Timeline for Finite nonabelian groups with abelian maximal subgroups

6 events

when toggle format	what		by	license	comment
Aug 9, 2017 at 8:15	vote	accept	H.Shahsavari
Aug 9, 2017 at 7:57	answer	added	Derek Holt		timeline score: 4
Aug 9, 2017 at 4:44	comment	added	H.Shahsavari		The above two comments are what I got about this question and I am not sure about them.
Aug 9, 2017 at 4:41	comment	added	H.Shahsavari		then $LR$ is a proper subgroup $G$ containing properly $L$, a contradiction. So $R\ntrianglelefteq G$. Since $R\leqslant N\trianglelefteq G$, then all Sylow $r$-subgroups of $G$ are contained in $N$. Therefore number of Sylow $r$-subgroups of $G$ is equal to those that are Sylow $r$-subgroups of $N$ too. Thus we have $[G:N_{G}(R)]=[N:N_{N}(R)]$. But this is impossible since $p$ divides the left side of the equation while obviously it does not divide the right side.
Aug 9, 2017 at 4:27	comment	added	H.Shahsavari		It is clear that every Sylow $p$-subgroup of $G$ is of order $p$. Otherwise the subgroup $L$ is contained properly in a Sylow $p$-subgroup of $G$ which is obviously not equal to $M$, a contradiction. So $N_{G}(L)=M$ and by by Burnside's normal $p$-complement theorem we see that there exists a proper normal subgroup of $G$ say $N$ such that $(\vert N\vert,p)=1$. We claim that $N$ is a $q$-group. Obviously $q\mid\vert N\vert$. Contrary suppose that $r$ is a prime number other than $q$ such that $r\mid\vert N\vert$ and let $R$ be a Sylow $r$-subgroup of $G$. If $R\trianglelefteq G$, then $LR$ ...
Aug 9, 2017 at 4:22	history	asked	H.Shahsavari	CC BY-SA 3.0