So I know: $$[\sigma_{I},\sigma{j}] = 2i \epsilon_{ijk} \sigma_{k}$$ So two products of this should give us the Lorentz group: $SO(4) = SU(2) \times SU(2)$

Where $SO(4)$ has 3 Lie algebra which can be expressed as one by denoting: $$J_{I}^{\pm} = \frac{1}{2} (J_{I} \pm ik_{I})$$

This gives two Lie algebra: $$[J_{I}^{\pm}, J_{j}^{\pm}] = i\epsilon_{ijk}J_{k}^{\pm}$$

But how is this related to our original product of SU(2)? I was under the impression if you have two Lie groups, $(g_{1},h_{1})$ and $(g_{2}, h_{2})$, then the product would be $(g_{1}g_{2},h_{1}h_{2})$. However, I tried that with the Pauli matrices and it doesn't work to be the same as $SO(4)$ Lie algebra. How should this be done?