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The following is from Hatcher's Algebraic Topology:

Suppose one is given a collection of groups $G_{\alpha}$ and one wishes to construct a single group containing all these groups as subgroups. One way to do this would be to take the product group $\prod_{\alpha}G_{\alpha}$, whose elements can be regarded as the functions $\alpha\mapsto g_{\alpha}\in G_{\alpha}$. Or one could restrict to functions taking on nonidentity values at most finitely often, forming the direct sum $\bigoplus_{\alpha}G_{\alpha}$. Both these constructions produce groups containing all the $G_{\alpha}$'s as subgroups, but with the property that elements of different subgroups $G_{\alpha}$ commute with each other. In the realm of nonabelian groups this commutativity is unnatural, and so one would like a nonabelian version of $\prod_{\alpha}G_{\alpha}$ or $\bigoplus_{\alpha}G_{\alpha}$. Since the sum $\bigoplus_{\alpha}G_{\alpha}$ is smaller and presumably simpler than $\prod_{\alpha}G_{\alpha}$, it sould be easier to construct a nonabelian version of $\bigoplus_{\alpha}G_{\alpha}$, and this is what the free product achieves.

Can someone explain what he means by "but with the property that elements of different subgroups $G_{\alpha}$ commute with each other."?

Koda
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1 Answers1

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If $g\in G$ and $h\in H$, then these correspond to $(g,1_H)$ and $(1_G,h)$ in the direct product or sum $G\oplus H$. We have $$ (g,1_H)\cdot(1_G,h)=(g\cdot_G1_G,1_H\cdot_H h)=(1_G\cdot_Gg,h\cdot_H 1_G)=(1_G,h)\cdot(g,1_H)$$

Hagen von Eitzen
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  • Thank you. I see what he means now. By "so one would like a nonabelian version of $\prod_{\alpha}G_{\alpha}$ or $\bigoplus_{\alpha}G_{\alpha}$", he does not mean these two groups are abelian for nonabelian $G_{\alpha}$'s, correct? Does this mean when $G$ and $H$ are nonabelian, the commutativity of elements $(g,1_G)$ and $(1_H, h)$ are unnatural, so we would like something else in which $(g,1_G)\dot(1_H, h)≠(1_H, h)\dot(g,1_G)$? – Koda Jan 05 '19 at 16:35