First recall the Jordan-Hölder theorem for groups: 



> 
> **Theorem** (Jordan-Hölder): Let $G$ be a group, and let $$ G=G\_1 \supset G\_2 \supset \dots \supset G\_r = \{ e \} $$ be a normal tower such that each group
>  $G\_i /G\_{i+1}$ is simple, and $G\_i \neq G\_{i+1}$ for $0<i<r$.
>  Then any other normal tower of $G$ having the same properties is
>  equivalent to this one (i.e. the sequence of factor groups in our two
>  towers are the same up to isomorphisms, and a permutation of the
>  indices). 
> 
> 
> 


[This paper](http://www.imsc.res.in/~sunder/paha.pdf) of Kodiyalam-Landau-Sunder contains the basic definition of a planar algebra, of a group (subfactor) planar algebra, the definitions (p16) of planar algebra morphism, planar ideal and quotient.  

We note that the planar ideals are precisely the kernel of the planar algebra morphisms !  

We call a planar algebra **simple** if it has no non-trivial planar ideal. 



> 
> **Question 1** : Do the planar ideals of a group planar algebra correspond to the normal subgroups ?
> 
> 
> 


If yes, a group planar algebra is simple iff the group is simple.  

If no, how adapt the concept of planar ideal for having a positive answer ? 



> 
> **Question 2** : Let $\mathcal{P}$ be a planar algebra, and let $$ \mathcal{P}=\mathcal{J}\_1 \supset \mathcal{J}\_2 \supset \dots \supset \mathcal{J}\_r = (0) $$ be an ideal tower such that each quotient
>  $\mathcal{J}\_i /\mathcal{J}\_{i+1}$ is simple, and $\mathcal{J}\_i \neq \mathcal{J}\_{i+1}$ for $0<i<r$.
>  Then any other ideal tower of $\mathcal{P}$ having the same properties is
>  equivalent to this one (i.e. the sequence of quotient planar algebras in our two towers are the same up to isomorphisms, and a permutation of the
>  indices) ? 
> 
> 
> 



Jordan-Hölder theorem for planar algebras?
The answer is no. In fact, the group planar algebra has no non-trivial planar ideals.


If $P\_\bullet$ is a spherical planar algebra with non-zero modulus, and if $P\_{0,\pm}$ are one dimensional, then every planar ideal is contained in the planar ideal $N\_\bullet$ of negligible elements, i.e., those $x\in P\_{n,\pm}$ for which tr$(xy)=0$ for all $y\in P\_{n,\pm}$. (See Proposition 3.5 of the extended Haagerup paper of Bigelow-Morrison-Peters-Snyder MR2979509.)


Now the group planar algebra is a subfactor planar algebra. In particular, the negligible ideal is the zero ideal, since the form $\langle x,y\rangle = \text{tr}(y^\*x)$ is positive definite.


I'll think about the followup question.


