If $K$ is a normal subgroup of $N$, prove that $K$ is normal in $G$.

Question

Let $K$ be a Sylow p-subgroup of a finite group $G$ and $N$ be a normal subgroup of $G$. If $K$ is a normal subgroup of $N$, prove that $K$ is normal in $G$.

I'm trying to solve this problem and I'm pretty new to Sylow Theorem. So, I tried to use the second Slow Theorem to show that there exists $x\in G$ such that $N=x^{-1} K x$. Since $N$ is normal, $N=x^{-1} K x=K$...

But that doesn't get me anywhere and I think I'm on the completely different route.

I'm just trying to learn this Sylow Theorems and it would be wonderful if anyone can prove the statement for me.

Answer 1

One of the Sylow theorems states that all Sylow $p-$groups are conjugate, so a second $p-$ group has the form $g^{-1}Kg, g \in G$. So if $K \subseteq N$ then $g^{-1}Kg \subseteq g^{-1}Ng = N$ since $N$ is normal. Now forget $G$ for a while. $N$ now has two Sylow $p-$ subgroups, $K$ and $g^{-1}Kg$, but now applying the same Sylow theorem inside $N$ we have that these subgroups are $N-$ conjugate but since $K$ is normal in $N$ these two subgroups coincide whence $K = g^{-1}Kg$, in other words $K$ is normal in $G$.

Answer 2

Since $K$ is a normal subgroup of $N$, $K$ is the only Sylow $p$-subgroup of $N$. So for any $g\in G, gKg^{-1} \subset gNg^{-1}=N$ so $gKg^{-1}$ is a subgroup of $N$ with the same order as $K$. Therefore $gKg^{-1}=K$. Since $g$ is arbitrary in $G$, $gKg^{-1}=K$.