Let $(H \subset G)$ be an inclusion of finite groups.  

Let $\{ g\_i \ \vert \ i \in I=[1, \dots ,n] \}$ a subset of $G$ of double coset representatives, i.e. $$G = \coprod\_{i \in I} Hg\_iH$$ 
On the algebra $\mathbb{C}G$ we consider the elements $$e\_i = \frac{1}{\vert H \vert} \sum\_{g \in Hg\_iH} g$$ 
They generate the double coset subalgebra of $\mathbb{C}G$, we have the following constants structure $c\_{i j}^{k} \in \mathbb{N}$:
$$e\_{i} \cdot e\_{j} = \sum\_{k \in I} c\_{i j}^{k} e\_{k}$$
*Remark*: we get the first examples of [finite hypergroups](https://www.imsc.res.in/~sunder/act.pdf) (up to a renormalisation), noted $G//H$.


For example, with $(H \subset G) = (\mathbb{Z}/2 \subset \mathbb{Z}/5 \rtimes \mathbb{Z}/2)$ we get the following table (with $g\_1 = e$ the neutral):  

$$\begin{array}{c|c|c|c|c|c|c} 
 \cdot & e\_1 & e\_2 & e\_3 \\ \hline 
 e\_1 & e\_1 & e\_2 & e\_{3} \\ \hline
 e\_2 & e\_2 & 2e\_1+e\_3 & e\_2+e\_3 \\ \hline
 e\_{3} & e\_{3} & e\_{2}+e\_3 &2e\_1+e\_2
\end{array} $$ 


*Definition*: a basic element $e\_{i}$ is called *generating* if $\forall j \in I, \exists r\_j$ such that $(e\_{i}^{r\_j}, e\_j) >0$.  

*Remark*: It exists a generating basic element iff $\exists g \in G$ such that $\langle H,g \rangle =G$.  

It is true for every primitive inclusion (and also for every "distributive" inclusion, see [here](https://mathoverflow.net/a/195331/34538)). 


Let the matrix $M\_i = (m\_{ij}^{(r)})\_{rj}$ defined by $e\_{i}^{r} = \sum\_{j \in I}m\_{ij}^{(r)}e\_j $. 


For the example above: $M\_1 = \left(\begin{matrix} 1 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 0 & 0\end{matrix}\right) $, $M\_2 = \left(\begin{matrix} 0 & 1 & 0 \\ 2 & 0 & 1 \\ 0 & 3 & 1\end{matrix}\right)$ and $M\_3 = \left(\begin{matrix} 0 & 0 & 1 \\ 2 & 1 & 0 \\ 0 & 1 & 3\end{matrix}\right)$ 


*Definition*: a basic element $e\_{i}$ is called *separating* if the matrix $M\_i$ is invertible.  

*Remark:* "separating" implies "generating", but the converse is false.  

*Definition*: $(H \subset G)$ is called *separating* if it exists a *separating* basic element $e\_{i}$. 


**Question**: Is a prime index inclusion of finite groups, separating?  

*Remark*: It's checked by GAP for $[G:H] \le 500$ and $\vert G \vert \le 4000$. 


The first non-separating primitive inclusions are given by: 


gap> G:=PrimitiveGroup(d,r);  

gap> H:=Stabilizer(G,1);


with $(d,r) = (16,1), (16,2), (25, 1 ),( 25, 4 ),( 25, 6 ),( 25, 11 ), \dots$ 


For example with $(d,r) = (16,1)$ we get the following table and $M\_i$ matrices:  

 $$\begin{array}{c|c|c|c|c|c|c} 
 \cdot & e\_1 & e\_2 & e\_3 & e\_4 \\ \hline 
 e\_1 & e\_1 & e\_2 & e\_{3} &e\_4 \\ \hline
 e\_2 & e\_2 & 5e\_1+2e\_3+2e\_4 & 2e\_2+2e\_3+e\_4 & 2e\_2+e\_3+2e\_4 \\ \hline
 e\_3 & e\_3 & 2e\_2+2e\_3+e\_4 &5e\_1+2e\_2+2e\_4 & e\_2+2e\_3+2e\_4 \\ \hline
e\_4 & e\_4 & 2e\_2+e\_3+2e\_4 &e\_2+2e\_3+2e\_4 & 5e\_1+2e\_2+2e\_3
\end{array} $$ 
$M\_1 = \left(\begin{smallmatrix} 1 & 0& 0& 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0\end{smallmatrix}\right)$, $M\_2 = \left(\begin{smallmatrix} 0 & 1& 0& 0 \\ 5 & 0 & 2 & 2 \\ 0 & 13 & 6 & 6 \\ 65 & 24 & 44 & 44\end{smallmatrix}\right)$, $M\_3 = \left(\begin{smallmatrix} 0& 0& 1& 0 \\ 5& 2& 0& 2 \\ 0& 6& 13& 6 \\ 65& 44& 24& 44 \end{smallmatrix}\right)$, $M\_4 = \left(\begin{smallmatrix} 0& 0& 0& 1 \\ 5& 2& 2& 0 \\ 0& 6& 6& 13 \\ 65& 44& 44& 24\end{smallmatrix}\right)$


*Bonus question*: How to improve the "separating" assumption for becoming true for every primitive inclusion?  

