Now, assume $G,H$ abelian and let $X$ be arbitrary abelian, $f_1,f_2$ the same. Take the defining diagram of $G\coprod H$. Abelianization takes this diagram to a diagram in the category of abelian groups, where $G,H,X,f_1,f_2$ stay the same (up to irrelevant isomorphy). This diagram plus uniqueness then asserts that $G\coprod_{Ab}H=G\coprod H/[G\coprod H,G\coprod H]$ (up to irrelevant isomorphy). There's a map $\varphi\colon G\coprod H\rightarrow G\times H$ sending $g_1h_1...g_nh_n\mapsto(g_1...g_n,h_1...h_n)$. Reduction happens in accordance with the group operations, so this is a homomorp…