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For each $i \in \mathbb{N}$, let $G_{i}$ be a finite abelian group and $\widehat{G_{i}}$ the $\overline{\mathbb{Q}}$-valued character group of $G_{i}$. Suppose that $|G_{i}| \rightarrow \infty$ as $i \rightarrow \infty$. Let $S_{i} \subset \widehat{G_{i}}$ be a subset of characters satisfying the following.

(i) For given $\epsilon > 0$, we have $\langle s | s \in S_{i} \rangle \gg |G_{i}|^{1-\epsilon}$ as $i \rightarrow \infty$. (ii) $S_{i}$ is $G_{\mathbb{Q}}$-stable i.e. if $s\in S_{i}$, then $s^{\sigma} \in S_{i}$ for each $\sigma \in Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ with $s^{\sigma}$ being the conjugate character. (iii) There exists a finite set $T$ independent of $(G_{i})_{i=1}^{\infty}$ and non-zero scalars $a_{i,j} \in \overline{\mathbb{Q}}$ dependent on $G_{i}$ such that $$Im(\sum_{j=1}^{|S|}a_{i,j}s_{j}) \subset T.$$ Here we view the linear combination $\sum_{j=1}^{|S|}a_{i,j}s_{j}$ a function on $G_{i}$.

Can we obtain an estimate for a lower bound on $|S|$? Is it reasonable to hope for a bound of the form $|G|^{\delta}$ for some $\delta > 0$?

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  • $\begingroup$ Could you remind us what $G_{\mathbb Q}$-stable means? $\endgroup$
    – Yemon Choi
    Commented Jul 3, 2017 at 3:34
  • $\begingroup$ It is included now. $\endgroup$
    – user111815
    Commented Jul 3, 2017 at 3:39
  • $\begingroup$ I don't understand the parsing, particularly of (iii). Do the scalars depend on $|G|$ which is going to infinity?! Do you mean something like (given $\epsilon$?): Suppose that an infinite sequence of $(G,S)$ satisfying (i)-(iii) is given, in particular with there being an absolute constant $B$ such that for each $S$ there are some $a_i$ with the linear combination taking on no more than $B$ values? And then you want to know how large the $S$ must be in terms of the $|G|$? $\endgroup$ Commented Jul 3, 2017 at 6:51
  • $\begingroup$ Sorry for the ambiguity. The question is updated. The example arising from $G_{n}=(\mathbb{Z}/2)^n$does not satisfy the third condition as a linear combination of the characters takes many values on $G_{n}$ as $n \rightarrow \infty$. $\endgroup$
    – user111815
    Commented Jul 3, 2017 at 8:07

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