Q:

Any subgroup of free abelian group $\mathbb{Z}^r$ of index $n$ contains subgroup $n\mathbb{Z}^r$.

Any subgroup of index $n$ of free abelian group $G= \mathbb{Z}^r$
  contains subgroup $n\mathbb{Z}^r$.

My attempt to prove this is as follows:
If I have any subgroup $F$ of index $n$ of G, then restriction of F to any copy of $\mathbb{Z}$ has to give subgroup of index dividing $n$. Hence, in case of integers, has to contain n.
Premise of this solution is that of a subgroup of $\mathbb{Z}$ is generated by $gcd$ of elements. 
I'm not convinced with my solution, I would appreciate if someone could point out a mistake or give some more details.

A:

Your proof is correct. Maybe one could restate it this way: if $F \le \mathbb{Z}^r$ has index $n$, then the quotient $\mathbb{Z}^r/F$ is an abelian group of order $n$. Let $e_i$ generate the $i$th $\mathbb{Z}$ factor and let $\pi\colon \mathbb{Z}^r \to \mathbb{Z}^r/F$ be the quotient map. The order of $\pi(e_i)$ in $\mathbb{Z}^r/F$ divides $n$, as you say. In particular, $\pi(ne_i) = n\pi(e_i) = 0$. Thus for each $i$, $1\le i\le r$, $ne_i$ is in the kernel of $\pi$, which implies that $ne_i$ is contained in $F$. But then the subgroup they generate, namely $n\mathbb{Z}^r$, is also contained in the kernel. 

