I am interested in finitely generating dense subgroups of the unitary group $U(d)$ of order $d$ (with respect to the operator norm topology), and am considering the following question.
Let $I_d$ be the $d \times d$ identity matrix. Do there exist finitely generated dense subgroups $S_1$ and $S_2$ of $U(d)$ such that for all $g_1 \in S_1 \backslash \{I_{d}\}$ and all $g_2 \in S_2 \backslash \{I_{d}\}$, the matrix $g_1 - g_2$ is invertible?
I feel this should be true because the subgroups $S_1$ and $S_2$ need not overlap, but I am unsure where to start for a proof.