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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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