Can the group $\mathbb Z \times \mathbb Z$ be written as union of finitely many proper subgroups of it?

Let $$H_{1}=\{(x,y)\in\mathbb{Z}\times\mathbb{Z}\,:\,x,y\,\text{have same parity}\}$$ $$H_{2}=\{(x,y)\in\mathbb{Z}\times\mathbb{Z}\,:\, 2\mid x\}$$ $$H_{3}=\{(x,y)\in\mathbb{Z}\times\mathbb{Z}\,:\, 2\mid y\}$$ It's easy to see that $\mathbb{Z}\times\mathbb{Z}=H_{1}\cup H_{2} \cup H_{3}$.

For more look up

• Finitely Generated Abelian Groups as Unions of Proper Subgroups by A Rosenfield. The American Mathematical Monthly, Vol. $70$, No. $10$ (Dec $1963$), pp. $1070-1074$

Any group $G$ with a non-cyclic finite quotient $\pi:G\to Q$ is union of finitely many proper subgroups.

Indeed, write $Q$ as union of its cyclic subgroups $Q_i$. Then $G$ is union of the $\pi^{-1}(Q_i)$, which are proper subgroups.

Note: a 1956 theorem of B.H. Neumann says that whenever a group $G$ is finite non-redundant union of subgroups $H_i$, then all $H_i$ have finite index. Hence, the only way in general is to pull from a finite quotient.

Therefore the converse of the above statement holds:

If a group has all its finite quotients cyclic, then it is not union of finitely many proper subgroups.