Suppose that $G$ is a finite minimal non-solvable group (i.e. a non-solvable group all of whose proper subgroups are solvable) whose Frattini subgroup $\Phi(G)$ is abelian. If for every subgroup $H < G$, $H'$ is core-free, then is it true that $G$ is simple (or equivalently $\Phi(G)$ is trivial)?
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1$\begingroup$ It might be helpful if you supplied some motivation for questions like this, and it might also lead to more people thinking about the problem. At first sight it looks like "Does this variegated collection of conditions on a group imply this conclusion?". Why do you think it might do? $\endgroup$– Derek HoltCommented Apr 30, 2022 at 12:20
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2$\begingroup$ @DerekHolt Maybe you are right as far as it concerns providing motivation -- but in my question I have just a condition on derived subgroups of a minimal non-solvable group whose Frattini subgroup is abelian. If this is a "variegated collection of conditions", then I think likely most questions on this site have a "variegated collection of conditions". GAP calculations suggest that the answer to the question may be positive. Besides -- haven't you asked yourself some time why so few female mathematicians contribute to this site under their real name? $\endgroup$– Leyli JafariCommented Apr 30, 2022 at 14:54
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7$\begingroup$ No I am afraid that I haven't asked myself that question. But I can honestly say that I hadn't given any thought at all (even subconscious) to your gender and I also have to admit that I have no idea whether "Leyli" is normally a male or a female name, or whether (like many Anglo-Saxon forenames) it could be either. Anyway, I will think about your question! $\endgroup$– Derek HoltCommented Apr 30, 2022 at 15:40
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1$\begingroup$ Here's why I asked. Notation: $\bar G=G/\Phi(G)$. Also $\hat G=G/N$, the unique nonabelian simple quotient of $G$. Then $\Phi(G)\le N$ so $\hat G$ is a quotient of $\bar G$. Putting no requirement on subgroups $H$ such that $H'\le \Phi(G)$ does not really strengthen the hypothesis. For if $\bar H$ is abelian, then $\hat H$ is abelian and therefore (by transfer results) $\hat H$ can't be maximal in $\hat G$; hence, $H<M<G$ for some $M$ such that $\hat M$ is not abelian. Therefore $\bar M$ is not abelian, so by your hypothesis $M'$ is core-free. A fortiori, $H'$ is then core-free as well anyway. $\endgroup$– Richard LyonsCommented Apr 30, 2022 at 23:02
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2$\begingroup$ It seems to me that you are looking for an example as in your previous question but with more conditions. The simple quotient group needs to be minimal simple (so the example $J_1$ from the previous question will not work), and the extension needs to be non-split. This means that it will be difficult to find examples, but it does not necessarily mean that there are no examples. $\endgroup$– Derek HoltCommented May 1, 2022 at 7:47
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