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).

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    $\begingroup$ Non finitely generated example: $\mathbb{Z}^{(\mathbb{N})}$. $\endgroup$ – Luc Guyot Mar 3 at 15:37
  • $\begingroup$ Ah, of course, yes, thanks. Holding out hope for a finitely generated example too.... $\endgroup$ – Matt Zaremsky Mar 3 at 15:51
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    $\begingroup$ A fairly easy finitely generated example is $F\times \mathbb{Z}$, where $F$ is Thompson's group. The center is nontrivial, and clearly $(F\times\mathbb{Z})\times (F\times \mathbb{Z})$ embeds into $F\times \mathbb{Z}$ since it embeds into $F$. $\endgroup$ – Jim Belk Mar 4 at 1:54
  • $\begingroup$ Oh yeah look at that! Very easy, this one I think I should have thought of. I still like your example of ribbon braided V better though, seems cleaner. $\endgroup$ – Matt Zaremsky Mar 4 at 2:36

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 $RV$.

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    $\begingroup$ Aha, beautiful, thank you! And yes, it is indeed $F_\infty$ (see the very last sentence of Thumann's paper arxiv.org/pdf/1409.1085.pdf). $\endgroup$ – Matt Zaremsky Mar 3 at 23:25

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:

enter image description here

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

enter image description here

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:

enter image description here

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:

enter image description here

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.

  • $\begingroup$ But doesn't the intersection of those subgroups contain $z_{pk}$ for all $k >0$? $\endgroup$ – Derek Holt Mar 3 at 17:11
  • $\begingroup$ Oh, you're right. Answer deleted. $\endgroup$ – AGenevois Mar 3 at 17:27
  • $\begingroup$ @DerekHolt: answer rewritten. $\endgroup$ – AGenevois Mar 4 at 13:38
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    $\begingroup$ Very nice! Note that the group $V(G)$ is isomorphic to the semidirect product $C(\mathfrak{C},G)\rtimes V$, where $C(\mathfrak{C},G)$ is the group of continuous (i.e. locally constant) functions from the Cantor set $\mathfrak{C}=\{0,1\}^\omega$ to $G$, and $V$ acts on $C(\mathfrak{C},G)$ by composition of functions. That is, $V(G)$ is something like a wreath product $G \wr_{\mathfrak{C}} V$, except that it uses $C(\mathfrak{C},G)$ instead of the full direct product $G^\mathfrak{C}$. $\endgroup$ – Jim Belk Mar 4 at 13:51
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    $\begingroup$ Aha, thanks! Another excellent example I'm a little embarrassed to have not thought of myself! By the way these groups were first (?) constructed by Slobodan Tanushevski (tandfonline.com/doi/abs/10.1080/00927872.2015.1087552), and in addition to $V(G)$ being fin. gen. once $G$ is, the converse is true, and in fact for any $n$ we have that $V(G)$ is type $F_n$ iff $G$ is. $\endgroup$ – Matt Zaremsky Mar 4 at 14:11

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.]

  • $\begingroup$ Wait, but $H_1$ isn't a subgroup of $G$, is it? If something in $G$ pointwise stabilizes $[1/2,1[$ then it's the identity (?). $\endgroup$ – Matt Zaremsky Mar 4 at 0:51
  • $\begingroup$ @MattZaremsky you're right, I want the supports to be invariant under the given involution. I therefore changed the involution. $\endgroup$ – YCor Mar 4 at 6:19
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    $\begingroup$ I think any involution, or indeed any element of finite order, will work. The key point here is that the centralizer $G$ of an element of finite order in $V$ always has nontrivial center and always contains $V$. Since $V$ contains $V\times V$, it follows that $G$ contains $G\times G$. $\endgroup$ – Jim Belk Mar 4 at 10:22
  • $\begingroup$ OK yes, this all makes sense. $\endgroup$ – Matt Zaremsky Mar 4 at 11:05

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