Let G be a group and $\{H_i\}_{i<\omega}$ be a countable family of subgroups of $G$, each of them of uncountable index. Can $G=\bigcup_{i<\omega} H_i$?
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$\begingroup$ I would like someone to confirm but I tend to think that $G=\{+1,-1\}^{\mathbb{N}}$ , with $H_i$ being each of the $\{+1,-1\}$ terms of the product, is a counterexample. $\endgroup$– Eric BrierCommented Jun 7 at 15:37
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$\begingroup$ @EricBrier Their union is not a subgroup, and if by $F_i$ you call the subgroup generated by $H_1,\cdots,H_i$, then $\bigcup_{i\in\omega} F_i=\{+1,-1\}^{\oplus \Bbb N}\subsetneqq G$. $\endgroup$– Sassatelli GiulioCommented Jun 7 at 15:42
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$\begingroup$ Union of subgroups is indeed in general not a subgroup. The question is to know if union equals the whole group. In my example, it is stricly smaller than the whole group and it is not a subgroup. $\endgroup$– Eric BrierCommented Jun 7 at 15:45
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$\begingroup$ @EricBrier Obviously there are instances where $\bigcup_{i\in \omega} H_i\ne G$: the question asks if there are instances where equality holds. There are no counterexamples to claims of existence. $\endgroup$– Sassatelli GiulioCommented Jun 7 at 15:53
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3$\begingroup$ Consider $G=(\Bbb R^{\oplus\Bbb N},+)$, i.e. $G=\{f\in\Bbb R^{\Bbb N}\,:\, \lvert \{x\in\Bbb N\,:\, f(x)\ne 0\}\rvert<\aleph_0\}$ with componentwise sum, and call $H_n=\{f\in G\,:\, \forall x>n, f(x)=0\}$. Then $\bigcup_{n\in\Bbb N} H_n=G$ and $G/H_n\cong G$. $\endgroup$– Sassatelli GiulioCommented Jun 7 at 16:06
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