$K \unlhd H\unlhd G$, K is sylow in H, proof $K \unlhd G$

Question

$K \unlhd H\unlhd G$, K is sylow in H, proof $K \unlhd G$

I know in general it's false, I wonder how should I use the condition that K is sylow? is it that the K is the unique sylow subgroup? and how could that help?

Answer 1

Another slightly more elementary approach:

$K$ is the unique subgroup of $H$ with order $|K|$. Now $H$ is normal in $G$, and $K \subset H$, so $gKg^{-1}$ is a subgroup of $H$ for any $g \in G$. Since $|gKg^{-1}| = |K|$, it follows that $gKg^{-1} = K$, so $K$ is normal in $G$.

Answer 2

If $K$ is a Sylow $p$ subgroup and normal, then it is necessarily the unique Sylow $p$ subgroup. This implies that $K$ is in fact characteristic in $H$. From there it's a well known exercise to show that $K$ is normal in $G$.(*)

[* Any inner automorphism of $G$ will fix $H$, because $H$ is normal. Thus it induces an automorphism of $H$, but since $K$ is characteristic in $H$, this automorphism of $H$ must fix $K$ as well. So every inner automorphism of $G$ fixes $K$.]