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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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    Every element of $\left<2x+3x,x-y\right>$ is of the form $a(2x+3y)+b(x-y)$ where $a,b\in\mathbb Z$. Now just prove that if $a(2x+3y)+b(x-y)=0$ then $a=b=0$.2012-10-12
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    @Thomas Andrews that was clear but I am saying that $2x+3y$ and $x-y$ doest span <$x$,$y$>2012-10-12
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    Maybe you should post the problem verbatim, rather than paraphrasing it, because the part where you wrote "the question is..." says nothing about showing that it doesn't span $$. Do you want us to prove that it is a free group *and* that it is a proper subgroup?2012-10-12
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    Ok, apologies for an unclear question. The question verbatim from Munkres is this: "If $G$ is free abelian with basis {x,y}, show that {2x+3y,x-y} is also a basis for $G$"2012-10-12
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    It seems like Munkres is wrong, then since $A(2x+3y)+B(x-y)=x$ has no solution. First $3A-B=0$ so $B=3A$ then $2A+B=5A=1$, so $A$ there cannot be a solution for $A2012-10-12
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    Exactly! That was the whole point of this question. Thanks for confirming it.2012-10-12
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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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    Yes that's what I wrote in the question but will they 'generate' the group ? I think no. But the question in Munkres was to prove that <2x+3y> and are also a basis for ''.2012-10-12
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    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 $A, B$.2012-10-12
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    Yes. Thats what I wanted to confirm for myself. Thanks!2012-10-12
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    That's not what Munkres claims. He just asserts a one-way implication, not equivalence. As everybody appears to agree that the claimed implication is true, this is more about reading the question properly (if A basis => B basis).2012-11-26
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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?