Give an example of an abelian group $G$ and an infinite family of proper subgroups $\{H_i\}_{i\in \mathcal{I}}$ none of which contains all the others where $\bigcup \limits_{i\in \mathcal{I}}H_i$ is a subgroup of $G$ and another example of such family where $\bigcup \limits_{i\in \mathcal{I}}H_i$ is not a subgroup of $G$.

Proof: I was able to solve the first part of problem, namely taking the group $(\mathbb{Z},+)$ and subgroups $4\mathbb{Z}, 6\mathbb{Z}, 8\mathbb{Z}, \dots$. And it's easy that none of them contains all the others, namely $(2n-2)\mathbb{Z}\subsetneq 2n\mathbb{Z}$ and it's quite to show that the union of these subgroups if also subgroup!

But how to come up with the example when the union is NOT subgroup?

I was trying to do something like that but no results!

Would be very grateful for ane help!

  • $\begingroup$ The union of $4\mathbb Z,6\mathbb Z,8\mathbb Z,\dots$ is not a subgroup, because $4+(-6)=-2$ is not in the union. $\endgroup$ – Mike Earnest Sep 18 '18 at 2:06
  • $\begingroup$ @Mike Earnest, but what if I add the subgroup $2\mathbb{Z}$? $\endgroup$ – ZFR Sep 18 '18 at 2:17
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    $\begingroup$ That's right. For the first example consider the abelian group $(\mathbb{R}^2,+)$ and the collection of all vector subspaces of dimension $1$. $\endgroup$ – Albert Sep 18 '18 at 3:13
  • $\begingroup$ @Carlos Ajila, what do you mean by vector subspaces of dimension 1? $\endgroup$ – ZFR Sep 18 '18 at 17:40
  • $\begingroup$ @RFZ I mean the sets of the form $E_{a,b}=\{(x,y)\in\mathbb{R}^2 : (x,y)=(ka,kb) \text{ for some } k\in\mathbb{R}\}$ where $a,b\in\mathbb{R}$ are such that $a^2+b^2=1$ $\endgroup$ – Albert Sep 18 '18 at 19:44

Consider the abelian group $\mathbb{Z}$ and the family $\{p\mathbb{Z}\}_{p\in P}$ where $P\subseteq\mathbb{Z}$ is the set of positive prime numbers. If $p$ and $q$ are prime numbers, then $p\mathbb{Z}\not\subseteq q\mathbb{Z}$. Now, every $n\in\mathbb{Z}\setminus\{-1,1\}$ has a factorization as a product of prime numbers. This implies that $$ \bigcup_{p\in P}p\mathbb{Z} = \mathbb{Z}\setminus\{-1,1\}, $$ but $\mathbb{Z}\setminus\{-1,1\}$ is not a subgroup of $\mathbb{Z}$, because $3+(-2)=1\not\in \mathbb{Z}$ but $3,-2\in\mathbb{Z}$.

  • $\begingroup$ Very nice example! Thanks a lot for that! $\endgroup$ – ZFR Sep 18 '18 at 2:19

The set of points of the plane is an abelian group (add x and y co-ordinates separately to make it a group; this is same as vector addition by completing parallelograms).

For each positive integer $k$, let the subset $L_k$ be defined as those points on the line $y=kx$. This is easily seen to be a subgroup. The union of $L_k$'s consists of points of all lines passing through origin having slope an integer value.

This union is not closed under addition: For example take $(1,1)\in L_1$ and $(1,2)\in L_2$ their sum $(2,3)$ does not belong to any line through origin with integer slope.


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