\lambda}\otimes p_{i},     \label{defcoppi} \\
 \Delta p_{0} &=& p_0\otimes1+1\otimes p_0. \label{defcopp0}\end{aligned}$$

It is worth mentioning that due to the twist deformation being expressed in terms of the similarity transformation, the homomorphic property of the undeformed coproduct $\Delta_0 $ naturally extends onto the coproduct $\Delta, $ so that it is also a homomorphism, $$\label{homomorphism}
  \Delta (h_1 h_2) = (\Delta h_1) (\Delta h_2), \qquad  h_1, h_2 \in H.$$ In analogy to the concluding remark of section 2, choosing $h_1=p_\mu, \,h_2=x_\nu$ in , together with relations $\mathcal{R}$, gives us four expressions for $\Delta(p_\mu x_\nu)$. Repeating the procedure for $\Delta(x_\nu p_\mu)$, we find that $\Delta\Bigl( [p_\mu, x_\nu ]\Bigr) =[\Delta p_\mu ,\Delta x_\nu]$, i.e. $\Delta h,\, h\in H$ is compatible with relations $\mathcal{R}$ ($\mathcal{R}\Delta h= \Delta h\mathcal{R}$).

The basic idea in this paper is that the coproduct for any element in $H$ can be calculated by using only two things. The first one is the homomorphism of the coproduct $\Delta $ and the second one are the relations ${\mathcal{R}},$ expressed in terms of the identities, Eqs.(\[idxi\]) and (\[idx0\]). Particularly, the coproducts for the Lorentz generators can be calculated by using this method, which we show in the next section. This is made possible since according to 