# Are the distributive permutation groups linearly primitive?

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 $24$.

• I don't have much feeling for the distibutivity of lattices, but I think that TransitiveGroup(8,22) (which is an extraspecial group of order $32$) might be a counterexample to your final question. It is certainly linearly primitive, and I don't think that it admits a faithful distributive actions. It has only three essentially different faithful transitive permutation actions, and these have degrees $8$, $16$ and $32$. – Derek Holt Jan 3 '15 at 15:11
• @DerekHolt: The degree $8$ action is not distributive. What is the number of the degree $16$ action? and $32$? – Sebastien Palcoux Jan 3 '15 at 18:17
• It's TransitiveGroup(16,23). The degree $32$ representation is the regular representation, so that cannot be distributive. – Derek Holt Jan 3 '15 at 19:17
• @DerekHolt: The degree 16 action is also not distributive (and the group is linearly primitive). The smallest counter-example for the bonus question is the quaternion group. – Sebastien Palcoux Jan 4 '15 at 7:50

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, 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. $\square$