This is from Munkres' Section 67, but there is a typo in the original question which is as follows:
Show that if $G$ is free abelian with basis $\{x, y\}$, show that $\{2x+3y, x-y\}$ is also a basis for $G$.
This is indeed a typo and has been disucssed here: Let $x$ , $y$ be the basis of a free abelian group of rank 2, prove that $2x+3y$ and $ x-y$ generate a free subgroup
However, this link https://dbfin.com/topology/munkres/chapter-11/section-67-direct-sums-of-abelian-groups/problem-3-solution/ gives an alternative version, which he claims doable:
Show that if $G$ is free abelian with basis $\{x, y\}$, show that $\{2x+3y, x+y\}$ is also a basis for $G$.
However, I did not follow the proof given in this link.
$2x+3y$ and $x+y$ are clearly independent, since if $$A(2x+3y)+B(x+y)=0,$$ then we have $$(2A+B)x+(3A+B)y=0,$$ since $\{x,y\}$ is a basis, we must have $2A+B=0$ and $3A+B=0$, which implies $A=B=0$.
Thus, $\{2x+3y, x+y\}$ clearly generates a group with them as basis.
However, how could I show this group generated by them is indeed $G$?
I tried to identify $g=AX+BY=C(2X+3Y)+D(X-Y)$ for $g\in G$, and in the end I have $A=2C+D$ and $B=3C-D$, but it seems that the link suggest another set of solutions, but I don't really understand where it comes from..
Any idea? Thank you!
Best Answer
Let $u=2x+3y$ and $v=x+y$. You want to show that $u$ and $v$ generate $G$. It suffices to show that there exist integers $A$, $B$, $C$, and $D$ such that \begin{align*} Au + Bv &= x &\implies A(2x+3y) + B(x+y) &= x &\implies (2A+B)x + (3A+B)y &= x\\ Cu + Dv &= y &\implies C(2x+3y) + D(x+y) &= y &\implies (2C+D)x + (3C+D)y &= y\\ \end{align*} The first line gives equations \begin{align*} 2A+B &= 1 \\ 3A+B &= 0 \end{align*} Apparently $A=-1$ and $B=3$ satisfy the system. The second line gives equations \begin{align*} 2C+D &= 0 \\ 3C+D &= 1 \end{align*} This time $C=1$ and $D=-2$ work.
What's really going on here is that the matrix which expresses $u$ and $v$ in terms of $x$ and $y$ is invertible over $\mathbb{Z}$: $$ \begin{bmatrix}u \\ v \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \implies \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 1 & 1 \end{bmatrix}^{-1} \begin{bmatrix}u \\ v \end{bmatrix} = \begin{bmatrix} -1 & 3 \\ 1 & -2 \end{bmatrix} \begin{bmatrix}u \\ v \end{bmatrix} $$