# Free product of groups as coproduct

Wikipedia says

"The coproduct in the category of sets is simply the disjoint union with the maps ij being the inclusion maps. Unlike direct products, coproducts in other categories are not all obviously based on the notion for sets, because unions don't behave well with respect to preserving operations (e.g. the union of two groups need not be a group), and so coproducts in different categories can be dramatically different from each other. For example, the coproduct in the category of groups, called the free product, is quite complicated."

I am wondering, is it correct to say that the free product of groups is the group generated by the disjoint union of the summands with no relation? Therefore it is similar to the coproduct in the category of sets, where the coproduct is the set generated by the disjoint unions of the summands (the disjoint union itself). And in the category of abelian groups, is it correct to say the coproduct (the weak direct sum) is the abelian group generated by the disjoint union of the groups?

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Well, with no relations apart from those already internal to the summands. –  Tobias Kildetoft Feb 4 '14 at 10:01
Not quite sure what "generated by" is supposed to mean in the category of sets, unless you're saying we're supposed to interpret the phrase vacuously in order to see the analogy all the way through. –  anon Feb 4 '14 at 10:05

The union of the two groups involved in a free product is not quite disjoint $-$ they have to share the same identity element. But otherwise yes, $G*H$ is the free group on $G\cup H$ modulo $1_G=1_H$ and the multiplication tables already in place for $G$ and $H$. Indeed if $G=\langle X|R\rangle$ and $H=\langle Y|S\rangle$ are presentations with $X,Y$ disjoint (by fiat) then $G*H=\langle X\cup Y|R\cup S\rangle$.
In the category of abelian groups the coproduct is the direct sum. This can be obtained from the free product by imposing commutativity: (assuming $G,H$ commutative) we can interpret $G,H$ themselves as subgroups of $G*H$, and we have $G\oplus H= (G*H)/[G,H]$. This is indeed the most free abelian group generated by $G\cup H$ (modulo $1_G=1_H$).
If $G,H$ are arbitrary groups, then still $(G*H)/[G,H] = G \times H$. By the way, I wouldn't emphasize as much $1_G = 1_H$. What really matters is that the inclusions from $G$ and $H$ are homomorphisms, which is probably what you mean by "the multiplication tables already in place for G and H.". –  Martin Brandenburg Feb 4 '14 at 10:44
@Martin It is equivalent to what I meant by the phrase, which is quotienting the aforementioned free group by the relations encoded in the multiplication tables for $G$ and $H$. I am speaking to the concept of "generated by" in the context of symbols and operations in group theory rather than the perspective of arrows and universal properties in category theory. –  anon Feb 4 '14 at 10:53