TITLE: Group with non-trivial center containing trivially-intersecting copies of itself
QUESTION [16 upvotes]: 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).

REPLY [5 votes]: An immediate variant of $V$ works.
View Thompson's $V$ with its natural action on $[0,1[$.
Let $G$ be the centralizer of the involution $s:x\mapsto x+c(x)$ with $c(x)=1/4$ if $x\in [0,1/4\mathclose[\cup [1/2,3/4\mathclose[$ and $c(x)=-1/4$ otherwise.
Let $H_1$ (resp. $H_2$) be the intersection of $G$ and the pointwise stabilizer of $[1/2,1\mathclose[$ (resp. of $[0,1/2\mathclose[$).
Clearly $H_1,H_2\simeq G$ and $H_1\cap H_2=\{\mathrm{id}\}$.
[Edit: I initially defined things in a clumsy way, fixed after Matt Zaremsky's comment.]