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S May 21, 2022 at 14:06 history bounty ended CommunityBot
S May 21, 2022 at 14:06 history notice removed CommunityBot
S May 13, 2022 at 12:11 history bounty started Leyli Jafari
S May 13, 2022 at 12:11 history notice added Leyli Jafari Draw attention
May 1, 2022 at 14:45 comment added Leyli Jafari @GeoffRobinson I think no difference between proving each of the equivalent conditions. Anyway, I added the equivalent conclusion in the question. Thank you!
May 1, 2022 at 14:44 history edited Leyli Jafari CC BY-SA 4.0
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May 1, 2022 at 14:15 comment added Geoff Robinson Well, OK , I thought that might be what you intended to ask, but why not just ask directly what you really want?
May 1, 2022 at 14:08 comment added Leyli Jafari @GeoffRobinson Actually what we aim to prove is that with the condition on the derived subgroups, G is simple, which is equivalent to prove that $\Phi (G)=1$.
May 1, 2022 at 12:27 comment added Geoff Robinson As the question is now worded, we either have that $G$ is simple ( so there is nothing to do), or else $G$ has a non-trivial Abelian minimal normal subgroup $M$ which is necessarily contained in $\Phi(G)$, as per @RichardLyons comment, so we definitely have $\Phi(G) \neq 1.$
May 1, 2022 at 7:47 comment added Derek Holt 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.
May 1, 2022 at 2:20 comment added Leyli Jafari @RichardLyons You are right. Actually since G is minimal non-solvable, so $G / \Phi(G)$ is minimal simple. I edited the question, thank you!
May 1, 2022 at 2:19 history edited Leyli Jafari CC BY-SA 4.0
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Apr 30, 2022 at 23:02 comment added Richard Lyons 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.
Apr 30, 2022 at 20:27 comment added Leyli Jafari @RichardLyons No, I mean $H' \leq \Phi(G)$.
Apr 30, 2022 at 20:14 comment added Richard Lyons Do you mean $H'\geq \Phi(G)$?
Apr 30, 2022 at 15:40 comment added Derek Holt 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!
Apr 30, 2022 at 14:54 comment added Leyli Jafari @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?
Apr 30, 2022 at 12:20 comment added Derek Holt 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?
Apr 30, 2022 at 11:03 history asked Leyli Jafari CC BY-SA 4.0