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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

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    Yeah, the problem with your question is that you started with your objections to the question, without telling us what the question was, so people tried to answer the only part of your question that looked like a question.2012-10-12

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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 $\pi_1$.

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    I'm guessing, although the problem comes from a topology book, that it comess from an early "pre-requisites" section, before you get to $H_1$ and $\pi_1$.2012-10-12
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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 $$. 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 $A,B$

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    It is true that $2x+3y, x-y$ are a basis for a free abelian subgroup of $$. It is of index 5 (the absolute value of the determinant).2012-11-26
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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?