Question: I am trying to prove the following:
Let $G$ be a finite (non $p$-group) group such that the maximal subgroup in every Sylow $p$-subgroup $P$ of $G$ is normal in $G$ (for each prime $p$). Then either $P$ is abelian, or, $P$ has a unique maximal subgroup.
My attempt as follows:
If $P$ is abelian (non-cyclic) then the theorem holds. So, assume that $P$ is non-abelian and has at least two maximal subgroups say $M_1$ and $M_2$. Suppose that $|P|=p^k$, then $|M_1|=|M_1|=p^{k-1}$. For $M_1$ and $M_2$ we consider two cases:
1. $M_1 \cap M_2 \ne 1$. Then we let $M_1 \cap M_2 =N$ which is normal in $G$ as intersection of two normal subgroups. and using induction on $G/N$ we get the result.
2. $M_1 \cap M_2 = 1$. Then $P= M_1 M_2$. Moreover,$|P|= |M_1 | |M_2|/|M_1 \cap M_2|$. Which implies, $p^k=p^{2k-2}$. Hence, $k=2$ and $P$ is abelian. Cotradiction.
I was wondering if I missed some thing! It is helpful to remind that groups with the prescribed property is supersolvable. Please advice. Thank you!.