The property of conjugate subgroup

Question

$\DeclareMathOperator\Syl{Syl}$$G$ is a finite group. $N\unlhd G$, $P\in \Syl_{p}(G)$, and $M$ is a proper subgroup of $P$. Suppose that for any $h\in N_G(P)$, if $MM^h=M^hM$, then $M=M^h$.

For any $g\in N_G(P)N$, if $(MN)(MN)^g=(MN)^g(MN)$, can we get that $MN=(MN)^g$ ?

Answer 1

Yes we can. The question is deceptive, because even weaker hypotheses actually imply a much stronger conclusion: that $M$ is normal in $N_G(P)$. The argument is valid for any nilpotent subgroup $P$ of $G$ and any subgroup $M\le P$ such that for any $h\in N_G(P)$, $MM^h=M^hM$ implies $M=M^h$.

Proof: First, let $R=N_P(N_P(M))$. Now $M\le N_P(M)$. Therefore for any $h\in R$, $M^h\le N_P(M)$, so $MM^h=M^hM$. Hence by assumption $h\in N_P(M)$. Thus $N_P(M)=R$ is its own normalizer in $P$, whence $N_P(M)=P$ by nilpotence of $P$. Hence, $M$ is normal in $P$. Finally, for any $h\in N_G(P)$, $M^h\le P$ so again $MM^h=M^hM$, whence $h\in N_G(M)$. Thus, $M$ is normal in $N_G(P)$.

It is then elementary that $MN$ is normal in $N_G(P)N$, which implies your conclusion.

Answer 2

Richard Lyons gave a proof of a more general statement, but I mention an approach to the $p$-group case asked in the question which is (I think) essentially the way Sylow proved (the existence part of) his theorem (given Cauchy's theorem). If $G$ is a finite group, $p$ is a prime, and $Q$ is a non-trivial $p$-subgroup of $G$ with $[G:N_{G}(Q)]$ divisible by $p$, then there is some $g \in G \setminus N_{G}(Q)$ such that $gQg^{-1} \leq N_{G}(Q)$. For if we let $Q$ act by right translation on the right cosets of $N_{G}(Q)$ in $G$, then (since, by hypothesis, the number of such cosets is divisible by $p$, while $Q$ fixes the coset $N_{G}(Q)$) there must be another coset $N_{G}(Q)g$ which is fixed by $Q$: hence $g \notin N_{G}(Q)$ but $N_{G}(Q)gQ = N_{G}(Q)g$, which means that $gQg^{-1} \leq N_{G}(Q)$.