Hi everybody,

I am dealing with the O' Nan Scott theorem, the classification of finite primitive groups (I am reading "Classes of Finite Groups", by Adolfo Ballester-Bolinches and Luis M. Ezquerro). Here for "primitive group" I mean a finite group $G$ endowed with a core-free maximal subgroup $U$. I say that a primitive group $(G,U)$ is monolithic if it has only one minimal normal subgroup, $N$, which is of the type $S^n$ for some simple group $S$ and positive integer $n$. I know that a monolithic primitive group can have two types of action: diagonal action or product action, and this is recognizable looking at the intersection of $U$ with $N$ (because $U$ turns out to be the normalizer in $G$ of $U \cap N$). This intersection can be either the direct product of some conjugates of a maximal subgroup of $S$ or a direct product of diagonals (a diagonal of $S^m$ is a subgroup containing the elements of the form $(x,x^{\alpha_2},...,x^{\alpha_m})$ for some $\alpha_2,...,\alpha_m \in Aut(S)$).

Now my question is: what can I say about the primitive groups $G$ for which every maximal core-free subgroup $U$ is of product type (i.e. $(G,U)$ has product action)?

I have an example of such a group: take the semidirect product $(A_5 \times A_5) \langle (1,\tau) \varepsilon \rangle$, where $\tau=(12)$ and $\varepsilon$ is the exchange of the two coordinates (I think of it as a subgroup of $Aut(A_5 \times A_5) = S_5 \wr \langle \varepsilon \rangle$): the action is $(x,y)^{(1,\tau)\varepsilon} = (y^{\tau},x)$. Here every subgroup of diagonal type is contained in the maximal subgroup $(A_5 \times A_5)\langle(\tau,\tau)\rangle$.

I looked for such groups in some article (by Kovacs, "Primitive permutation groups of simple diagonal type" and "Primitive subgroups of wreath products in product action"), but I didn't find a precise answer.