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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).

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1 Answer 1

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Yes, it's residually finite.

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 $y_i$, $0<i<n$, as $2$-elementary abelian group. In particular, $y_J$ is not trivial in this quotient.

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