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Let $G$ be a finite cyclic 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 $|H_{S_{i}}| \gg |G_{i}|^{1-\epsilon}$. Then, is it necessary to have $|S_{i}| \gg |G_{i}|^{\epsilon}$? Can we say something about optimal lower bound?

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  • $\begingroup$ If think $\{ \chi^\sigma, \sigma \in Gal\}$ has $\phi(k)$ elements if the order of $\chi$ is $k \ge 3$ $\endgroup$ – reuns Jul 2 '17 at 23:07
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No. Take $G = \mathbb F_2^i$, $S$ to be the set of characters given by projection onto the $j$th factor composed with the sign character of $\mathbb F_2$, for $j$ from $1$ to $i$. Then $S_i$ is Galois-stable, $|S_i|= i$, $|H_{S_i}| = |G_i| =2^i$.

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Since $G$ is a finite abelian group we have $\widehat{G}\cong G$. For $\chi\in\widehat{G}$ to be Galois-stable, or just stable under complex conjugation, its order must be at most 2. However $\widehat{G}$ is cyclic, so it has at most one element of order 2. Thus a Galois-stable subset $S\subseteq \widehat{G}$ has $|S|\leqslant2$, and $H_S=\langle S\rangle$ has $|H_S|\leqslant2$. Hence for each $\varepsilon$ there is no sequence of finite cyclic groups $G_i$ and subsets $S_i\subseteq\widehat{G_i}$ with $|G_i|\to\infty$ and $2\geqslant |H_{S_i}|>|G_i|^{1-\varepsilon}$. Have I misunderstood your question? Did you did want $G$ to be abelian (and noncyclic)? I suggest you clarify the meaning of the symbol $\gg$, see https://math.stackexchange.com/questions/641618/what-does-this-symbol-gg-mean.

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  • $\begingroup$ There seems to be a key misunderstanding. The hypothesis is $S$ being $G_{\mathbb{Q}}$-stable i.e. $\chi^{\sigma} \in S$ for $\chi \in S$ and $\sigma \in G_{\mathbb{Q}}$. This need not (and typically does not) mean $\chi^{\sigma}$ equals $\chi$. $\endgroup$ – user111815 Jul 10 '17 at 1:15

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