# If $K\rtimes \mathbb{Z}$ is a finitely generated group but $K$ isn't, must the fixed points of $1_\mathbb{Z}$ be a finitely generated group?

I've copied over this question from what I asked on StackExchange, in the hope that an expert here can readily answer the question.

Is there an example of a group $$G=K\rtimes \mathbb{Z}$$ satisfying the following three conditions?

• $$G$$ is finitely generated;
• $$K$$ is not finitely generated;
• the fixed points of $$\phi(1)$$, which is the automorphism on $$K$$ corresponding to $$1_\mathbb{Z}$$, are not a finitely generated group.

I suspect there is an example, but I don't have enough experience with infinite groups to come up with one right away.

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One idea, which may or may not work:

finding matrices $$M_1,\ldots,M_k\in\text{GL}_n(\mathbb{C})$$ and $$g\in\text{GL}_n(\mathbb{C})$$

such that $$S=\langle g^{-n}\,M_j\,g^n\rangle_{1\leq j\leq k,\;n\in\mathbb{Z}}$$ is not finitely generated

but such that $$g$$ commutes with surprisingly many matrices in $$S$$.

• If you use the trivial action, so you're looking at the direct product, then the fixed points are all of $K$, hence non-finitely generated. But probably you want the action to be sufficiently "robust" in some sense? Mar 26 at 10:15
• @MattZaremsky If you use the trivial action i.e. the direct product, "$K$ is not finitely generated" implies "$K\oplus\mathbb{Z}$ is not finitely generated" Mar 26 at 10:21
• Oh, ha, right. Not paying attention. Mar 26 at 10:34

No. Fix $$p\ge 2$$. Take the group $$G=\{M(x,y,z;n):(x,y,z)\in\mathbf{Z}[1/p],n\in\mathbf{Z}\}$$where $$M(x,y,z;n)=\begin{pmatrix}1 & x & z \\ 0 & p^n & y\\ 0 & 0 & 1\end{pmatrix}$$

and $$K$$ the set of such $$M(x,y,z;n)$$ for $$n=0$$, and identify $$\mathbf{Z}$$ to powers of $$M(0,0,0,1)$$.

Then $$G$$ is finitely generated (namely by $$\{M(0,0,0;1),M(1,0,0;0),M(0,1,0;0)\}$$), $$K$$ is not finitely generated, and indeed the centralizer of $$M(0,0,0;1)$$ in $$K$$ is not finitely generated (isomorphic to the abelian group $$\mathbf{Z}[1/p]$$).

• Merci! I did suspect that an example would come from matrices... Mar 26 at 11:11

Here is another example, which I am fond of, because it played a role in a PhD thesis of a student that I supervised a long time ago.

It is constructed as a central extension of $$C_p \wr {\mathbb Z}$$ for a prime $$p$$.

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 $$\begin{eqnarray*} x_i^p &=& y_j^p= 1\ \mbox{for all}\ i,j,\\ [x_j,x_i] &=& y_{j-i}\ \mbox{for}\ j>i,\\ [y_k,x_i] &=& 1\ \mbox{for all}\ i,k, \end{eqnarray*}$$ so it is nilpotent of class $$2$$ with $$K$$ and $$K/Z(K)$$ both infinite elementary abelian, where $$Z(K)$$ is generated by the elements $$y_k$$.

Now we define the action of $$({\mathbb Z},+)$$ on $$K$$ to make the element $$1 \in {\mathbb Z}$$ map $$x_i$$ to $$x_{i+1}$$ for all $$i \in {\mathbb Z}$$. So the induced action on $$Z(K)$$ is trivial.

Now let $$G$$ be the semidirect product $$K \rtimes {\mathbb Z}$$. Then $$G$$ is generated by $$x_0$$ and $$1 \in {\mathbb Z}$$, but $$K$$ and $$Z(G) = Z(K)$$ are not finitely generated.