If $N$ is any normal subgroup of $G$, then the factor group $G/N$ is abelian if and only if $G' \subseteq N$.

In the proof I don't understand why $G/N$ is the homomorphic image of $G/G'$

$G\subseteq N \subseteq G'$

What is the explicit homomorphism? Does homomorphic image mean $G/N \cong G/G'$?



$G/N$ being the homomorphic image of $G/G'$ means that there is a surjective homomorphism $\varphi : G/G' \to G/N$. The existence of such homomorphism is an immediate result of the fundamental theorem of homomorphisms, since the kernel of the canonical projection $G \to G/N$ (which is $N$) contains $G'$.

If you're not familiar with this theorem, you can construct the homomorphism as follows: Let $a \in G$, and define the image of $\overline{a} \in G/G'$ to be the class of $a$ in $G/N$. It's an easy exercise to show that this is a well-define homomorphism. Again, the fact that $G' \subset N$ is crucial to the proof.


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