Transitive subgroups of a product Given a finite group $G$ acting transitively on a finite set $X$, we can talk about transitive subgroups of $G$, i.e. subgroups of $G$ that also act transitively on $X$.
Now suppose $G=H_1\times H_2$ is a direct product of groups, and assume that $H_2$ acts freely on $X$, so that all its orbits have the same size. Clearly $H_1$ acts transitively on the orbit space $X/H_2$.
Is it true that the transitive subgroups of $G$ are exactly those of the form $K\times H_2$ for $K$ a transitive subgroup of $H_1$?
If not, can they be described in general?
 A: I haven't thought about the general question, but the answer to your main question is no in general.
Let $H=H_1=H_2$ be any group with trivial centre, such as $S_3$.
Now consider the action of $G = H_1 \times H_2$ on the set $X$ of cosets of a diagonal subgroup $\{(h,h) : h \in H \}$ of $G$.
Then both $H_1$ and $H_2$ act regularly (i.e. freely an transitively) on $X$, and of course $H_1$ does not contain $H_2$.
These examples can be viewed as the group generated by the left and right regular representation of the group $H$, which centralize each other and, in general, intersect  in  $Z(H)$.
A: Following my comment in Derek's answer, the transitive groups are of the form $K×H_2$ if the action of $H_1$ on $X/H_2$ is faithful.
Let $H$ be a transitive subgroup of $G$. Then, for any $x,y\in X/H_2$, we may choose representatives $\bar{x},\bar{y}$ of the orbits in $X$ and an element $(h_1,h_2)$ of $H$ such that
$$(h_1,h_2)(\bar{x})=\bar{y}$$
and hence
$$h_1(x)=y$$
and so $H$ acts transitively on the orbit space.
Now, the action of $G$ on $X/H_2$ factors through $H_1/Stab(X/H_2)$, where $Stab(X/H_2)$ is the subgroup of elements of $H_1$ acting trivially on the orbit space.
Hence, so does the action of $H$. In particular, we have projections $H\to H_1$ (projecting onto the first factor) and $H_1\to H_1/Stab(X/H_2)$, both of which are compatible with the actions on the orbit space, and the composition determines the action of $H$ on the orbit space. The image of $H$ under the composition must be a transitive subgroup of $H_1/Stab(X/H_2)$.
Now, if the action of $H_1$ is faithful, then $Stab(X/H_2)=\{1\}$ and so the projection of $H$ onto $H_1$ must be a transitive subgroup $K$ of $H_1$.
Thus, $H$ must be contained in $K\times H_2$. This does not ensure equality, but note that, if $H$ were a strict subgroup, then there would exist an $x\in X$ and distinct $k_1, k_2\in K$ such that $k_1k_2^{-1}x$ lies in the $H_2$ orbit of $X$. If the action of $G$ on $x$ is free, then this gives $k_1=k_2$. I'm not sure if freedom can be weakened
