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

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.

• Hint: Since $N$ is a normal subgroup of $G$, that means that $g^{-1}Ng = N$ for all $g\in G$. In particular, this means that $g^{-1}Kg\subseteq N$ for all $g\in G$. So $g^{-1}Kg$ is a subgroup of $N$--what is its size, and what do the Sylow theorems say about subgroups of $N$ of that size? – Joey Zou Apr 7 '17 at 4:03
• @JoeyZou I guess this indicates that $N = K$, which further implies that K is the unique subgroup of G? But that doesn't really get me anywhere with the initial condition that $K$ is a normal subgroup of $N$. – Ya G Apr 7 '17 at 4:22
• No, it doesn't imply that $N=K$. Again, $g^{-1}Kg$ is a subgroup of $N$--if you know the size of $K$, do you know the size of $g^{-1}Kg$? – Joey Zou Apr 7 '17 at 23:31
• @JoeyZou I don't think so... I'm very new to this... – Ya G Apr 8 '17 at 0:09

## 2 Answers

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 $G$.

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 $G$, $gKg^{-1}=K$.