We'll give a proof of the third isomorphism theorem using the Fundamental homomorphism theorem.

Let $G$ be a group, and let $K\subseteq H$ be normal subgroups of $G$ . Define $p,q$ to be the natural homomorphisms from $G$ to $G/H$ , $G/K$ respectively:$$p(g)=gH, q(g)=gK\;\forall\;g \in G$$ $K$ is a subset of $\ker(p)$ , so there exists a unique homomorphism $\varphi\colon G/K \to G/H$ so that $\varphi \circ q=p$ .

$p$ is surjective, so $\varphi$ is surjective as well; hence $\operatorname{im}\varphi=G/H$ . The kernel of $\varphi$ is $\ker(p)/K=H/K$ . So by the first isomorphism theorem we have$$(G/K) / \ker(\varphi)=(G/K) / (H/K) \approx \operatorname{im}\varphi=G/H$$