Product quotient of $SL_2(\mathbb{R})$ by non-commensurable discrete subgroups

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This question is related to Kronecker's approximation theorem: fix a positive integer $N$ and consider $N$ real numbers $x_1,\cdots,x_N$ which are linear independent over $\mathbb{Q}$. Then the lattices $\mathbb{Z}x_1,\cdots,\mathbb{Z}x_N$ are non-commensurable one to another (two subgroups $H_1,H_2$ in a group $G$ are commensurable if $H_1\cap H_2$ have finite indices in both $H_1$ and $H_2$). Then one form of the Kronecker's/Weyl's density theorem says that the product quotient map $$\pi\colon\mathbb{R}\rightarrow\prod_{j=1}^N\mathbb{R}/(\mathbb{Z}x_j)$$ has dense image. There is an adelic version of this theorem: let $\mathbb{A}$ be the Adele ring of $\mathbb{Q}$ and $x_1,\cdots,x_N$ are elements in $\mathbb{A}$ that are linearly independent over $\mathbb{Q}$. Then the product quotient map $$\pi\colon\mathbb{A}\rightarrow\prod_{j=1}^N\mathbb{A}/(\mathbb{Q}x_j)$$ again has dense image (see, for example, Cantor, 'On the elementary theory of Diophantine approximation over the ring of Adeles-1').

Now my question is about generalisation of these results to non-abelian groups: let $G=SL_2/\mathbb{Q}$. If $x_1,\cdots,x_N$ are elements in $G(\mathbb{R})$ such that the conjugates $x_jG(\mathbb{Z})x_j^{-1}$ are non-commensurable one to another. Again we have the product quotient map $$\pi\colon G(\mathbb{R})\rightarrow\prod_{j=1}^NG(\mathbb{R})/(x_jG(\mathbb{Z})x_j^{-1})$$ Can we show that $\pi$ has dense image? What about its adelic version? And what if we replace $G$ by some other simply connected and simple algebraic groups?