# Union of inifinitely many subgroups not subgroup

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?

Would be very grateful for ane help!

• 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. – Mike Earnest Sep 18 '18 at 2:06
• @Mike Earnest, but what if I add the subgroup $2\mathbb{Z}$? – ZFR Sep 18 '18 at 2:17
• That's right. For the first example consider the abelian group $(\mathbb{R}^2,+)$ and the collection of all vector subspaces of dimension $1$. – Albert Sep 18 '18 at 3:13
• @Carlos Ajila, what do you mean by vector subspaces of dimension 1? – ZFR Sep 18 '18 at 17:40
• @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 $a^2+b^2=1$ – 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 $3,-2\in\mathbb{Z}$.
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.