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!
Best Answer
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 $3,-2\in\mathbb{Z}$.