mathmari
Gold Member
MHB
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Hey! 
Let $P$ be a $p$-Sylow subgroup in $G$ and $N=N_G(P)$.
I want to show that there is no proper normal subgroup $H$ of $G$ that contains $N$.
We suppose that there is a proper normal subgroup $H$ of $G$ that contains $N$, $$N\leq H<G$$
Then $[G:N]=[G
][H:N]$, with $[G
]>1$.
How can we find a contradicion? (Wondering)
Do we use the definition of a normal subgroup? (Wondering)

Let $P$ be a $p$-Sylow subgroup in $G$ and $N=N_G(P)$.
I want to show that there is no proper normal subgroup $H$ of $G$ that contains $N$.
We suppose that there is a proper normal subgroup $H$ of $G$ that contains $N$, $$N\leq H<G$$
Then $[G:N]=[G


How can we find a contradicion? (Wondering)
Do we use the definition of a normal subgroup? (Wondering)