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Say $\operatorname{Ab}(G)$ is the abelianization of $G$. Let $G_1$ and $G_2$ be two groups, $G_1\times G_2$ is the free product, then $G_1$ can be viewed as a subgroup in it. $j:G_1\times G_2\rightarrow \operatorname{Ab}(G_1\times G_2)$ is the natural homomorphism. $[G_1]$ is the normal subgroup generated by $G_1$ in $G_1\times G_2$. It is claimed that $j([G_1])=j(G_1)$. Can anyone explain it more explicitly?

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Since it is not clear exactly which point of the question you don't understand, I'll try and over-explain and hopefully catch your sticking point.

The normal subgroup in $G$ generated by a subset $S$ of $G$ is the subgroup generated by the set of conjugators: $$ \left[ S \right] = \left< \bigcup_{g \in G} \left\{ g^{-1}sg \mid s \in S \right\} \right> $$ Then, using your notation, I hope it is clear that since $G_{1} \subseteq \left[ G_{1} \right]$ we have $j\left(G_{1}\right) \subseteq j\left(\left[ G_{1} \right]\right)$, so it remains to show the reverse inclusion. That is we need to show that $$ j\left(\left[ G_{1} \right]\right) \subseteq j\left(G_{1}\right) $$ I hope it is clear that it suffices to show that $$ j\left( g^{-1}hg \right) \in j\left(G_{1} \right) \ \text{for all} \ g \in G_{1} * G_{2}, h \in G_{1}. $$ But given $g \in G_{1} * G_{2}$, and $h \in G_{1}$, since $j$ is a group homomorphism into an Abelian group we have $$ j\left(g^{-1}hg \right) = j(g)^{-1}j(h)j(g) = j(g)^{-1}j(g)j(h) = j(h) \in j\left( G_{1}\right). $$ This concludes the proof. I hope this helps.


Just as a comment, notice that we haven't really used the specific structure of the free-product, or of the homomorphism $j$. In fact we have proven the following statement:

Let $\phi : G \rightarrow H$ be a group homomorphism with $H$ an Abelian group. Then for any subset $S \subseteq H$, if $\left[ S \right]$ is the Normal subgroup in $G$ generated by $S$, then $\phi\left( \left[S \right] \right) = \phi(S)$.

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