An action of a group $G$ on a set $X \neq \emptyset$ is called *transitive* if $\forall x,y \in X$, $\exists g \in G$ such that $g.x = y$.

It is called *primitive* 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.

[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 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).

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, 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?

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 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 $24$.