I’m reading an old paper, which is really short. I got totally stuck why it is claimed that $N_G(H)\subseteq N^y$.
Theorem.$~~$If the finite group $G$ has exactly $2$ conjugacy classes of maximal subgroups, then $G=PQ$ where $P$ and $Q$ are Sylow $p$-subgroup and Sylow $q$-subgroup of $G$, $P\triangleleft Q$ and $Q$ is cyclic. Further, $Q$ acts irreducibly on $P/\Phi(G)$.
Proof. Let $G$ be a minimal counterexample to the theorem. By the remark following lemma 4 [1], it suffices to show that $G$ is not simple. Let $M$ and $N$ be non-conjugate maximal subgroups of $G$. For $g\in G-M$, let $p\in\pi(M\cap M^g)$ be a prime. By lemma 5 [1], $p\in\pi(M\cap N)$. Now choose $x\in G-M$ such that the Sylow $p$-subgroup $H$ of $M\cap M^x$ has maximal order. By lemma 1 [1], the simplicity of $G$ and the maximality of $H$, one has ${\color{red}{ N_G(H)\subseteq N^y}}$ for some $y\in G$. We conclude that $|M\cap M^g|\le|M\cap N|$ for all $g\in G-M$. Similarly $|N\cap N^z|\le |M\cap N|$ for all $z\in G-N$. It is clear now that either $|MM^g|>|G|$ or $|NN^z|>G$, which is impossible.
I know we can get $N_G(H)\subseteq N^y$ if $H$ is a Sylow subgroup of $G$. Since $G$ is simple, $N_G(H)$ is a proper subgroup of $G$ and is hence contained in a maximal subgroup of $G$. If $N_G(H)\subseteq M^a$ for some $a\in G$, then by Lemma 1 we have $M=M^a=M^x$, which is a contradiction to the assumption that $x\in G-M$. Thus we must have $N_G(H)\subseteq N^y$ for some $y\in G$.
I’m really confused about how to get $N_G(H)\subseteq N^y$ when $H$ is no longer a Sylow subgroup of $G$ but only a Sylow subgroup of $M\cap M^x$. I don’t seem to use the maximality in my argument above when $H\in {\rm Syl}_p(G)$. But I really don’t know how to use that.
Could you give me some help? I got stuck here for quite a long time. Any help is appreciated. Thanks!
Notice that what you are really trying to show that is that $N_G(H)$ is not contained in a conjugate of $M$.
Suppose that $H$ is not a Sylow $p$-subgroup of $G$. Notice that the $p$-part of $N_G(H)$ is larger than that of $H$. By assumption, $N_G(H)$ is not contained in a conjugate of $N$, therefore it must be contained in a conjugate $M^y$ of $M$. Also, it cannot be contained in two different conjugates of $M$, as then it would lie in their intersection, which would have larger $p$-part. But now $H$ is not a Sylow $p$-subgroup of $M$ either, since the $p$-parts of $M$ and $M^y$ are the same. Thus $M$ contains a $p$-subgroup $L$ normalizing $H$. This is, by definition, also in $M^y$, a contradiction.
Thus $H$ is a Sylow $p$-subgroup of $G$. Suppose that $N_G(H)$ lies in a conjugate $M^y$ of $M$. Then $H\leq M\cap M^y$ and $N_G(H)^{y^{-1}}\leq M$. There exists $m\in M$ such that $H^{y^{-1}m}=H$ (as both are Sylows of $M$) and so $N_G(H)^{y^{-1}m}=N_G(H)$. Thus $y^{-1}m$ normalizes $N_G(H)$, thus $y^{-1}m\in N_G(H)\leq M$ (as $N_G(N_G(H))=N_G(H)$). Thus $y\in M$, and $M=M^y$.
Thus $N_G(H)$ lies in a conjugate of $N$, as needed.