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In the proof of the coproduct of groups being the free product, it seems that we haven't made any assumption that the groups are non-Abelian. However, we know that for Abelian groups, the coproduct is the direct sum. And I think the free product cannot be isomorphic to the direct sum. For example, the direct sum of two finite groups is finite but the free product is in general infinite.

So the question is I don't understand why the proof of coproduct being free product can't be applied to Abelian groups.

A sketch of the proof in another post of this site can be found here:

Coproducts exist in the category of groups

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  • $\begingroup$ You speak of "the" proof. You should include this proof for people who aren't familiar with it. $\endgroup$ – Matt Samuel Jan 27 '15 at 13:25
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The free product $G\ast H$ of two groups $G,H$ is defined and is the coproduct of $G$ and $H$ in the category of groups regardless of whether the groups are abelian or not. Of course, the free product $G\ast H$ a fortiori also satisfies the universal property of the coproduct when restricted to abelian target groups $A$, that is $$\text{Hom}_{\textsf{Grp}}(G\ast H,A)\cong\text{Hom}_{\textsf{Grp}}(G,A)\times\text{Hom}_{\textsf{Grp}}(H,A),\quad\quad(\ddagger)$$ but even if $G,H$ are abelian and hence $\text{Hom}_{\textsf{Grp}}(G,A)=\text{Hom}_{\textsf{Ab}}(G,A)$, $\text{Hom}_{\textsf{Grp}}(H,A)=\text{Hom}_{\textsf{Ab}}(H,A)$, this does not imply that $G\ast H$ is also the coproduct in the category of abelian groups - simply because it is not abelian if $G,H$ are nontrivial.

However, you can now use the fact that every group $G$ has an abelianization $G_{\text{ab}}$, and that $G\mapsto G_{\text{ab}}$ is left adjoint to the embedding $\textsf{Ab}\hookrightarrow\textsf{Grp}$, to rewrite the left hand side of $(\ddagger)$ as $\text{Hom}_{\textsf{Ab}}((G\ast H)_{\text{ab}},A)$, proving that the abelianization $(G\ast H)_{\text{ab}}$ of $G\ast H$ is the coproduct in the category of abelian groups. And indeed, if you look at the explicit constructions of the free group, the direct sum and the abelianization, you'll see that the abelianization of the free product is naturally isomorphic to the direct sum.

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  • $\begingroup$ Thank you @Hanno! In fact I also have a question about free product being the coproduct. In the free product $G*H$, $e_G$ and $e_H$ are identified. So there is only one $e=i_G(e_G)=i_H(e_H)$ in the product (correct me if I am wrong). But $f(e_G)$ may not equal $g(e_H)$. How can we construct a homomorphism $\phi$ so that $\phi(e)=f(e_G)=g(e_H)$? $\endgroup$ – velut luna Jan 27 '15 at 13:51
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    $\begingroup$ Dear Kyson - if you want to invoke the universal property of $G\ast H$, you have to be given group homomorphisms $f:G\to K$ and $g: H\to K$, which in particular satisfy $f(e_G)=e_K=g(e_H)$. Does this help? $\endgroup$ – Hanno Jan 27 '15 at 13:53
  • $\begingroup$ Dear @Hanno, thank you very much for your help! Sorry I am new to category theory. $\endgroup$ – velut luna Jan 27 '15 at 13:55
  • $\begingroup$ Dear @Kyson, nothing to apologize for, you're welcome :) $\endgroup$ – Hanno Jan 27 '15 at 13:58
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    $\begingroup$ Yes that's true, I shouldn't have been sloppy about that: The coproduct of two groups $G,H$ is a group $G\ast H$ together with maps $\iota_G: G\to G\ast H$ and $\iota_H: H\to G\ast H$ such that for any other group $K$ the induced map $\text{Hom}(G\ast H,X)\to\text{Hom}(G,X)\times\text{Hom}(H,X)$ is a bijection. Then, the triple $(G\ast H,\iota_G,\iota_H)$ is unique up to unique isomorphism. $\endgroup$ – Hanno Jan 27 '15 at 14:02

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