Idea to check: widen the notion to the existence of a separating (non-necessarily basic) element. 



Is a prime index inclusion of finite groups, separating?
I think the answer for prime index inclusions is Yes. I see the question from the following point of view: For each double coset $D= HgH$, you form the element
$$ e\_D := \frac{1}{\lvert H \rvert} \sum\_{x\in D} x \in \mathbb{Q} G . $$ 
The product of two such elements is an $\mathbb{N}$-linear combination of such elements. It follows that the $e\_D$'s form the basis of a $\mathbb{Q}$-algebra $C$ (say), and $e\_H$ is the $1$ of this algebra. This algebra $C$ is isomorphic to the centralizer algebra of the corresponding permutation representation (over $\mathbb{Q}$).  

(Later edit: That is, the centralizer of the permutation matrices in the ring of quadratic $\mathbb{Q}$-matrices of size $\lvert G:H\rvert$. Possibly the most elegant way to see this is to realize the permutation module as the left ideal $\mathbb{Q}G e\_H$ with $G$ acting from the left. Then the centralizer can be identified with the endomorphism ring $\operatorname{End}\_{\mathbb{Q}G} (\mathbb{Q}G e\_H)$, which is isomorphic to the ring
$ e\_H \mathbb{Q}G e\_H$, acting from the *right*. And the $e\_D$'s (with $D$ running over the $H$-$H$-double cosets) form a basis of $e\_H \mathbb{Q}Ge\_H$.) 


The rows of a particular matrix $M\_i$ contain the coefficients of $e\_D$, $(e\_D)^2$, $\dots$, $(e\_D)^n$ in terms of the natural basis of $C$, where $D = Hg\_iH$. Thus $e\_D$ is a separating element if and only if the powers of $e\_D$ form a basis of the centralizer ring $C$. This is the case if and only if $C= \mathbb{Q}[e\_D]$ and $e\_D$ is invertible. In particular, if there is a separating element (maybe not basic), then the centralizer ring must be commutative, or equivalently, the permutation representation must be multiplicity free. Conversely, if this is the case, then the centralizer ring is a direct product of fields, and at least some (non-basic) separating element exists. 


Now let $G $ be a permutation group of prime degree $p$. By a theorem of Burnside, $G$ is either $2$-transitive, or is isomorphic to a proper subgroup of the affine group $\operatorname{AGL}(1,p)$. The case of $G$ being $2$-transitive is clear. In the other case, we may assume that $G$ is a subgroup of $\operatorname{AGL}(1,p)$, viewed as permutation group on $\mathbb{F}\_p$, and $H$ is the stabilizer of $0\_{\mathbb{F}\_p}$. Let $\widehat{H}$ be the stabilizer in $\operatorname{AGL}(1,p)$ of $0\_{\mathbb{F}\_p}$. Then the $e\_D$'s with $D$ not the double coset $H$ get permuted transitively under the conjugation action of $\widehat{H}$. 


On the other hand, the centralizer algebra is isomorphic to the sum of $\mathbb{Q}$ and another field $\mathbb{K}$ contained in the cyclotomic field $\mathbb{Q}(\varepsilon\_p)$. The group $\widehat{H}$ acts on $\mathbb{K}$ as field automorphisms. Let $\alpha$ be the projection of one $e\_D$ onto $\mathbb{K}$, where $D\neq H$. It follows from the previous paragraph that the Galois conjugates of $\alpha$ generate $\mathbb{K}$. As all subfields of $\mathbb{K}$ are Galois, we see that $\mathbb{K} = \mathbb{Q}(\alpha)$. The projection of $e\_D$ onto $\mathbb{Q}$ is non-zero (namely, $\lvert D \rvert / \lvert H \rvert $). Thus $e\_D$ is an invertible element of the centralizer algebra. Thus to see that $e\_D$ is separating, it suffices to show that $(e\_D)^0 = e\_H$, $e\_D$, $(e\_D)^2$, $\dots$, $(e\_D)^{n-1}$ are linear independent over $\mathbb{Q}$. Suppose $\sum\_{r=0}^{n-1} q\_r (e\_D)^r = 0$ with $q\_r\in \mathbb{Q}$. Then the minimal polynomial of $\alpha$ must divide $p(x) = \sum\_r q\_r x^r$. The minimal polynomial of $\alpha$ has degree $n-1$ and $p(x)$ at most degree $n-1$, however, so $p(x)$ is the minimal polynomial up to a scalalr multiple (possibly zero). The projection of $e\_D$ onto $\mathbb{Q}$ can not be a zero of the minimal polynomial of $\alpha$. Thus $p(x) = 0$. It follows that the centralizer ring is generated by the powers of $e\_D$, so $e\_D$ is separating.


