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In Theorem 3.1 in this paper, part (1) of the proof, $X$ may not be a subgroup of $N$, so how did the authors apply the induction method?

Also, I do not understand this: In the proof of Theorem 3.2 he applied Theorem 3.1 with the possibility that $X$ may not be a subgroup of $H$.

This is a link for the paper in reference [3] which has the proof of Lemma 2.1.

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    Could you please clarify what you mean by "If $G$ satisfies $H$-some statement". Asaf wants me to elaborate on my answer, and I'll be able to do that better if I fully understand your question.2012-04-18
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    I guess $H$ is not fixed. You want to show $G$ is such that for "any" $H$ of some sort (normal in $G$), the $H$-statement is true. Then, regarding to $N$, you know that for any $H$ (normal subgroup of $N$), the $H$-statement is true for $N$.2012-04-18
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    Actually H is fixed.2012-04-18
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    @user28083: It really depends on the nature of the "$H$-statement" in question. Could you provide the specifics? Usually in this kinds of "minimal criminal" proofs, one uses the fact that $G$ satisfies the property to show that the proper normal subgroup also satisfies the property (often $H$ itself!), then pass to the quotient (which by the minimality of $G$ satisfies the property) and lift back. But the exact mechanics depend on the exact nature of the condition.2012-04-18
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    So... this questions turns out not to be about "induction methods", but about understanding a particular line in a particular argument in a particular research paper... And connected to your [previous question](http://math.stackexchange.com/q/132648/742) and random-seeming comment there. Why *hide* all of this information from your question?!2012-04-18
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    I did not want to trouble the people by the entire thing, as you can see the paper is long and confusing. I tried to simplify my question, but it did not work.2012-04-18
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    @user28083: You want to provide *context*, precisely to avoid this sort of thing. Note that when one is confused about an argument, it is often the case that one cannot tell what *is* important information and what is not. After all, if you were entirely clear on just what information you will need and which information you will not, chances are you are not actually confused about the argument!2012-04-18
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    It seems clear that the intention is to claim that $N$ satisfies the hypothesis of the theorem relative to $\mathcal{H}=H\cap N$ and $\mathcal{X}=X\cap N$. The normality, solvability, and supersolvability conditions follow easily, the only nonobvious part is to show that you can find $x\in\mathcal{X}$. It is clear you can find $x\in X$, and it is also easy to show you can find $y\in N$ that works, but I see no immediate reason why you can assume you can find $x\in\mathcal{X}$. I'll think about it.2012-04-18
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    In theorem 3.2. proof in the last page, he writes that all maximal subgroups of the Sylow subgroups $F^{*}(H)$ are $X-s-$permutable in H. So, he did not use $X \cap H$2012-04-18
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    @user28083: Different theorem, different argument (the proof of Theorem 3.2 does **not** proceed by contradiction with a minimal counterexample), so I see no reason to assume that *they* would be trying to do the same argument.2012-04-18
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    I meant that he applied theorem 3.1. with the possibility that $X$ may not be a subgroup of $H$ in the proof of theorem 3.2.2012-04-18
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    @user28083: Stop. Go back to Lemma 2.1.1. No hypothesis require $X\subseteq N$ in order to say "$H$ is $X$-s-permutable in $N$". I would suggest looking up the proof of that statement in the reference, since it is clear that they are applying **that** Lemma in both instances.2012-04-18
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    I know that lemma 2.1.1. does not require $X \subseteq N$, but theorem 3.1. requires it2012-04-18
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    @user28083: (i) if you don't "ping" me, I don't get notifications of your comments. (ii) You miss my point. The *definition* of "$X$-s-permutable in $G$" assumes $X\subseteq G$. If in Lemma 2.1.1 they are saying "$X$-s-permutable in $N$" without assuming $X\subseteq N$, then there is something at play which is not obvious from the definition, and that will **no doubt** have a bearing on the argument on Theorem 3.12012-04-18
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    This is my proof for Lemma2.1.1. Since $H$ is $X-s-$permutable in G, then for $P$ Sylow of $G$ there exists $x \in X$ such that $P^{x}H=HP^{x}$. The Sylow of $N$ are of the form $P \cap N$. Thus,$(P\cap N)^{x}H=H(P\cap N)^{x}$. Hence, H is $X-s-$permutable in N. This is only need that $H \subseteq N$ Sorry, but I do not have the paper in reference [3]2012-04-18
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    Maybe someone could edit the title and question so as to accurately reflect the actual situation?2012-04-20
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    @Gerry: That would be lovely. It's quite a mess at the moment.2012-04-20
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    I don't know how to clarify it more than this.2012-04-20
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    @user28083: You need to write `@Arturo` if you want me to be notified of your responses. I almost missed your comment here.2012-04-20
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    @user28083: Look at the definition of $X$-s-permutable in $G$. The **assumes** $X\subseteq G$. Your proof does **not** assume that the $x$ you get is contained in $N$, which means it does not satisfy the definition *as literally given*. That is my point. To satisfy the definition as literally given, you need $x\in X\cap N$. If the authors are not proving this, then Lemma 2.1.1 is insufficient by itself for their purposes in Them 3.1. (Don't ignore the possibility that the argument is *wrong*! To be frank, I've seen a *lot* of shoddy papers with incorrect arguments come from China lately)2012-04-20
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    @Arturo: I get your point about the definition and my proof. I found the paper [3] and I want to put a link for it. How to put a link in a comment? I forgot one thing, THANK YOU for trying so hard with me.2012-04-20
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    I posted the link for the paper in [3] within the question.2012-04-20
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    @ArturoMagidin : I posted the link for the paper in [3] within the question.2012-04-20
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    @user28083: Well, there you go. They claim Lemma 2.1 in that reference as authority for their Lemma 2.1.1. *But their reference does **not** prove their stated result.* (All conclusions in their reference are for $X$-s-permutability in either the original $G$, or in a quotient of $G$, never in a subgroup). If I were refereeing this, I would ask for clarification; I suggest contacting the authors and asking, pointing out that their reference does not justify their claim.2012-04-20
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    @ArturoMagidin: I was also surprised when I did not found the proof of the Lemma as the claimed. I will try to send them an email and see what happens. Thank very very much.2012-04-20
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    @ArturoMagidin: I contacted one of the authors and he told me that there was a mistake. He just altered his definition to make things work. I do not know if there are more things that need to be fixed. I just wrote this comment to let you know. Thank you very much.2012-05-05
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    @user28083: Thanks; good to know.2012-05-05

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