Let $G$ be a topological group and $H$ its closed subgroup. $K$ and $L$ are open subgroups of $G$ and $H$ respectively. Let $g_{1}, g_{2}\in G$. We assume $L\cap g_{1}K\neq \emptyset$ and $L\cap g_{2}K\neq \emptyset$. Then is there an element $x\in H$ (also $x\in L$) such that $L\cap g_{1}K=x(L\cap g_{2}K)$ ?
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$\begingroup$ The question makes when $G$ is discrete, and trivially a positive answer in the discrete case implies a positive answer in general (just use the underlying group with the discrete topology). Hence, it's just a question about abstract groups. $\endgroup$– YCorCommented Apr 7, 2021 at 18:24
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$\begingroup$ There exist $k_{1}, k_{2}\in K$ such that $g_{1}k_{1}, g_{2}k_{2}\in L$. Then $L\cap g_{i}K=g_{i}k_{i}(L\cap K), (i=1, 2)$ are both left cosets of $L\cap K$ in $L$. So I could take $x\in L$ such I needed in my Question. $\endgroup$– M masaCommented Apr 8, 2021 at 0:35
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