Are the distributive permutation groups linearly primitive?

Question

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 03 '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 03 '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 03 '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 04 '15 at 07:50

Sebastien Palcoux · Accepted Answer · 2017-09-10T11:00:11.677

1

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, 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. $\square$

edited Sep 10 '17 at 11:00

answered Jan 07 '15 at 07:37

Sebastien Palcoux

25,922

in the theory of permutation groups, the usual Galois correspondence is between subgroups and subrings of the centraliser ring of the permutation representation, due to Schur (?). Wielandt's book has details, IIRC. – Dima Pasechnik Jan 07 '15 at 12:21
I've asked the following historical question about the Galois correspondence for finite groups: http://mathoverflow.net/q/198819/34538 – Sebastien Palcoux Jan 27 '16 at 10:04

Are the distributive permutation groups linearly primitive?

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