I have the following: A group $X$ and a family $(X_n)_{n \in \mathbb{Z}}$ of subgroups of $X$, such that the following holds:
$X = \langle X_n \;\vert\; n \in \mathbb{Z} \rangle$
$X_n \times X_{n+1} \times \dots \times X_m \to X_{n,m} := \langle X_i \;\vert\; n \leq i \leq m \rangle, \\ (a_n, a_{n+1}, \dots, a_m) \mapsto a_n a_{n+1} \dots a_m \text{ is a bijection}$
for $n < m \in \mathbb{Z}$ we have $[X_n, X_m] \leq \langle X_i \;\vert\; n < i < m\rangle =: X_{n+1,m-1}$.
there exists an automorphism $t$ of $X$ such that $t(X_n) = X_{n+2}$ for all $n$ (however I do not think this is needed here)
I'd like to show that $X_{n,m}$ is nilpotent if all the $X_n$ are nilpotent.
I wanted to use the fact that in this case $X_n \times \dots \times X_m$ is nilpotent for all $n \leq m \in \mathbb{Z}$, however I think (2.) is of not much use as I only have a bijection and not an isomorphism.
Can someone point me in the right direction?
It is enough to prove $X_{0,n}$ is nilpotent for all $n$, so suppose inductively that $X_{0,n}$ is nilpotent for some $n$, with lower central series $X_{0,n} = L_1 > L_2 > \cdots L_{c+1}=1$ for some $c$.
For $1 \le i \le c$ and $0 \le j \le n$, let $K_{i,j} = (X_{j,n} \cap L_i)L_{i+1}$. Then, since $[X_{n+1},X_{j,n}] \le X_{j+1,n}$, we have $[X_{n+1},K_{i,j+1}] \le K_{i,j}$, and so $X_{n+1}$ stabilizes the central series
$$ 1 \le K_{c,n} \le K_{c,n-1} \le \cdots \le K_{c,0}=L_c \le K_{c-1,n} \le\cdots \le K_{2,0}=L_2 \le K_{1,n} \le \cdots \le K_{1,0}=L_1 $$
of $X_{0,n}$, and so by adjoining a central series of $X_{n+1}$ to the top of this series, we get a central series for $X_{0,n+1}$, which is therefore nilpotent, completing the induction.