Any subgroup of an abelian group is undistorted. I need some help with the following math problem.  I am studying some notes on Geometric Group Theory and I came across the following problem.  
Prove:
Any subgroup of an abelian group is undistorted.
A subgroup $H$ of a group $G$ is said to undistorted if the inclusion map of $H$ into $G$ is a quasi-isometric embedding.  That is we think of $G$ and $H$ as being identified their unique up to quasi-isomorphism Cayley graphs.
 A: Here is an expansion of Derek Holt's comment, with added details regarding how to deal with various finite index issues that arise along the way. I will use your terminology "undistorted", although the more standard terminology is "quasi-isometry" and "quasi-isometric embedding".
The group $G$ has a finite index free abelian subgroup which I will denote $A$. 
The inclusion maps $i_1 : A \to G$ and $i_2 : A \cap H \to H$ are both undistorted. Also, there are undistorted maps $j_1 : G \to A$, $j_2 : H \to A \cap H$ such that for $m=1,2$ the two maps $i_m,j_m$ are coarse inverses, meaning that $i_m \circ j_m$ moves each group element by a uniformly bounded amount, as does $j_m \circ i_m$. These statements hold in general for any finite index subgroup of any finitely generated group.
If we can prove that the inclusion $i_3 : A \cap H \to A$ is undistorted then it will follows that the composed map 
$$H \xrightarrow{j_2} A \cap H \xrightarrow{i_3} A \xrightarrow{i_1} G
$$
is undistorted, and then we will be done because this composed map differs from the inclusion $H \to G$ by a uniformly bounded amount.
So by replacing $G$ with $A$ and $H$ with $A \cap H$, the problem has been reduced to the case that $G \approx \mathbb{Z}^n$ is free abelian of some rank $n \ge 1$. The subgroup $H$ is free abelian of some rank $k$ with $0 \le k \le n$. In this case you can prove using linear algebra that there exist $v_1,…,v_n \in \mathbb{Z}^n$ which form a basis over $\mathbb{Z}$, and there exist integer constants $a_1,\ldots,a_k \ge 1$, such that $H$ has basis $a_1 v_1, …, a_k v_k$. So the inclusion $H \to G$ is a composition of two undistorted maps: the inclusion map into $\mathbb{Z}^n$ of the space spanned by the vectors $a_1 e_1,…,a_k e_k$; and the change of basis map from the standard basis $e_1,…,e_n$ to the basis $v_1,…,v_k$.
