An action of a group $G$ on a set $X \neq \emptyset$ is called [*transitive*](http://en.wikipedia.org/wiki/Group_action#Types_of_actions) if $\forall x,y \in X$, $\exists g \in G$ such that $g.x = y$.  

It is called [*primitive*](http://en.wikipedia.org/wiki/Group_action#Types_of_actions) if it is transitive and preserves no non-trivial partition of $X$. 


There is the following natural type of action (of $G$ on $X$) between "primitive" and "transitive":  

An action is called *distributive* if it is transitive and the lattice of preserved partitions is [distributive](http://en.wikipedia.org/wiki/Distributive_lattice).  

[A primitive action is obviously distributive] 


A transitive permutation group is a subgroup $G \subset S\_n$ whose action on $X=\{ 1, \dots , n \}$ is transitive.  

This group $G$ is called primitive (resp. distributive) if its action is primitive (resp. distributive). 


Let $G \subset S\_n$ be a transitive permutation group (of degree $n$) and $G\_1 := \{g \in G \ \vert \ g.1=1 \}$.  

Then $[G:G\_1] = n$ and $G$ is a primitive (resp. distributive) permutation group iff $G\_1 \subset G$ is a maximal subgroup (resp. the lattice of intermediate subgroups $\mathcal{L}(G\_1 \subset G)$ is distributive). 


A finite group $G$ is [linearly primitive](http://groupprops.subwiki.org/wiki/Linearly_primitive_group) if it has a faithful complex irreducible representation. 


**Question:** Are the distributive permutation groups linearly primitive?  

*Remark*: I've checked by a GAP computation that it's true for $n=[G:G\_1] \le 31$ and $\vert G \vert \le 10^4$ (and also checked for $\vert G \vert \le 10^5$ except indices $24$ and $30$).  

Moreover, all the primitive permutation groups are linearly primitive (see [this post](https://math.stackexchange.com/q/1088519/84284)). 




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Jan 3, 2014


Every finite group $G$ admits a faithful transitive action on $X = G$.  

There are groups without faithful primitive action, for example the cyclic group (of non-prime order). 


*Remark:* An abelian finite group $G$ is linearly primitive iff it admits a faithful distributive action.  

*Proof:* On one hand, $G$ admits a faithful complex irreducible representation iff it is a subgroup of $\mathbb{C}$, iff it is cyclic. On the other hand, $\mathcal{L}(G\_1 \subset G)$ is distributive, but $G\_1 = 1$ because it is [core-free](http://groupprops.subwiki.org/wiki/Core-free_subgroup), so the result follows by a theorem of Øystein Ore: a finite group $G$ is cyclic iff $\mathcal{L}(G)$ is distributive. $\square$ 


Generalization of the above remark to the non-abelian case:  

*Bonus question:* Is a group linearly primitive iff it admits a faithful distributive action? 




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Jan 4, 2014 


*Answer* of the bonus question: No, see the first comment of Derek. In fact, the first counterexample is the quaternion group $Q$: it is linearly primitive but has no faithful distributive action. 


A lattice is called *endistributive* if the sublattice generated by the [join prime](http://en.wikipedia.org/wiki/Lattice_(order)#Important_lattice-theoretic_notions) elements, is distributive.  

An action of a group $G$ on a set $X$ is called *endistributive* if it is transitive and the lattice of preserved partitions is endistributive. [A distributive action is obviously endistributive]  

The lattice $\mathcal{L}(Q)$ is endistributive, so $Q$ has a faithful degree $8$ endistributive action. 


*Bonus question 2:* Is a group linearly primitive iff it admits a faithful endistributive action? 


*Remark:* the second condition means, in others words, that $G$ admits a core-free subgroup $H$ such that $\mathcal{L}(H \subset K)$ is distributive, with $K$ the group generated by the minimal overgroups of $H$ in $G$.  

 It is immediately verified if $G$ admits a unique minimal subgroup, or a core-free maximal subgroup. 


*Remark*: Here are GAP checks about the bonus question 2:  

 $(\Leftarrow)$: $\vert G \vert \le 10^4$ and degree $\le 31$, **or** $\vert G \vert \le 10^5$ except degrees $24$ and $30$.  

 $(\Rightarrow)$: $\vert G \vert < 512$ (except $256$) for any degree, **or** $\vert G \vert \le 10^3$ and degree $\le 31$, except $24$. 



Are the distributive permutation groups linearly primitive?
**Yes**, a distributive permutation group is linearly primitive.


*Proof*: Let $G \subset S\_n$ be a distributive permutation group, then $G\_1 \subset G$ is a core-free subgroup and the lattice $[G\_1, G]$ is distributive. Then, by the [dual Ore's theorem](https://mathoverflow.net/q/195366/34538), there is an irr. complex rep. $V$ such that $G\_{(V^{G\_1})} = G\_1$. Now, $V^{G\_1} \subseteq V$, so $G\_{(V)} \subseteq G\_{(V^{G\_1})} $. But $K:=ker(\pi\_V)=G\_{(V)}$. It follows that $K \subseteq G\_1$, but $G\_1$ is a core-free subgroup of $G$, and $K$ a normal subgroup, so $K= \{ e \}$ which means that $V$ is faithful. So $G$ is linearly primitive. $\square$ 


