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We know that a group can not be written as a union of two proper subgroups and obiously a finite group can be written as a finite union of proper subgroups.So I want to ask if a infinite group be written as a finite union of proper subgroups?Moveover,Can a field be a finite union of proper subfields?

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How are you going to write $\Bbb Z/3\Bbb Z$ or $\Bbb Z/4\Bbb Z$ as a union of proper subgroups? –  Brian M. Scott Aug 15 '12 at 4:37
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Or $\mathbb Z$, for that matter. –  Robert Israel Aug 15 '12 at 6:02
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3 Answers 3

Sometimes an infinite group can be written as a finite union of proper subgroups, e.g., every element of ${\bf Z}\oplus{\bf Z}$ is of at least one of the forms $(2a,b)$, $(a,2b)$, or $(a+b,a-b)$ for integers $a,b$, and that gives you 3 proper subgroups whose union is the whole group.

As for fields, note that a field and its subfields are all vector spaces over the prime field, so if a vector space can't be a finite union of proper subspaces, then a field can't be a finite union of proper subfields. The vector space question for infinite fields is handled here. For finite fields, it's easy to see the answer is no. That leaves infinite fields of finite characteristic to think about.

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Complete answer! –  B. S. Aug 15 '12 at 7:09
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As a side note, it can be shown that a group $G$ can be written as the union of three proper subgroups if and only if it has $H = \mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z$ as a quotient, in which case the three subgroups in question are the inverse images (under the quotient map from $G$ to $H$) of the three two-element subgroups of $H$. Gerry's example fits this characterization. –  Greg Martin Aug 15 '12 at 8:24
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As you say, obviously a finite (noncyclic!) group $G$ can be written as a finite union of proper subgroups - call them $G_1,\dots,G_f$. Now let $H$ be any infinite group. The direct product $G\times H$ can then be written as a finite union of the proper subgroups $G_1\times H,\dots,G_f\times H$.

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There is a lot of research on groups covered by a finite set of proper subgroups. It all started with a seminal paper of Bernhard Neumann, Groups covered by finitely many cosets, Publ. Math. Debrecen 3 (1954), 227-242, in which he proves that if a group is the union of a finite number of subgroups (or even cosets), you only have to take into account those subgroups having finite index. This implies that a group has a finite covering by subgroups if and only if it has a finite non-cyclic homomorphic image. See also recent work of Mira Bhargava (the mother of the brilliant number theorist Manjul Bhargava) http://www.math.uga.edu/~pete/MiraBhargava07.pdf

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+1 Very nice, neat and interesting answer –  DonAntonio Aug 15 '12 at 12:21
    
Is it also true that any way of writing a group as a proper union of subgroups union is the pullback of a way of writing a finite group as a proper union of subgroups along an epimorphism onto the finite group? –  HenrikRueping Aug 16 '12 at 12:23
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