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Is there a group $G$ and subgroups $H_1, \dots, H_n \leq G$ for some $n \in \mathbb{N}$, such that $[G : H_i] = \infty$ for each $1 \leq i \leq n$, and $$G = \bigcup_{i=1}^n H_i \ \ ?$$

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It may well be a lot easier than this, but it follows from the answer to the weaker question Can a group be a finite union of (left) cosets of infinite-index subgroups? that it's not possible.

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  • $\begingroup$ Looks good to me. $\endgroup$ – Pete L. Clark Sep 7 '15 at 16:51
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    $\begingroup$ More precisely, the claim follows from Lemma 4.2 of Newman's paper in math.uga.edu/~pete/Neumann54.pdf , which has a half-page proof (basically by induction on the number of distinct subgroups, the right-cosets of which one is trying to use to cover the whole group). $\endgroup$ – Terry Tao Sep 7 '15 at 21:08

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