Return to Answer

The cyclic subfactors theory defines a too large class$^{\star}$ of subfactors for being an good quantum arithmetic (i.e. quantum generalization of the natural numbers theory).
Nowadays, my better attempt for this is the natural subfactors theory (see the optional part here).

$^{\star}$Up to equivalence, more than $70$% of the inclusions of groups of index $≤30$ are distributive lattice! Nevertheless the percentage is decreasing for the following indices (see this post).

Question 1: No, $(R^{A_7} \subset R^{S_4})$ is a counterexample.
Thanks to the subgroups lattice of $A_7$ (see here), there is a subgroup of $A_7$ called $2^2:S_3$ isomorphic to $S_4$, such that $(S_4 \subset A_7)$ has exactly two non-trivial intermediate subgroups: $K=L_2(7)$ and $L=A_4:S_3$, so that $(R^{A_7} \subset R^{S_4})$ is cyclic.
Now if we consider the two maximal chains $(S_4 \subset K \subset A_7)$ and $(S_4 \subset L \subset A_7)$, then the intermediate indices are $(7,15)$ and $(3,35)$, so these chains can't be equivalent.

See the questions: Jordan-Hölder theorem for subfactors? and Abelian subfactors, a relevant concept?

Question 3: No, the maximal subfactor $(R^{\mathbb{Z}_3 \rtimes \mathbb{Z}_2} \subset R^{\mathbb{Z}_2})$ gives counter-examples.
It is depth $4$ and it admits several compositions with itself which are also cyclic subfactors of depth $4$,
for example $(R^{(\mathbb{Z}_3 \rtimes \mathbb{Z}_2) \times (\mathbb{Z}_3 \rtimes \mathbb{Z}_2)} \subset R^{\mathbb{Z}_2 \times \mathbb{Z}_2})$ and $(R^{\mathbb{Z}_9 \rtimes \mathbb{Z}_2} \subset R^{\mathbb{Z}_2})$.
Their intermediate subfactors lattices are as $\mathcal{L}_6$ and $\mathcal{L}_4$, so they are not isomorphic.
($\mathcal{L}_n$ is the lattice of divisors of $n$).

The cyclic subfactors theory defines a too large class$^{\star}$ of subfactors for being an good quantum arithmetic (i.e. quantum generalization of the natural numbers theory).
Nowadays, my better attempt for this is the natural subfactors theory (see the optional part here).

$^{\star}$Up to equivalence, more than $70$% of the inclusions of groups of index $≤30$ are distributive lattice! Nevertheless the percentage is decreasing for the following indices (see this post).

Question 1: No, $(R^{A_7} \subset R^{S_4})$ is a counterexample.
Thanks to the subgroups lattice of $A_7$ (see here), there is a subgroup of $A_7$ called $2^2:S_3$ isomorphic to $S_4$, such that $(S_4 \subset A_7)$ has exactly two non-trivial intermediate subgroups: $K=L_2(7)$ and $L=A_4:S_3$, so that $(R^{A_7} \subset R^{S_4})$ is cyclic.
Now if we consider the two maximal chains $(S_4 \subset K \subset A_7)$ and $(S_4 \subset L \subset A_7)$, then the intermediate indices are $(7,15)$ and $(3,35)$, so these chains can't be equivalent.

See the questions: Jordan-Hölder theorem for subfactors? and Abelian subfactors, a relevant concept?

Question 3: No, the maximal subfactor $(R^{\mathbb{Z}_3 \rtimes \mathbb{Z}_2} \subset R^{\mathbb{Z}_2})$ gives counter-examples.
It is depth $4$ and it admits several compositions with itself which are also cyclic subfactors of depth $4$,
for example $(R^{(\mathbb{Z}_3 \rtimes \mathbb{Z}_2) \times (\mathbb{Z}_3 \rtimes \mathbb{Z}_2)} \subset R^{\mathbb{Z}_2 \times \mathbb{Z}_2})$ and $(R^{\mathbb{Z}_9 \rtimes \mathbb{Z}_2} \subset R^{\mathbb{Z}_2})$.
Their intermediate subfactors lattices are as $\mathcal{L}_6$ and $\mathcal{L}_4$, so they are not isomorphic.
($\mathcal{L}_n$ is the lattice of divisors of $n$).

The cyclic subfactors theory defines a too large class$^{\star}$ of subfactors for being an good quantum arithmetic (i.e. quantum generalization of the natural numbers theory).
Nowadays, my better attempt for this is the natural subfactors theory (see the optional part here).

$^{\star}$Up to equivalence, more than $70$% of the inclusions of groups of index $≤30$ are distributive lattice,
so we can interpolate that in the general subfactors theory, the cyclic subfactors are majority !

Question 1: No, $(R^{A_7} \subset R^{S_4})$ is a counterexample.
Thanks to the subgroups lattice of $A_7$ (see here), there is a subgroup of $A_7$ called $2^2:S_3$ isomorphic to $S_4$, such that $(S_4 \subset A_7)$ has exactly two non-trivial intermediate subgroups: $K=L_2(7)$ and $L=A_4:S_3$, so that $(R^{A_7} \subset R^{S_4})$ is cyclic.
Now if we consider the two maximal chains $(S_4 \subset K \subset A_7)$ and $(S_4 \subset L \subset A_7)$, then the intermediate indices are $(7,15)$ and $(3,35)$, so these chains can't be equivalent.

See the questions: Jordan-Hölder theorem for subfactors? and Abelian subfactors, a relevant concept?

