We know that we can build an irreducible subfactor realizing a finite single chain lattice containing any finite index irreducible maximal subfactors, by using the free composition (see [here](https://mathoverflow.net/a/167461/34538)).  

Now about infinite single chain lattice: 


*Question*: Is there an irreducible subfactor with an infinite homogeneous single chain lattice? 



Is there an irreducible subfactor with an infinite homogeneous single chain lattice?
Yes, for example $(R \subset R \rtimes \mathbb{Z}(p^{\infty}))$ the [Prüfer $p$-group](http://en.wikipedia.org/wiki/Pr%C3%BCfer_group) subfactor.


More precisely:  

Let $p$ be a prime number and let the chain of embeddings $\mathbb{Z}/p\mathbb{Z} \hookrightarrow \mathbb{Z}/p^2\mathbb{Z} \hookrightarrow \mathbb{Z}/p^3\mathbb{Z} \hookrightarrow \cdots$  

given by multiplication by $p$, then $\bigcup\_n \mathbb{Z}/p^n\mathbb{Z}$ is $\mathbb{Z}(p^{\infty})$ the [Prüfer $p$-group](http://en.wikipedia.org/wiki/Pr%C3%BCfer_group).  

(Note that $\mathbb{Z}(p^{\infty})$ is the Pontryagin dual of the group of $p$-adic integers $\mathbb{Z}\_p$) 


$\mathbb{Z}(p^{\infty})$ admits no extra subgroups so its lattice is an infinite homogeneous single chain.  

Now it is a discrete group so by Galois correspondence for discrete group (theorem 3.13 p 48 of [this paper](http://www.sciencedirect.com/science/article/pii/S0022123697932286) of Izumi-Longo-Popa) for a given outer action of $\mathbb{Z}(p^{\infty})$ on the hyperfinite ${\rm II}\_1$ factor $R$ then the subfactor $(R \subset R \rtimes \mathbb{Z}(p^{\infty}))$ admits an infinite homogeneous single chain lattice. 


*Remark*: The planar subalgebra of $P(R \subset R \rtimes \mathbb{Z}(p^{\infty}))$ generated by all the biprojections "should be" $TLJ\_{p^{1/2}}^{\* \infty}$. So for any $\delta$ we should have the example $TLJ\_{\delta}^{\* \infty}$.  

 More generally, any infinite free composition of a maximal subfactor should be an example. 


