Let $\mathbb{Z}^n, n\in\mathbb{N}$ be a free abelian group.
Prove the intersection of index $h$ subgroups is the subgroup $(h\mathbb{Z})^n \space \forall h\in \mathbb{N}$
I think I have to prove that $(h\mathbb{Z})^n$ lies in each subgroup with index $h$ and it's the maximal subgroup on the intersection, is it correct ?
I am really not sure how to approach this problem, any help is welcome.
Instead of subgroups of index $h$, we consider surjective maps $\rho:\mathbb Z^n\rightarrow A$ where $A$ is an abelian group of order $h$, then $\rho(ha)=h \rho(a)=0$, hence $h\mathbb Z^n\subset \ker \rho$. This shows that $h\mathbb Z^n$ is included in the intersection.
If $(a_1, \cdots, a_n)\in \mathbb Z^n$ with $h\nmid a_j$. Consider the homomorphism from $\mathbb Z^n$ to $\mathbb Z/h\mathbb Z$ defined by $\rho(e_i)=0$ for $i\not=j$, and $\rho(e_j)=1+h \mathbb Z$. By the universal property of free abelian groups, such a homomorphism exists, and it's clearly surjective as $\rho(e_j)$ is a generator, but the kernel doesn't include $a$ since $\rho(a) = a_j\rho(e_j)\not=0$.