# project normal subgroup generated by a subgroup to its abelianization

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?

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 $$\phi\left( \left[S \right] \right) = \phi(S)$$.