induction method 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.
 A: 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."
