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.

  • $\begingroup$ 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? $\endgroup$ – Joey Zou Apr 7 '17 at 4:03
  • $\begingroup$ @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 $N$. $\endgroup$ – Ya G Apr 7 '17 at 4:22
  • $\begingroup$ 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 $g^{-1}Kg$? $\endgroup$ – Joey Zou Apr 7 '17 at 23:31
  • $\begingroup$ @JoeyZou I don't think so... I'm very new to this... $\endgroup$ – Ya G Apr 8 '17 at 0:09

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.