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).
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
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)
$\begingroup$
$\endgroup$
4
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.)
answered Oct 20, 2017 at 21:40
-
$\begingroup$ Just to clarify the latter sentence: the linked paper is from 1960 and it was unknown at that time (at least to the authors) whether every simple group has a maximal subgroup. $\endgroup$– YCorCommented Oct 20, 2017 at 22:27
-
1$\begingroup$ So of course it can look obsolete to state this equivalence since simple groups without maximal subgroups are now known to exist. But actually their Theorem 3 is more precise, and implies: let $\mathcal{C}$ be a class of groups stable under taking quotients, and in which every simple group has a maximal subgroup (examples: the class of finitely generated groups or the class of solvable groups); then $\Phi(G\times H)=\Phi(G)\times\Phi(H)$ for all $G,H\in\mathcal{C}$. $\endgroup$– YCorCommented Oct 20, 2017 at 22:35
-
$\begingroup$ @YCor, thanks for clarifying dates. I edited slightly to reflect. Theorem 3 is certainly easier to apply than Theorem 1. $\endgroup$ Commented Oct 20, 2017 at 23:03
-
1$\begingroup$ The reference is: Dlab, Vlastimil ; Kořínek, Vladimír The Frattini subgroup of a direct product of groups. (English). Czechoslovak Mathematical Journal, vol. 10 (1960), issue 3, pp. 350-358, link dml.cz/handle/10338.dmlcz/100419 $\endgroup$– YCorCommented Oct 29, 2017 at 23:01