A theorem from the theory of groups

Let $H< K$ with $H\lhd G$. Then $G/H$ acts on $(G/H)/(K/H)$ by left multiplication. From the isomorphism theorems $(G/H)/(K/H)\approx G/K$ and the original action factors as $$\mathbb L\colon G\to G/H\to \operatorname{Aut}_{\mathbf{Set}}(G/K)$$ (because we talk about the same multiplication after all). Especially, $H\subseteq \ker\mathbb L$. Hence if $K$ contains a normal subgroup of $G$, then $\mathbb L$ is not a monomorphism.

On the other hand, let $H=\ker\mathbb L$. Then $H<K$ because we need $hK=K$ for all $h\in H$. And as kernel of a homomorphism, clearly $H\lhd G$.