A finite group-subgroup subfactor is a subfactor $(N \subset M)$ isomorphic to $(R^G \subset R^H)$ with $(H \subset G)$ an inclusion of finite groups acting as outer automorphism on the hyperfinite II$_1$ factor $R$.

Recall the equivalence on the inclusions of (finite) groups:

$(A \subset B) \sim (C \subset D)$ if $(A/A_B \subset B/A_B) \simeq (C/C_D \subset D/C_D)$ with $A_B=core_B(A)$

In this paper, M. Izumi gives a purely group-theoretic reformulation of the weaker equivalence:

$(A \subset B) \sim_{w} (C \subset D)$ if $(R^B \subset R^A) \simeq (R^D \subset R^C)$

**Exemple** (Kodiyalam-Sunder p47) : $(\langle (1234) \rangle \subset S_4) \sim_w (\langle (13),(24) \rangle \subset S_4)$,

whereas these two (core-free) inclusions are **not** $\sim$-equivalent.

So a finite group subfactor $(R^G \subset R)$ remembers the isomorphism class of the group $G$, because $Gal(R^G \subset R) \simeq G$, but a finite group-subgroup subfactor $(R^G \subset R^H)$ does **not** remember the $\sim$-equivalence class of the inclusion $(H\subset G)$, only its $\sim_w$-equivalence class.

Now the $\sim$-equivalence class of $(H \subset G)$ (**core-free**) is the same thing that the conjugacy class of a transitive permutation group of degree $n=[G:H]$, i.e. up to identifying the sets $G/H$ and $X_n=\{1,2,\dots,n\}$ (with $H=1$), $G$ is a subgroup of $S_n$ acting transitively on $X_n$ and $H=G_1$.

The question is about a "**section**" from group-subgroup subfactors to transitive permutation groups:

Question: Given finite group-subgroup subfactor $(N \subset M)$, how build a transitive permutation group $G$ of degree $[M:N]$ such that $(R^G \subset R^{G_1}) \simeq (N \subset M)$ ?

If such a natural process exists, it would be fruitful to generalize it to all the irreducible subfactors.

**Generalization to irreducible subfactors**

For the group-subgroup subfactors of index $n$, the group $S_n$ (acting on $X_n$) always underlies:

If $\mathcal{P}$ is the *spin model planar algebra* then $\mathcal{P}^{S_n}$ and $\mathcal{P}^{G}$ are the subfactor planar algebras of $(R^{S_n} \subset R^{S_{n-1}})$ and $(R^G \subset R^{G_1})$, and $\mathcal{P}^{S_n} \subset \mathcal{P}^{G}$ (see here p13-14).

Each group-subgroup subfactor of index $n$ is given by a subgroup of $S_n$ acting transitively on $X_n$.

**Remark** : For $k=1,2,3$, $(R^G \subset R^{G_1})$ is $k$-supertransitive iff $G$ is $k$-transitive (see this comment).

For the irreducible subfactors $(N \subset M)$ of index $\alpha (> 4)$, $\mathcal{P}^{S_n}$ is replaced by the Temperley-Lieb planar algebra $TL_{\delta}$ with $\delta^2=\alpha$, and $TL_{\delta} \subset \mathcal{P}(N \subset M)$, $(R^{S_n} \subset R^{S_{n-1}})$ is replaced by the $TL_{\delta}$-subfactor of index $\alpha$ (with principal graph $A_\infty$). The $TL_{\delta}$-algebra gives a representation $V_{\alpha}$ of the Braid group $B_{\infty}$, so in some sense $S_n$ and $X_n$ are replaced by $B_{\infty}$ and $V_{\alpha}$.

Question: Does a subgroup of $B_{\infty}$ acting transitively on a basis of $V_{\alpha}$ gives an index $\alpha$ irreducible subfactor? Is the converse true ? Else is there a link with the braided fusion categories ?

**Remark**: For subfactors with non-trivial intermediate subfactors, the Fuss-Catalan algebras plays the same role than the Temperley-Lieb algebras.

$S_n=\langle s_1, \dots , s_{n-1} \vert s_is_{i+1}s_i= s_{i+1} s_i s_{i+1}, s_i s_{j} = s_j s_{i} \text{ for } \vert i-j \vert \ge 2, s_i^2=1 \rangle$

$B_n=\langle s_1, \dots , s_{n-1} \vert s_is_{i+1}s_i= s_{i+1} s_i s_{i+1}, s_i s_{j} = s_j s_{i} \text{ for } \vert i-j \vert \ge 2 \rangle$

$B_{\infty}=\langle s_1, s_2, \dots \vert s_is_{i+1}s_i= s_{i+1} s_i s_{i+1}, s_i s_{j} = s_j s_{i} \text{ for } \vert i-j \vert \ge 2 \rangle$

$TL_{\delta} = C^{\star} \langle e_1, e_2, \dots \vert e_ie_{i \pm 1}e_i= \delta e_i, e_i e_{j} = e_j e_{i} \text{ for } \vert i-j \vert \ge 2, e_i^2= e_i^{\star} = e_i \rangle$

Representation of the Braid groups: $s_i \to te_i-(1-e_i)$, iff $\delta = \frac{t}{1-t^2}$