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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.

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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

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  • $\begingroup$ I hope it can be extended to $\mathbb Z^n$ for $n>1,$ as I can't access the paper you mentioned. $\endgroup$
    – user371231
    Sep 26 '19 at 8:26
  • $\begingroup$ Not sure about Z^n. Added a link for the reference $\endgroup$
    – C.S.
    Sep 26 '19 at 8:56
  • $\begingroup$ If you define the same thing on the $n-$tuples it appears that $\mathbb Z^n$ is union of $n+1$ proper subgroups. $\endgroup$
    – user371231
    Sep 26 '19 at 9:18
  • $\begingroup$ Perhaps Yes. Defining the same thing might not work though. Some subtle modifications might be required $\endgroup$
    – C.S.
    Sep 26 '19 at 9:24
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    $\begingroup$ $\mathbb{Z}^n=\mathbb{Z}^2\times\mathbb{Z}^{n-2}$ 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? $\endgroup$
    – freakish
    Sep 26 '19 at 9:46
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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.

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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.

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