Free abelian group. Transformation

Question

I have a question. If I have two free abelian groups $U \subset G$, so that $U \cong \mathbb{Z}^n \cong G$. Why then exists a base for $U$, a base for $G$ and a quadratic diagonal matrix $M$, which entries are integers $\neq 0$, so that $M$ is the transformation matrix of the two bases.

Answer 1

One has even more than that: the diagonal entries $(d_1,\dots,d_n)$ can be chosen such that $$d_1\mid\dots\mid d_i\mid d_{i+1}\mid\dots\mid d_n.$$ These integers are called the invariant factors of the quotient group $G/U$,and result from the Structure theorem for finitely generated modules over a P.I.D., which isn't exactly a trivial theorem.

Answer 2

Choose bases $B_G = \{g_1, \dots, g_n\}$ and $B_U = \{u_1, \dots, u_n\}$ for $G$ and $U$ respectively. As $U$ is contained in $G$, we can write each $u_i$ as some $\mathbb{Z}$-linear combination of the $g_j$s: $u_i = \lambda_{i1} g_1 + \dots + \lambda_{in} g_n$. Hence, if

$\begin{pmatrix} a_1\\ \vdots \\ a_n\end{pmatrix}_{B_G},\begin{pmatrix} b_1\\ \vdots \\ b_n\end{pmatrix}_{B_U}$

represent the same element of $U$, but written with respect to $B_G$ and $B_U$ respectively, then

$\begin{pmatrix} b_1\\ \vdots \\ b_n\end{pmatrix} = \begin{pmatrix} \lambda_{11} &\dots &\lambda_{1n}\\ \vdots &\ddots &\vdots \\ \lambda_{n1} &\dots &\lambda_{nn}\end{pmatrix} \begin{pmatrix} a_1\\ \vdots \\ a_n\end{pmatrix}.$

Call that square matrix $\Lambda$. Its job: left-multiplication by it "translates" vectors from $B_G$-form into $B_U$-form.

Now suppose we can find invertible matrices $S$ and $T$ with the property that $S, T, S^{-1}, T^{-1}$ all have integer entries, and such that $S\Lambda T$ is diagonal. Then obviously left-multiplying by the diagonal matrix $S\Lambda T$ will translate vectors from $T^{-1}B_G$-form to $SB_U$-form.

These $S$ and $T$ are found by computing the Smith normal form of $\Lambda$, which I'll leave you to look up.