# Is this semi-direct product residually finite?

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).

This group maps onto the residually finite wreath product $$C_2\wr\mathbf{Z}$$ and the kernel is central, free over the $$y_k$$, $$k\ge 0$$, as 2-elementary abelian group.
So one needs to show that for every non-empty subset $$J$$ of the set $$J$$ of positive integers, the element $$y_J=\prod_{j\in J}y_j$$ survives in some finite quotient.
Let $$t$$ be the generator of $$\mathbf{Z}$$ and kill $$t^n$$. This kills $$y_n$$ and identifies $$y_i$$ to $$y_{i+n}$$ for all $$n$$ (and identifies $$x_i$$ to $$x_{i+n}$$. At the level of the quotient, the central kernel of the homomorphism onto $$C_2\wr C_n$$ is freely generated by the $$y_i$$, $$0, as $$2$$-elementary abelian group. In particular, $$y_J$$ is not trivial in this quotient.