Question 3: No, the maximal subfactor $(R^{\mathbb{Z}_3 \rtimes \mathbb{Z}_2} \subset R^{\mathbb{Z}_2})$ gives counter-examples.
It is depth $4$ and it admits several compositions with itself which are also cyclic subfactors of depth $4$,
for example $(R^{(\mathbb{Z}_3 \rtimes \mathbb{Z}_2) \times (\mathbb{Z}_3 \rtimes \mathbb{Z}_2)} \subset R^{\mathbb{Z}_2 \times \mathbb{Z}_2})$ and $(R^{\mathbb{Z}_9 \rtimes \mathbb{Z}_2} \subset R^{\mathbb{Z}_2})$.
Their intermediate subfactors lattices are as $\mathcal{L}_6$ and $\mathcal{L}_4$, so they are not isomorphic.
($\mathcal{L}_n$ is the lattice of divisors of $n$).

The cyclic subfactors theory defines a too large class$^{\star}$ of subfactors for being an good quantum arithmetic (i.e. quantum generalization of the natural numbers theory).
Nowadays, my better attempt for this is the natural subfactors theory (see the optional part here).

$^{\star}$Up to equivalence, more than $70$% of the inclusions of groups of index $≤30$ are distributive lattice! Nevertheless the percentage is decreasing for the following indices (see this post).

Question 1: No, $(R^{A_7} \subset R^{S_4})$ is a counterexample.
Thanks to the subgroups lattice of $A_7$ (see here), there is a subgroup of $A_7$ called $2^2:S_3$ isomorphic to $S_4$, such that $(S_4 \subset A_7)$ has exactly two non-trivial intermediate subgroups: $K=L_2(7)$ and $L=A_4:S_3$, so that $(R^{A_7} \subset R^{S_4})$ is cyclic.
Now if we consider the two maximal chains $(S_4 \subset K \subset A_7)$ and $(S_4 \subset L \subset A_7)$, then the intermediate indices are $(7,15)$ and $(3,35)$, so these chains can't be equivalent.

See the questions: Jordan-Hölder theorem for subfactors? and Abelian subfactors, a relevant concept?

Question 3: No, the maximal subfactor $(R^{\mathbb{Z}_3 \rtimes \mathbb{Z}_2} \subset R^{\mathbb{Z}_2})$ gives counter-examples.
It is depth $4$ and it admits several compositions with itself which are also cyclic subfactors of depth $4$,
for example $(R^{(\mathbb{Z}_3 \rtimes \mathbb{Z}_2) \times (\mathbb{Z}_3 \rtimes \mathbb{Z}_2)} \subset R^{\mathbb{Z}_2 \times \mathbb{Z}_2})$ and $(R^{\mathbb{Z}_9 \rtimes \mathbb{Z}_2} \subset R^{\mathbb{Z}_2})$.
Their intermediate subfactors lattices are as $\mathcal{L}_6$ and $\mathcal{L}_4$, so they are not isomorphic.
($\mathcal{L}_n$ is the lattice of divisors of $n$).

Question 1: No, $(R^{A_7} \subset R^{S_4})$ is a counterexample.
Thanks to the subgroups lattice of $A_7$ (see here), there is a subgroup of $A_7$ called $2^2:S_3$ isomorphic to $S_4$, such that $(S_4 \subset A_7)$ has exactly two non-trivial intermediate subgroups: $K=L_2(7)$ and $L=A_4:S_3$, so that $(R^{A_7} \subset R^{S_4})$ is cyclic.
Now if we consider the two maximal chains $(S_4 \subset K \subset A_7)$ and $(S_4 \subset L \subset A_7)$, then the intermediate indices are $(7,15)$ and $(3,35)$, so these chains can't be equivalent.

A more relevant related open question is: Jordan-Hölder theorem for subfactors?

Remark: It could be relevant to add the assumptions abelian and Dedekind for being a cyclic subfactor.
(see Abelian subfactors, a relevant concept?)

The cyclic subfactors theory defines a too large class$^{\star}$ of subfactors for being an good quantum arithmetic (i.e. quantum generalization of the natural numbers theory).
Nowadays, my better attempt for this is the natural subfactors theory (see the optional part here).

$^{\star}$Up to equivalence, more than $70$% of the inclusions of groups of index $≤30$ are distributive lattice,
so we can interpolate that in the general subfactors theory, the cyclic subfactors are majority !

Question 1: No, $(R^{A_7} \subset R^{S_4})$ is a counterexample.
Thanks to the subgroups lattice of $A_7$ (see here), there is a subgroup of $A_7$ called $2^2:S_3$ isomorphic to $S_4$, such that $(S_4 \subset A_7)$ has exactly two non-trivial intermediate subgroups: $K=L_2(7)$ and $L=A_4:S_3$, so that $(R^{A_7} \subset R^{S_4})$ is cyclic.
Now if we consider the two maximal chains $(S_4 \subset K \subset A_7)$ and $(S_4 \subset L \subset A_7)$, then the intermediate indices are $(7,15)$ and $(3,35)$, so these chains can't be equivalent.

See the questions: Jordan-Hölder theorem for subfactors? and Abelian subfactors, a relevant concept?

Question 3: No, the maximal subfactor $(R^{\mathbb{Z}_3 \rtimes \mathbb{Z}_2} \subset R^{\mathbb{Z}_2})$ gives counter-examples.
It is depth $4$ and it admits several compositions with itself which are also cyclic subfactors of depth $4$,
for example $(R^{(\mathbb{Z}_3 \rtimes \mathbb{Z}_2) \times (\mathbb{Z}_3 \rtimes \mathbb{Z}_2)} \subset R^{\mathbb{Z}_2 \times \mathbb{Z}_2})$ and $(R^{\mathbb{Z}_9 \rtimes \mathbb{Z}_2} \subset R^{\mathbb{Z}_2})$.
Their intermediate subfactors lattices are as $\mathcal{L}_6$ and $\mathcal{L}_4$, so they are not isomorphic.
($\mathcal{L}_n$ is the lattice of divisors of $n$).