This post was inspired by an exchange with the indian woman mathematician [Ajit Iqbal Singh](https://en.wikipedia.org/wiki/Ajit_Iqbal_Singh).


The chinese remainder theorem can be stated as follows:  

Let $n\_1, \dots, n\_r \ge 2$ be positive integers such that $n\_i \wedge n\_j = 1$ iff $i \neq j$, then the system of equations $$ x \equiv a\_i [n\_i] \text{, for } i=1, \dots , r$$ admits a solution $x$ which is unique modulo $\prod\_i n\_i$.


There is a natural way to generalize this statement to the cyclic subfactor planar algebras, as follows  

(I warn the reader who don't know the subfactor planar algebra theory, that I've reformulated the statement in the case of finite index inclusions of groups, just after). 


Let $P$ be a finite index irreducible subfactor planar algebra, and assume that $P$ is cyclic, i.e. admits a *distributive* biprojections lattice. The statement is the following:  

Let $b\_1, \dots , b\_r \not \in \{ e\_1, id \}$ be a subset of biprojections satisfying $ b\_i \vee b\_j = id$ iff $i \neq j$, then the system:
$$ b\_i \* x \* b\_i \sim b\_i \* a\_i \* b\_i \text{, for } i=1, \dots ,r $$
(with $a\_i, x$ positive operators) admits a solution $x$ which is unique (up to $\sim$) as $B \* x \* B$, with $B = \bigwedge\_i b\_i$.


Now in the group framework: let $(H \subset G)$ be a finite index inclusion of groups, such that the interval lattice $[H,G]$ is *distributive*, then the statement is the following:  

Let $K\_1, \dots , K\_r$ be strict intermediate subgroups (i.e. $H \subsetneq K\_i \subsetneq G$, $\forall i$), such that $\langle K\_i , K\_j \rangle = G$ iff $i \neq j$, then the system of double-coset equations: 
$$ K\_ixK\_i = K\_ig\_iK\_i \text{, for } i=1, \dots ,r $$
admits a solution $x$ which is unique as $KxK$ with $K = \bigcap\_i K\_i$. 


*Remark*: This group statement above with $G = \mathbb{Z}$, $K\_i = n\_i\mathbb{Z}$ and $H = K = \prod\_i n\_i \mathbb{Z}$ is equivalent to the classical statement of the chinese remainder theorem. 


**Question**: Is the chinese remainder statement true for the cyclic subfactor planar algebras? 


*Remark*: A counter-example for an inclusion of finite groups gives a counter-example for the subfactors planar algebras. I will edit a ckeck by GAP soon.


*Remark*: The *distributivity* assumption is very useful for generalizing the beginning of the proof [here](https://proofwiki.org/wiki/Chinese_Remainder_Theorem).


*Motivation*: Try to better understand in what sense the [cyclic subfactor theory](https://mathoverflow.net/q/133050/34538) is a *quantum arithmetic*.



Chinese remainder theorem for cyclic subfactor planar algebras
**No**, we show above an inclusion of finite groups counter-example. 


First of all, the statement for an inclusion of finite groups $(H \subset G)$ can be reformulate as follows: 


Let $K\_1, \dots , K\_r$ be strict intermediate subgroups (i.e. $H \subsetneq K\_i \subsetneq G$, $\forall i$), such that $\langle K\_i , K\_j \rangle = G$ iff $i \neq j$ (note that $r$ is assumed $\ge 2$). Without lose of generality, we can assume that $K:=\bigcap\_{i}K\_i = H$ (otherwise we replace $H$ by $K$). Let $\{ K\_i g\_{i,1} K\_i, \dots , K\_i g\_{i,r\_i} K\_i \}$ be the set of all the double cosets for $K\_i$ in $G$.  

Then the Chinese remainder statement is equivalent to: $$\forall g\_1, g\_2, \dots , g\_r \in G \ , \ \exists g \in G \ ; \ \bigcap\_{i}(K\_i g\_i K\_i) \subset HgH$$ 


The first counter-example having the interval lattice $[H,G]$ distributive is the following: 



```
gap> G:=TransitiveGroup(10,4);
1/2[F(5)]2
gap> H:=Stabilizer(G,1);
Group([ (2,10)(3,9)(4,8)(5,7) ])

```

First the lattice is distributive because: 



```
gap> IntermediateSubgroups(G,H).inclusions;
[ [ 0, 1 ], [ 0, 2 ], [ 1, 3 ], [ 2, 3 ] ]

```

Let $K\_1$ and $K\_2$ the two strict intermediate subgroups:



```
gap> K1:=IntermediateSubgroups(G,H).subgroups[1];
Group([ (2,10)(3,9)(4,8)(5,7), (1,6)(2,9,10,3)(4,5,8,7) ])
gap> K2:=IntermediateSubgroups(G,H).subgroups[2];
Group([ (2,10)(3,9)(4,8)(5,7), (1,9,7,5,3)(2,10,8,6,4) ])

```

Obviously $\langle K\_1, K\_2 \rangle = G$ and $K\_1 \cap K\_2 = H$.  

Let $D$ be the intersection $(K\_1 g\_{1,2} K\_1) \cap (K\_2 g\_{2,2} K\_2)$: 



```
gap> D:=Intersection(DoubleCosets(G,K1,K1)[2], DoubleCosets(G,K2,K2)[2]);
[ (1,2,5,4)(3,8)(6,7,10,9), (1,2,9,8)(3,6,7,4)(5,10), (1,4,5,2)(3,8)(6,9,10,7), (1,4,3,10)(2,7)(5,6,9,8), (1,8,9,2)(3,4,7,6)(5,10), (1,8,7,10)(2,5,6,3)(4,9), 
  (1,10,7,8)(2,3,6,5)(4,9), (1,10,3,4)(2,7)(5,8,9,6) ]

```

Then $\forall g \in G$, $D \not \subset HgH$: 



```
gap> List(DoubleCosets(G,H,H), S->IsSubset(S,D));
[ false, false, false, false, false, false ]

```

So $(H \subset G)$ is a counter-example with a distributive lattice. 


*Remark*: up to equivalent there are $166$ index $\le 10$ inclusions of finite groups: 



```
gap> Sum(List([1..10], d->NrTransitiveGroups(d)));
166

```

Among them (with distributive lattice or not), only $6$ do not check the Chinese remainder statement: 



```
gap> G:=TransitiveGroup(d,r);
gap> H:=Stabilizer(G,1);

```

with $(d,r) = ( 6, 2 ), ( 8, 3 ), ( 8, 4 ), ( 9, 5 ), ( 10, 2 ), ( 10, 4 )$.  

(The last one, i.e. $(d,r) = ( 10, 4 )$, is the first distributive lattice counter-example) 


It follows that in this case, more than $96 \%$ check the Chinese remainder statement! 


