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

I want to know if the group $$G=\mathbb Z \times \mathbb Z$$ can be written as union of finitely many proper subgroups of it ?

It is clear that $$\mathbb Z$$ can't be written as union of finitely many proper subgroups as the subgroups are of the form $$n \mathbb Z$$ for some integer $$n$$ and there are infinitely many primes in $$\mathbb Z.$$

My way to think: If possible $$G= H_1 \cup \cdots\cup H_r$$ where $$r>1$$ and $$H_i's$$ are proper subgroups of $$G.$$ Now considering the projection maps $$\pi_1$$ and $$\pi _2$$ on $$G$$ there exist $$i$$ and $$j$$ such that $$\mathbb Z=\pi_1(H_i)$$ and $$\mathbb Z=\pi_2(H_j).$$ I can't complete my arguments after that. Any helps will be appreciated. Thanks.

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. $$1070-1074$$
• I hope it can be extended to $\mathbb Z^n$ for $n>1,$ as I can't access the paper you mentioned. Sep 26 '19 at 8:26
– C.S.
Sep 26 '19 at 8:56
• If you define the same thing on the $n-$tuples it appears that $\mathbb Z^n$ is union of $n+1$ proper subgroups. Sep 26 '19 at 9:18
• Perhaps Yes. Defining the same thing might not work though. Some subtle modifications might be required
– C.S.
Sep 26 '19 at 9:24
• $\mathbb{Z}^n=\mathbb{Z}^2\times\mathbb{Z}^{n-2}$ can be written as union of three groups using the example from the answer. Just extend each subgroup: $H_1\times\mathbb{Z}^{n-2}, H_2\times\mathbb{Z}^{n-2}, H_3\times\mathbb{Z}^{n-2}$, right? Sep 26 '19 at 9:46

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.

You can use following facts, which strengthen your intuition/motivation for the question.

(1) A group $$G$$ can be written as union of proper subgroups if and only if it is non-cyclic.

(2) Now try to get non-cyclic finite quotient of $$\mathbb{Z}\times \mathbb{Z}$$; you can easily find (one appears in accepted answer). The quotient (being finite and non-cyclic) is union of finitely many proper subgroups; pull back them in $$G$$ to reach the destination.