If $A$ is an Abelian group and $B < A$, $ A \simeq B \simeq \mathbb Z^n$ for some natural $n$. Prove that $mA \subset B $ for some $m$. If $A$ is an Abelian group and $B < A$, $ A \simeq B \simeq \mathbb Z^n$ for some natural $n$. Prove that $mA \subset B $ for some $m$.
I know this has something to do with the fact that there exists bases for both A and B, but I don't understand what $mA$ denotes. Any help would be appreciated!
 A: Without loss of generality, we may assume that $A$ is a subgroup of $\mathbb Q^n$.
Then $A \simeq B \simeq \mathbb Z^n$ implies that the $\mathbb Q$-subspaces generated by $A$ and $B$ are both equal to $\mathbb Q^n$. In particular, both $A$ and $B$ contain bases for $\mathbb Q^n$.
Write the basis in $A$ in terms of the basis in $B$. Then $mA \subset B$ for $m$ equal to denominator of the determinant of the basis transformation matrix.
A: We have the following exact sequence
$$
0 \rightarrow B \rightarrow A \rightarrow A/B \rightarrow 0. 
$$
Since $B \cong A \cong \mathbb{Z}^n, A/B \cong \bigoplus_{i = 1}^r \mathbb{Z}/d_i\mathbb{Z},$ for some $d_1, d_2, \cdots , d_r \in \mathbb{N}.$ Take $m = \prod_{i = 1}^r d_i.$ Then $mA = 0$ in the quotient group, i.e. $mA \subseteq B.$
A: Apply the following theorem:
There exists a (free) basis $\;\{a_1,...,a_n\}\;$ of $\;A\;$ and integers $\;d_1,...,d_n\;$ s.t. $\;\{d_1a_1,...,d_na_n\}\;$ is a (free) basis of $\;B\;$
Now just take $\;m:=d_1\cdot\ldots\cdot d_n\;$
Remark: The above quoted theorem is  stronger and the integers can be chosen as to fulfill some divisibility properties, but we don't care about this now.
