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I want to find a group homomorphism $f: \mathbb{Z}^2 \to \mathbb{Z}^2$ which satisfies $f(1,0) = (2,6)$. Can any such homomorphism be made into an isomorphism?

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A homomorphism of what? Groups? – M.B. Oct 28 '12 at 23:03
Indeed. I will amend the statement at once. – user44069 Oct 28 '12 at 23:13
up vote 8 down vote accepted

A homomorphism $\mathbb{Z}^2\rightarrow \mathbb{Z}^2$ can be represented as a $2\times 2$ matrix with integer coefficients with respect to the canonical bases. For example saying that $f(1,0)$ should be $(2,6)$ ammounts to specifying the first column of the matrix as $\left(\begin{smallmatrix}2\\6\end{smallmatrix}\right)$. The inverse of such a homomorphism, if it exists, will be given by the inverse matrix, since this is true of homomorphisms $\mathbb{Q}^n\rightarrow \mathbb{Q}^n$. With this information, you should be able to answer your own question.

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Great hint! Thank you very much! – user44069 Oct 28 '12 at 23:26

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