(Small generating sets of $\mathbb{Z} \times \mathbb{Z}$) Find all generating sets of cardinality at most 2 of the group $\left(\mathbb{Z}^{2},+\right)$.

## Group Homomorphisms

Definition 71. (Group Homomorphism) Let $(G, *)$ and $(H, \circledast)$ be two groups, and $\varphi: G \rightarrow H$ be a function such that for all elements $x, y$ of $G$:

1. $\varphi(x * y)=\varphi(x) \circledast \varphi(y)$.

(LINEARITY)

We then say that $\varphi$ is a group homomorphism - or a group endomorphism if $(G, *)=$ $(H, \circledast)$.
Let $S=\{a, b\}$ be a subset of $\mathbb{Z}^{2}$ that is assumed to be generating, and let $x, y, z$, $t$ be integers such that $x a+y b=\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)$ and $z a+t b=\binom{0}{1}$. Then $(z y-x t) a=\binom{-t}{y}$ and $(z y-x t) b=\binom{z}{-x}$. Since we cannot have $x=y=z=t=0$ simultaneously, it follows that $z y-x t \neq 0$. Furthermore, $(z y-x t)^{2}\left(a_{1} b_{2}-a_{2} b_{1}\right)=x t-z y$, so $(x t-z y)\left(a_{1} b_{2}-a_{2} b_{1}\right)=1$. Since $x t-z y$ and $a_{1} b_{2}-a_{2} b_{1}$ are integers, it follows that $a_{1} b_{2}-a_{2} b_{1}= \pm 1$.

Conversely, if $a_{1} b_{2}-a_{2} b_{1}=\varepsilon \in\{-1,1\}$, then for all integers $u$ and $v,\binom{u}{v}=$ $\varepsilon\left(b_{1} v-b_{2} u\right) a+\varepsilon\left(a_{2} u-a_{1} v\right) b$.

Finally, if $a=b$, then it is clear that $a_{1} b_{2}-a_{2} b_{1}=0 \notin\{-1,1\}$. It follows that the generating subsets of $\mathbb{Z}^{2}$ of cardinality at most 2 are the sets $\{a, b\} \subseteq \mathbb{Z}^{2}$ such that $a_{1} b_{2}-a_{2} b_{1}= \pm 1$.