If \(G\) is free abelian with basis \(\ts{x, y}\), show that \(\ts{2x + 3y, x - y}\) is also a basis for \(G\).
Solution
Let \[ u=2x+3y,\qquad v=x-y. \] Relative to the basis \((x,y)\), the coordinate matrix with columns \(u,v\) is \[ M=\begin{pmatrix}2&1\\3&-1\end{pmatrix}, \qquad \det M=-5. \] Thus, as written, \(\{2x+3y,x-y\}\) is not a basis of the free abelian group \(G\cong\mathbb Z^2\): a change-of-basis matrix over \(\mathbb Z\) must have determinant \(\pm1\). Indeed the subgroup generated by \(u,v\) has index \(5\) in \(G\).
For example, solving for \(x,y\) gives \[ x=\frac{u+3v}{5},\qquad y=\frac{u-2v}{5}, \] which need not be integral combinations of \(u,v\). Hence the exercise statement is false as written.