This is from section 2 of the paper "Quantum algorithms for solvable groups" by J. Watrous, where $G$ is a solvable group, and $G = G^{(0)} \rhd G^{(1)} \rhd \cdots \rhd G^{(n)}$ is the derived series of $G$:
Now we return to the topic of solvable groups, and review some known facts about solvable groups in the context of black-box groups. First, with respect to any given group oracle, if we are given generators $g_1, . . . , g_m$ of encoding length $n$, it is possible to test whether $G = \langle g_1, . . . , g_m\rangle$ is solvable via a polynomial time (in $nm$) Monte Carlo algorithm. Moreover, the same algorithm can be used to construct (with high probability) generators $g_1^{(j)}, \cdots, g_k^{(j)}$, for $j = 0, \cdots, n$ and where $k = O(n)$, such that $G^{(j)} = \langle g_i^{(j)},\cdots g_k^{(j)}\rangle$ (so that testing solvability can be done by verifying that $g_1^{(n)}, \cdots, g_k^{(n)}$ are each the identity element). At this point we notice (under the assumption that G is solvable) that by relabeling the elements $$g_1^{(n−1)},\cdots, g_k^{(n−1)}, g_1^{(n−2)},\cdots, g_k^{(n−2)},\cdots, g_1^{(0)}, \cdots, g_k^{(0)}$$ as $h_1,\cdots, h_{kn}$ (in the order given) we have the following. If $H_j = \langle h_1,\cdots, h_j\rangle$ for $j = 0, . . . , kn$, then $\{1\} = H_0 \lhd H_1 \lhd\cdots\lhd H_{kn} = G$. This follows from the fact that $G^{(j)} \lhd G^{(j−1)}$ for each $j$, and further that $G^{(j−1)}/G^{(j)}$ is necessarily abelian.
My question is why do we have $\{1\} = H_0 \lhd H_1 \lhd\cdots\lhd H_{kn} = G$. As far as I can understand, $H_0, H_1, \cdots, H_{kn}$ is a chain of subgroups, but why are they normal? I'm not sure where the facts $G^{(j)} \lhd G^{(j−1)}$ and $G^{(j−1)}/G^{(j)}$ is abelian is used to show that this chain of subgroups is normal.
If $H_i = G^{(j)}$ for some $i,j$, then I can convince myself that $H_i \lhd H_{i+1}$, but what about the case where $H_i, H_{i+1} \subseteq G^{(j)}$?