The group $S O(N)$ is defined as the set of $N \times N$ matrices $R_{i j}$ that leave the $N$-dimensional Euclidean metric invariant, i.e., $$ \delta_{i j}=R_{i k} R_{j l} \delta_{k l},$$ which also has the determinant of 1.
Consider representing $S O(N)$ by unitary matrices $U(R)$. Show that the generators of the $S O(N)$ group satisfy the commutation relations $$ \left[J^{i j}, J^{k l}\right]=\frac{1}{2} i\left(\delta^{i l} J^{j k}-\delta^{j l} J^{i k}-\delta^{k i} J^{j l}+\delta^{k j} J^{i l}\right) $$
To show this, write an infinitesimal $S O(N)$ transformation as $U(R)=1+i \omega_{i j} J^{i j}$, where $J^{i j}=-J^{j i}$ and $\omega_{i j}=-\omega_{ji}$. Use the fact that, $U(R) U(1+\omega) U(R)^{-1}=U\left(R(1+\omega) R^{-1}\right)$ (Hint: Use infinitesimal $R_{i j}=\delta_{i j}+\omega_{i j}$)
My attempt: I have used the fact of $U(R) U(1+\omega) U(R)^{-1}=U\left(R(1+\omega) R^{-1}\right)$ to get $U(R) J^{l k} U^{+}(R)=R_{i l} R_{j k} J^{i k}$ and I plugged in $U(R)=1+i \omega_{i j} J^{i j}$ and $R_{i j}=\delta_{i j}+\omega_{i j}$ and by doing the procedure twice and using the fact that $\omega$ is antisymmetric, I was able to get: $$ \omega_{i j}\left[J^{i j}, J^{l k}\right]=\frac{i}{2}\left[w_{j l} J^{j k}+w_{i l} J^{i k}+w_{k j} J^{j l}+w_{k i} J^{i l}\right] $$.
I am not sure this equation is correct but I feel like I am really close if miraculously $\frac{\omega_{jl}}{\omega{ij}} = \delta_{jl}$. I can't find any reason for this to be true though. Any insights will be greatly appreciated!
Apologies if the question is framed improperly, please point out and I can make improvements to the question.