I've copied over this question from what I asked on Mathematics Stack Exchange, in the hope that some experts here can help me find a way to check the residual finiteness of this group.

Consider the group $G=K\rtimes \mathbb{Z}$ defined as follows:

The subgroup $K$ is generated by elements $x_i,y_k$ with $i,k \in {\mathbb Z}$ and $k > 0$, and it has defining relations \begin{eqnarray*} x_i^2 &=& y_j^2= 1\ \mbox{for all}\ i,j,\\ [x_j,x_i] &=& y_{j-i}\ \mbox{for}\ j>i,\\ [y_k,x_i] &=& 1\ \mbox{for all}\ i,k, \end{eqnarray*}

The action of $({\mathbb Z},+)$ on $K$ is defined by the automorphism $1 \in {\mathbb Z}$ maps $x_i$ to $x_{i+1}$ for all $i \in {\mathbb Z}$.

**Question**: Is group $G$ residually finite?

**The progress**: My idea is to check if $K$ is residually finite first (because if $K$ is not residually finite, then $G$ can't be). So far, if a word $w$ from $K$ satisfies the following condition, then there is a homomorphism from $K$ to a finite group that doesn't send $w$ to the identity.

if there exists $x_i$ in $w$, and the total power of $x_i$ is odd. (we can map $K$ to some direct product of $\mathbb{Z}_2$)

if $w= y_j$ and $j$ is odd. (We can map $K$ to the Heisenberg group over $\mathbb{Z}_2$)

I am not sure how to show such homomorphism exists for any general word. (e.g. a string of $y_i$'s).