Set Up: Let $ K $ be a totally real number field. Let $ \mathcal{O}_K $ be the ring of integers of $ K $. Let $ p $ be a prime in $ \mathcal{O}_K $. Consider the subring $ \mathcal{O}_K[1/p] $ of $ K $. Let $ G $ be a simple linear algebraic group. Suppose that $ G(\mathbb{R}) $ is a compact Lie group.
Question: Is it the case that $ G(\mathcal{O}_K[1/p]) $ is always dense in the Lie group $ G(\mathbb{R}) $ ?
Context: I think for the choice $ K=\mathbb{Q}(\cos(2 \pi/16)) $, $ \mathcal{O}_K=\mathbb{Z}[\cos(2 \pi/16)] $, $ p= \cos(2 \pi/16) $, $ \mathcal{O}_K[1/p]=\mathbb{Z}[\cos(2 \pi/16),1/\cos(2 \pi/16)] $, $ G=SU_{2^n} $ then $ G(\mathcal{O}_K[1/p]) $ is dense in $ G(\mathbb{R}) $ because for example $$ \frac{i}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} $$ and $$ \begin{bmatrix} \overline{\zeta_{16}} & 0 \\ 0 & \zeta_{16} \end{bmatrix} $$ generate a dense subgroup of $ SU_2 $. I wonder if this phenomenon generalizes?