Here is an idea to construct Thompson-like groups with non-trivial centers. The construction depends on an arbitrary group $G$ we fix once for all.

**Labelled strand diagrams.** In the same way that every element of $V$ can be represented by a strand diagram, every element of our new group $V(G)$ will be represented by a strand diagram all of whose wires are labelled by elements of $G$. Graphically:

**Equivalent diagrams.** Two labelled diagrams are equivalent if one can transform one into the other by applying the following elementary moves:

Observe that, up to equivalence, every diagram can be written in such a way that only the wires connecting to top and bottom trees are labelled by non-trivial elements of $G$.

From now on, we look at diagrams up to equivalence.

**Group law.** The product of two (equivalence classes of) diagrams is obtained by mimicing the product in Thompson's groups. I just give an example:

The first step is to move the non-trivial labels to the middle wires; the second step is to modify the diagrams so that the left bottom tree coincides with the right top tree; and the thid step is to remove the left bottom and right top trees and to glue the right diagram below the left diagram.

Of course, there is some work to do in order to verify that everything is well-defined, but the classical arguments used for Thompson's groups seem to apply similarly. I denote by $V(G)$ the group thus obtained.

**The center of $V(G)$ contains the center of $G$.** Our group $G$ naturally embeds into $V(G)$: for every $g \in G$, consider the strand diagram with a single wire labelled by $g$. If we identify $G$ with its image in $V(G)$, then $Z(G) \subset Z(V(G))$. It seems reasonable to think that there is actually an equality, but the point is that our group $V(G)$ has a non-trivial center as soon as $G$ has a non-trivial center.

**$V(G)$ is finitely generated if so is $G$.** Of course, $V(G)$ contains a natural copy of $V$, corresponding to diagrams all of whose wires have trivial labels. We identify $V$ with its image in $V(G)$. If $G_1$ denotes the copy of $G$ in $V(G)$ defined by:

then it is not too difficult to show that $V(G)$ is generated by $V$, $G$ and $G_1$. Therefore, if $G$ is finitely generated then so is $V(G)$.

**$V(G)$ contains many copies of itself.** The assertion is pretty clear. In particular, $V(G)$ contains a subgroup isomorphic to $V(G) \times V(G)$, so it is possible to find two copies of $V(G)$ having trivial intersection.