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.
Best Answer
Your proof is correct (I did not check what you wrote in your remark).
What you use in the last part is called Frattini's argument: