Let $G$ be a group of order $120$, let $H≤G$ with $|H|=40$. Prove that there exists $K$ such that $K\unlhd G$, $K≤H$, and $|K|≥20$.

Question

Let $G$ be a group of order $120$, let $H≤G$ with $|H|=40$. Prove that there exists $K$ such that $K\unlhd G$, $K≤H$, and $|K|≥20$.

I think this is associated with the action of the left coset of $H$. Maybe the kernel. But I don't know why the kernel of this action commutes with every element of $G$

Answer 1

Let $C$ be the set of left cosets of $H$, and consider the action of $G$ on $C$ by left multiplication, that is $g_1 * (g_2H) = (g_1g_2)H$. This induces a group homomorphism $\phi : G \rightarrow S_C$. Let $K = \ker \phi$. Observe that $K \leq H$ and $K \lhd G$. Since $|G| = 3|H|$ and $G/\ker \phi \simeq \phi(G) \leq S_C \simeq S_3$, we find $\dfrac{3|H|}{|\ker\phi|} = |\phi(G)|$ divides $6$, so $\dfrac{|H|}{|\ker\phi|}$ is either 1 or 2. Regardless, $|K| \geq 20$.