Primitive action of direct product

Question

The question could be simple, but this came to my mind when I saw an analogue of this in linear representations of group by Aschbacher.

Let $G_i$ acts on $\Omega_i$ for $i=1,2$. Consider then the induced action of $G_1\times G_2$ on $\Omega_1\times \Omega_2$. Is the following statement true?

$G_1\times G_2$ is primitive on $\Omega_1\times \Omega_2$ if and only if each $G_i$ is primitive on $\Omega_i$.

(An analogue of this is true for linear $\mathbb{C}$-representations, where product of sets is replaced by tensor product of vector spaces on which the groups act.)

Answer 1

Being primitive is equivalent to having no (proper, nontrivial) homomorphic images under equivariant maps. (I think this is a good exercise, and shows how primitivity in the category of $G$-sets is like being a group being simple in the category of groups.)

If $|\Omega_1|,|\Omega_2|>1$ then $\Omega_1\times\Omega_2$ can never be a primitive $G_1\times G_2$-set, because there is a homomorphism of $G_1\times G_2$-sets given by $\Omega_1\times\Omega_2\to\Omega_1$ (where $G_2$ acts trivially).

In terms of partitions, this means $\{\{\omega\}\times\Omega_2:\omega\in\Omega_1\}$ would be a nontrivial $G_1\times G_2$-stable partition of $\Omega_1\times\Omega_2$.

Answer 2

This is not true. Take $G_1 = Sym(3), \Omega_1 = \{1,2,3\}, G_2 = Sym(4)$ and $\Omega_2 = \{1,2,3,4\}$. Then $G_1$ and $G_2$ are solvable and so is $G_1 \times G_2$, but $|\Omega_1 \times \Omega_2| = 2^2 * 3$ is not a prime power. So your implication "$\Leftarrow$" cannot hold.

I haven't thought about the other direction since my time is limited right now but maybe I will add something later on.