I read that a direct product of a finite number of nilpotent groups is nilpotent. Here the definition of a nilpotent group is one that has a central series. A comment in my book following this claim says
If $G_{ij}$ is the $i^{th}$ term of a central series of the $j^{th}$ factor $H_j$, with $G_{ij}=G$ if the series has already terminated at $G$, then $\prod_j G_{ij}$ will be the $i^{th}$ term of a central series for $\prod_j H_j$.
My guess is that the central series for $\prod_j H_j$ is something like $ 1\unlhd \prod_j G_{1j}\unlhd\prod_j G_{2j}\unlhd\cdots\unlhd\prod_j G_{rj}=\prod_j H_j $ and additionally $ \prod_j G_{i+1,j}/\prod_j G_{ij}\subseteq Z(\prod_j H_j/\prod_j G_{ij}). $ I'm struggling to understand why the containment above is true. I think I need to show $ \begin{align*} \prod_j g_{i+1,j}\prod_j G_{ij}\cdot\prod_j h_j\prod_j G_{ij} &= \prod_j g_{i+1,j}\prod_j h_j\prod_j G_{ij} \\ &= \prod_j h_j\prod_j g_{i+1,j}\prod_j G_{ij} \\ &= \prod_j h_j\prod_j G_{ij}\cdot\prod_j g_{i+1,j}\prod_j G_{ij} \end{align*} $ but I just don't see why the second equality would be true. I'm sure there's a nice simple explanation, and I'd be glad to see it. Thanks.