Prove that a finite group $G$ is nilpotent if and only if every maximal proper subgroup of $G$ is normal.
Denote the normalizer of $H$ to be the group $N_{G}(H)=\{a\in G:aHa^{-1}=H\,\},$where $H\le G.$
Here's my working :
Assume $G$ is nilpotent and fix an maximal proper subgroup $M$ of $G$.Then as $G$ is nilpotent, one has $M\lneqq N_{G}(M).$By the maximality of $M$, we have $N_{G}(M)=G.$Therefore,$\,M$ is normal in $G$.
Conversely, we suppose to the contrary that $G$ is not a nilpotent. So,in other words, there exists a Sylow $p$-subgroup in G,call $\widehat{P},$ such that $N_{G}(\widehat{P})\ne G.$Thus there is some maximal proper subgroup $K$ in $G$ that $N_{G}(\widehat{P})\le K$ and $K\unlhd G$ by hypothesis.
Without loss of generality that we may assume the maximal $K\ne G.$
But $\widehat{P}\le N_{G}(\widehat{P})\le K\,.$ Then we conclude that $G=KN_{G}(\widehat{P})=N_{G}(\widehat{P})K=K$, a contradiction and the last equality holds by $N_{G}(\widehat{P})\le K.$
Remark:
here we use the two statements that $G$ is nilpotent iff each sylow $p-$subgroup in G is normal and a subgroup $T$ in $G$ is normal iff $~~N_{G}(T)=G.$
Now we claim $G=KN_{G}(\widehat{P})=N_{G}(\widehat{P})K$ under assumptions $\widehat{P}\le K\,$ and $K\unlhd G:$
Note that for each $a\in G$, $$a\widehat{P}a^{-1}\le aKa^{-1}=K~~~~and~~~~~~|a\widehat{P}a^{-1}|=|\widehat{P}|.$$ So,$~~a\widehat{P}a^{-1}$ is also a sylow $p-$subgroup of $K.$By Sylow theorem $\widehat{P}$ and $a\widehat{P}a^{-1}$ are conjugate, thus there is some $k\in K$ that $k\widehat{P}k^{-1}=~a\widehat{P}a^{-1}.$
So, $$\widehat{P}=k^{-1}a\widehat{P}a^{-1}k=k^{-1}a\widehat{P}(k^{-1}a)^{-1},$$then, $$k^{-1}a \in N_{G}(\widehat{P})\Longrightarrow a=k(k^{-1}a)\in KN_{G}(\widehat{P})=N_{G}(\widehat{P})K$$
,where the last equality holds by $k\unlhd G.$
Is there anybody checking my proof for validity ? Any advice or comment will be the greatest appreciated. Thanks for patient reading and considering my request.
