Third isomorphism theorem

Statement

Let $G$ be a group, and let $N$ and $K$ be normal subgroups of $G$ with $N \subseteq K$. Then $K/N$ is a normal subgroup of $G/N$, and the map $(gN)(K/N) \mapsto gK$ is an isomorphism from $(G/N)/(K/N)$ to $G/K$.

Remarks

The proof applies the first isomorphism theorem to the surjective homomorphism $G/N \to G/K$, $gN \mapsto gK$, whose kernel is $K/N$. For example, $(\mathbb{Z}/12\mathbb{Z})/(4\mathbb{Z}/12\mathbb{Z}) \cong \mathbb{Z}/4\mathbb{Z}$. For finite groups, it gives $[G : K] = [G/N : K/N]$.

Explanations

L1 human

If you divide by $N$ and then by the image of $K$, you get the same result as dividing by $K$ in one step. It looks like cancelling fractions: $(G/N)/(K/N) = G/K$. The theorem makes this cancellation precise.