Different actions of an affine primitive group?

Question

Fairly new to group actions and I'm having trouble finding answers to these in textbooks...

Say we have a primitive action of $G$ on $\Omega$, with regular elementary abelian socle $N$. Now suppose we have another set $\Delta$ which $G$ also acts on transitively.

First question: Is there any way of knowing if the action of $G$ on $\Delta$ is primitive without other information?

Second Question: Can we say $N$ acts regularly on $\Delta$?

Answer 1

The answer to your first question seems to be "no". The group $G = \operatorname{AGL}(2,3)$ is a primitive group of degree $9$ and order $432$. It also acts faithfully and transitively on the cosets of its Sylow $2$-subgroup $S$, but the action is not primitive, since $S$ is not a maximal subgroup of $G$.

Here is a Maple (17) session demonstrating this (which is how I found it in the first place):

> with( GroupTheory ):
> L := select( IsPrimitive, AllTransitiveGroups( 9 ) ):
> L := select( IsSoluble, L ): nops( L );
                               7

> G := L[ -1 ]: GroupOrder( G ); # G = AGL(2,3)
                              432

> S := SylowSubgroup( 2, G ): GroupOrder( S );
                               16

> GroupOrder( Core( S, G ) ); # action is faithful
                               1

> M := convert( SubgroupLattice( G ), 'maximalsubgroups' ):
> {op}( map( GroupOrder, M ) ); # maximal subgroups are all larger than S
                      {48, 108, 144, 216}

(The last line shows that the maximal subgroups of $G$ all have order strictly greater than the order of $S$, so $S$ is not itself a maximal subgroup of $G$.)

This also shows that the answer to your second question is "no", since an Abelian transitive group is regular, and the degree here is the index $[G:S] = 27$, while the socle has order $9$.

Answer 2

• You may have misstated your question as the first is almost always false (and so then is the second).

Take $\Delta=G$ to have the regular action. Unless $G=N$ has order $p$, this action on $\Delta$ is never primitive. Unless $G=N$, the action of $N$ on $\Delta$ is not transitive, so I would not call it regular, though it is called quasi-regular.

• Here is a related sufficient condition for the action to be primitive:

If $N$ is transitive on $\Delta$, then $G$ acts primitively on $\Delta$ (Wielandt's exercise 8.8 with some from chapter 11 to make this version of the claim more obvious).

• Here is an answer to the second question, more in line with the action on $\Delta$ is primitive:

A group (like $G$) that acts faithfully and primitively (like on $\Omega$) has at most one solvable minimal normal subgroup and every regular normal subgroup is a minimal normal subgroup. If $G$ acted faithfully and primitively on $\Delta$ as well, then $N$ is still a minimal normal subgroup, so still acts regularly (Wielandt's 11.4 and 11.5).

If $G$ is solvable, then it has at most one faithful, primitive permutation action up to isomorphism. Even if $G$ is not solvable, if $G$ has a nonidentity solvable normal subgroup $N$ and if $G$ acts faithfully and primitively on $\Omega$, then in any faithful primitive action of $G$, $N$ acts regularly and $N$ is the unique minimal normal subgroup of $G$.

• Wielandt, Helmut. Finite permutation groups. Academic Press, New York-London (1964). MR183775