$ for which co-cyclic subgroups $H_1,\dotsc,H_k<G$ with $H_1\cap\dotsb\cap H_k=\{0\}$ exist?

The answer is no, you need at least $r$ such subgroups.
Let $d(G)$ be the smallest size of a generating set of a group $G$. Since the $G$ in the question has an elementary abelian quotient of order $p^r$, where $p$ is a prime dividing $m_1$, we have $d(G)=r$.
For a cocyclic subgroup $H_1$ of $G$, we have $d(G) \le d(H_1) + d(G/H_1) \le d(H_1) + 1$, so $d(H_1) \ge d(G)-1 = r-1$.
Then since $H_2 \cap H_1$ is cocyclic in $H_1$, we get $d(H_2 \cap H_1) \ge d(H_1) - 1$, and by induction $d(H_1 \cap \ldots \cap H_k) \ge r-k$.
