There is no proper subgroup

[mathmari]

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)

 

[mathmari]

Do we maybe use Frattini Argument? (Wondering)

By Frattini Argument we have that $G=HN_G(P)$.

Since $H$ is a normal subgroup of $H$ and since $N_G(P)\leq H$, we have that $HN_G(P)\subseteq H \Rightarrow G\subseteq H$.

We have that $H\subseteq G$.

So, it holds that $G=H$. This is a contradiction, since $H$ is a proper subgroup of $G$. Is this correct? (Wondering)