The intersection of all subgroups with the finite index of additive group $\Bbb Q'$ is $\{0\}$ where $\{0\}\lt\Bbb Q'\lt\Bbb Q$. $$ G=\bigcap_{\Bbb H\leqslant\Bbb Q'\quad |\Bbb Q':\Bbb H|<\infty} \Bbb H $$ I tried to prove this statement from the opposite. Let $G$ is the intersection of all subgroups with the finite index of additive group $\Bbb Q'$ and $g\in G$ is nonzero element. $g\in G\Rightarrow \langle g\rangle\leqslant\Bbb H$ for all $\Bbb H\leqslant\Bbb Q'$ and $|\Bbb Q' :\Bbb H|<\infty$. But I couldn't find a contradiction. Then I tried to prove if $g\ne0$ then $\exists\Bbb H\leqslant\Bbb Q'$ and $|\Bbb Q' :\Bbb H|<\infty$ for which $g\notin \Bbb H$. I couldn 't prove this statement either.
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What is $\mathbb{Q}'$? Arbitrary subgroup of $\mathbb{Q}$? – freakish Aug 10 '23 at 10:50
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@DietrichBurde I think these are supposed to be subgroups of fixed $\mathbb{Q}'$, which itself is subgroup of $\mathbb{Q}$. – freakish Aug 10 '23 at 10:54
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$\Bbb Q'\ne {0}$ and $\Bbb Q'\ne \Bbb Q$ – Samvel Safaryan Aug 10 '23 at 10:55