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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. – Tara B Apr 18 '12 at 16:03
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 $N$. – André Caldas Apr 18 '12 at 16:19
Actually H is fixed. – user28083 Apr 18 '12 at 16:45
@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. – Arturo Magidin Apr 18 '12 at 16:58
@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! – Arturo Magidin Apr 18 '12 at 19:47

The condition you write makes no sense. Just take $H=G$ to get the 'conclusion' that all groups are solvable.

Moreover, the kinds of conditions that are amenable to "minimal criminal" proofs are those in which the condition in question is intrinsic to the group. You would want something along the lines of: "If there exists a proper subgroup $H$ of $G$ such that if $P$ and $Q$ are $p$-Sylow subgroups of $G$, then there exists $x\in H$ such that $P^x = Q$, then $G$ is solvable."

(That is, the condition is given in terms of something that the group $G$ satisfies within itself, not in relation to some "other" group). The condition you give, with $H$ "fixed", is not amenable to minimal counterexample proofs. So, not only is the given condition nonsensical, any attempt to make it fit into the kind of proof you are talking about is doomed.

If you have an actual example, I suggest using that, rather than trying to make something up yourself. Right now, the answer is "You can't, the entire thing is false as written."

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