1
$\begingroup$

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 $\mathbb{Z}_2$)

May I ask if this is the right direction? I am not sure how to show such homomorphism exists for any general word. (e.g. a string of $y_i$'s).

(This group was defined here https://mathoverflow.net/a/419028/479955, where Derek Holt calls this a central extension of $C_2 \wr {\mathbb Z}$, which is a residually finite group.)

$\endgroup$
3
  • $\begingroup$ What if the map from $K$ to the Heisenberg group when $w = y_j$ ($j$ odd)? $\endgroup$
    – Derek Holt
    Apr 8, 2022 at 8:01
  • $\begingroup$ I was thinking of this map: $x_i \mapsto \begin{bmatrix} 1&1&0 \\0&1&0\\0&0&1\end{bmatrix}$ when $i$ is even, $x_i \mapsto \begin{bmatrix} 1&0&0 \\0&1&1\\0&0&1\end{bmatrix}$ when $i$ is odd. $y_k \mapsto \begin{bmatrix} 1&0&1 \\0&1&0\\0&0&1\end{bmatrix}$ when $k$ is odd, $y_k \mapsto I$ when $k$ is even. $\endgroup$
    – ghc1997
    Apr 8, 2022 at 8:46
  • $\begingroup$ Yes, sorry, that works fine, and its kernel is invariant under the action of ${\mathbb Z}$, which is what you want for proving that $G$ is residually finite. Perhaps it generalizes to a homomorphism $\phi_k$ to a $k$-generated finite group for each $k \ge 2$. I will think about that later. $\endgroup$
    – Derek Holt
    Apr 8, 2022 at 9:28

0

You must log in to answer this question.