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# proof of third isomorphism theorem

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

$(G/K)/\ker(\varphi)=(G/K)/(H/K)\approx\operatorname{im}\varphi=G/H.$ |

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