I'm trying to think of an example of a group $G$ with non-trivial center such that there exist subgroups $H\_1,H\_2\le G$ both isomorphic to $G$ and satisfying $H\_1\cap H\_2=\{1\}$. Does such a group exist? Preferably an easy-to-state example that I'm just not thinking of? (Ideally finitely generated, but I won't insist on it.)


If the center is trivial then this is easy (free groups work, I think any centerless RAAG will work, various Thompson-like groups work), but requiring non-trivial center seems to complicate things. I would also be happy with an example that requires intersecting more than two copies of $G$ inside $G$ to ensure a trivial intersection (as long as it's finitely many copies).



Group with non-trivial center containing trivially-intersecting copies of itself
Here is a construction for a group similar to the braided Thompson group $BV$ that ought to have this property. Define the $n$th ***ribbon group*** to be the semidirect product
$$
R\_n = \mathbb{Z}^n \rtimes B\_n
$$
where the braid group $B\_n$ acts on $\mathbb{Z}^n$ by permuting the $n$ factors. Elements of $R\_n$ can be thought of like braids except that each strand is replaced by a "ribbon" that can twist on its own. That is, $R\_n$ is the mapping class group of a disk with $n$ disjoint, oriented arcs in its interior, rel the boundary. (You don't have to orient the arcs, but it's simpler if you do since this restricts each ribbon to full twists. If you don't orient the arcs you also get a group isomorphic to $R\_n$ but the generators of the $\mathbb{Z}^n$ correspond to half-twists.)


In the same way that there are natural functions $\sigma\_i\colon B\_n\to B\_{n+1}$ ($1\leq i\leq n$) that correspond to splitting the $i$th strand of a braid, there are also natural splitting functions $\sigma\_i\colon R\_n\to R\_{n+1}$, where splitting a ribbon that twists $n$ times results in two ribbons that each twist $n$ times and also braid $n$ times around each other. These splitting functions allow you to define a ***ribbon Thomspon group*** $RV$ that fits into a short exact sequence
$$
R\_{2^\infty} \hookrightarrow RV \twoheadrightarrow V
$$
where $R\_{2^\infty}$ is the ascending union of a sequence $\{R\_{2^n}\}$ of ribbon groups.


This group $RV$ ought to be finitely generated for the usual Thompson-like reasons (indeed it should have type $\mathrm{F}\_\infty$), and clearly it contains $RV\times RV$. However, unlike the braided Thompson group $BV$, the group $RV$ has nontrivial center. In particular, the center of $RV$ ought to be the infinite cyclic group generated by the full twist, i.e. the canonical copy of $R\_1$ contained in $RV$.


