Product of Frattini Groups Let be $G, H$ groups and $\Phi(G), \Phi(G)$ their Frattini groups. I'm looking for a conterexample that shows that in general $\Phi(G \times H)= \Phi(G) \times \Phi(H)$ doesn't hold, therefore that there exist $G, H$ such that $\Phi(G) \times \Phi(H)\not \subset \Phi(G \times H)$ (inclusion in other direction is indeed always true).
 A: Have you seen the paper The Frattini subgroup of a direct product of groups?
Theorem 1 of that paper gives a necessary and sufficient condition for failure of the equality $\Phi(G \times H) = \Phi(G) \times \Phi(H)$.  Perhaps more useful are the opening remarks of the paper: that equality holds for all finite groups or solvable groups; and that the question of whether equality holds in general is equivalent to the question of whether there are simple groups without maximal subgroups.  (This was unknown at the time of the paper, see @YCor's nice answer for more on the latter.)
A: The inclusion can indeed be proper. Indeed, there exist simple groups $S$ with no maximal subgroup. So $\Phi(S)=S$. But the diagonal in $S\times S$ is a maximal subgroup, and in particular contains $\Phi(S\times S)$. Since $\Phi(S\times S)$ is normal, it follows that $\Phi(S\times S)=\{1\}$, which is properly contained in $\Phi(S)\times\Phi(S)=S\times S$.
I'm not sure what's the simplest example of such $S$, but at least one comes from Shelah's construction of a simple Jonsson group (a Jonsson group an uncountable group in which every proper subgroup is countable). (Sciencedirect link to Shelah's 1980 article)
(added: there also exists a simple countable group without maximal subgroups: Theorem 35.3 in Olshanskii's book Geometry of defining relations in groups)
