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?
I was trying to do something like that but no results!
Would be very grateful for ane help!
 A: 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 $3,-2\in\mathbb{Z}$.
A: 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. 
