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Let $G$ be a finite abelian group and $\widehat{G}$ the character group. Let $S \subset \widehat{G}$ be a Galois-stable subset i.e. if $\chi \in S$, then the Galois conjugates $\chi^{\sigma} \in S$ for any $\sigma \in Gal(\overline{\mathbb{Q}}/\mathbb{Q})$. Let $H_{S} \subset \widehat{G}$ be the subgroup generated by $S$.

We now consider a sequence $(G_{i},S_{i})$ as above with $|G_{i}|\rightarrow \infty$. Suppose that there exists $\epsilon$ with $ 0 < \epsilon < 1$ such that $|H_{S_{i}}| \gg |G_{i}|^{\epsilon}$. Then, is it necessary to have $|S_{i}| \gg \ln(|G_{i}|)$? Can we say something about optimal lower bound?

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    $\begingroup$ Can you say more precisely what you mean by $|H_S|>>|G|^\epsilon$? Who varies here? $\endgroup$ – user1688 Jul 18 '16 at 9:24
  • $\begingroup$ Sorry, the question is edited now. Please let me know if it still looks unclear. $\endgroup$ – user95196 Jul 18 '16 at 10:41
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Suppose that $G_i=C_2^i$, and let $S_i$ be a minimal generating set of $\widehat{G_i}$. Since all characters are rational, $S_i$ is stable, but we have $|H_{S_i}|=|G_i|=2^i$, $|S_i|=i\asymp\log|G_i|$. So the bound $\log|G_i|$ is in fact optimal.

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  • $\begingroup$ Indeed! I am wondering if there is an elementary Galois theory argument to show the bound in general. $\endgroup$ – user95196 Jul 18 '16 at 12:47

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