Homomorphism $\mathbb{Z}^2 \to \mathbb{Z}^2$

Question

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?

Answer 1

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.