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

Question

The problem here is that the matrix $$\begin{pmatrix} 2 & 3 \\ 1 & -1 \\ \end{pmatrix}$$ is invertible but not unimodular and hence the elements $2x+3y$ and $x-y$ generate a free abelian group of rank 2 but still a proper subgroup of $\langle x,y\rangle$. But the question was to prove that $2x+3y$ and $x-y$ form a basis.

This question is problem 3 of section-67 in Topology by Munkres.

EDIT: The question verbatim from Munkres book is this: If $G$ is free abelian with basis $\{x,y\}$, show that $\{2x+3y, x-y\}$ is also a basis for $G$ Thanks

Answer 1

Think of the torus. Its fundamental group is ${\mathbb Z} \oplus {\mathbb Z}$. Now draw a $(2,3)$ curve and a $(1,-1)$ curve on it. You can do this using a square with the edges identified. These two curves should intersect transversely once. Thus they, too, are a symplectic basis for $H_1(T^2)$ and hence for $\pi_1$.

Answer 2

Certainly they are independent: If A(2x+3y)+B(x-y)=0 then $2A+B=0, 3A-B=0$ so $2A+3A=0$, and $A=0$, hence $B=0$.

Thus the generators $2x+3y, x-y$ span a free abelian group of rank 2.

No they are not a basis for $<x,y>$. Try to solve $x=A(2x+3y)+B(x−y)$, then $3A−B=0,2A+B=1$, so $5A=1$, which is impossible for integers $A,B$

Answer 3

If they form a basis for $G$, then you can find integers $A, B, C, D$ such that $x=A(2x+3y)+B(x-y)$ and $y=C(2x+3y)+D(x-y)$. There are unique rational numbers $A, B, C, D$ satisfying those equations. Are they integers?