Let $(N \subset M)$ be an irreducible finite index inclusion of hyperfinite ${\rm II}_1$ factors.

Let $K_1$ and $K_2$ be two *distinct* intermediate subfactors $N \subset K_i \subset M$, such that $|M:K_i| = 2$.

*Question*: Is there (at least) a third intermediate subfactor strictly between $K_1 \cap K_2$ and $M$?

*Remark*: It is true for the finite group-subgroup subfactors $(R \rtimes H \subset R \rtimes G)$ and $(R^G \subset R^H)$.

*Proof using Galois correspondence*: Let $H_1$, $H_2$ be two intermediate subgroups $H \subset H_i \subset G$.

- First, if $|G:H_i| = 2$ then $H_i$ is a normal subgroup of $G$. It follows that $L=H_1 \cap H_2$ is also normal and $[L,G] \simeq [1,G/L]$. If there is not a third intermediate, then $[1,G/L]$ is boolean, so by Ore's theorem $G/L$ is cyclic, but with two subgroups of index $2$, contradiction.

- Next if $|H_i:H| = 2$ then $H$ is a normal subgroup of $H_i$. It follows that $H$ is also a normal subgroup of $T=\langle H_1 , H_2 \rangle $. If there is no third, then by the same argument $T/H$ is cyclic with two subgroups of order $2$, contradiction. $\square$

Angles between two subfactors, Remark and Corollary 6.1 p230. $\endgroup$ – Sebastien Palcoux Feb 12 '17 at 21:15