# A natural example of f.g. group, with uncountably many normal subgroups, but is not SQ-universal

I am looking for "natural" examples of a finitely generated group which have uncountably many normal subgroups but are not SQ-universal. A group $G$ is SQ-universal if for any countable group $H$, $H$ embeds into some quotient of $G$. For example, non-elementary hyperbolic groups are known to be SQ-universal (in fact acylindrically hyperbolic groups are SQ-universal)

For the most part, by natural I just mean it is not difficult to prove that the group exists. A non example would be a Burnside group of sufficiently high exponent. Ideally it would be a reasonably well known group, maybe even a linear group or something like that.

To be SQ-universal, the group must contain a free group of rank 2, so maybe looking at groups which don't contain such free groups would be a good place to look.

(In writing this question I found an acceptable group, so answered, but I welcome more examples)

Let $$A= \begin{pmatrix} 1 & 0 & 0 \\ 0 & x & 0 \\ 0 & 0 & 1 \end{pmatrix}, B=\begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}$$ where these are matrices over $\mathbb{Q}(x)$, then the group $\langle A,B \rangle$ is 3-solvable, with center the countable free abelian group (this is sketched in the book and seems like a bit much to prove in a post like this). The countable free abelian group has uncountably many subgroups, so there are uncountably many normal subgroups. It turns out that these quotients correspond to pairwise nonisomorphic solvable groups.
• That the center contains a free abelian group of countable rank is easy to prove, even within a comment: write $C_n=A^nBA^{-n}$. Then $C_n=\begin{pmatrix}1&0&0\\x^n&1&0\\0&x^{-n}&1\end{pmatrix}$, and $D_n=BC_nB^{-1}C_n^{-1}=\begin{pmatrix}1&0&0\\0&1&0\\x^n-x^{-n}&0&1\end{pmatrix}$. So $D_n$ is central and $(D_n)_{n\ge 1}$ is an infinite $\mathbf{Z}$-free family. Since the center is contained in the group of matrices $\begin{pmatrix}1&0&0\\0&1&0\\*&0&1\end{pmatrix}$ with $*\in\mathbf{Z}[x^{\pm 1}]$, itself free abelian, it follows that the center itself is free abelian of countable rank. – YCor Jun 26 '16 at 10:22