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Background:

Let $G, H$ be groups. I am currently trying to show that $G*H$, the free product of $G$ and $H$, is the coproduct of $G$ and $H$ in $\text{Grp}$.

I have no resources for this, so I am probably making a great many mistakes, but my idea was to try and use the universal property of free groups to show $F(G\amalg H)=G*H$ satisfies the universal property of the coproduct in $\text{Grp}$. I'm not sure if this can work though. The problem I am having is that the universal property of free groups involves set functions from $\{G,H\}$ to groups, whereas the univeral property for coproducts involves homomorphisms from $G$ and $H$.

As Thomas Andrews points out in the comments $F(G\amalg H)\not \cong G*H$, so this approach is not a really approach at all. As such my question is:

How does one see that $G*H$ is the coproduct in group?

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I think you are confused about what a free product is. $G*H$ is not a free group, in general. Assuming $F(S)$ means the free group on the set $S$, your set $\{G,H\}$ has just two elements, so $F(\{G,H\})$ is just the free group on two elements. –  Thomas Andrews Dec 5 '12 at 17:36
    
How is the free product defined in the source book you are using? –  Thomas Andrews Dec 5 '12 at 17:39
    
What he means is probably this. –  Jesko Hüttenhain Dec 5 '12 at 17:53
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@thomasandrews I think I have really mangled the question. I mean to let $F(\{G,H\})$ represent the free group on the underlying sets of the groups $G$ and $H$. Perhaps I should have written $F(G\cup H)$. Again, I am probably mistaken, but I thought $G*H$ and $F(G\cup H)$ were the same thing based on the wiki entry for free products and their construction. Also, I am using Algebra: Chapter 0 by Aluffi. –  Holdsworth88 Dec 5 '12 at 18:40
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Nope, $G*H$ is an image of $F(G\sqcup H)$, but, for example, if $H=\{1\}$, then $G*H\cong G\not\cong F(G)$. (We use $G\sqcup H$ just to make sure we know we are talking about the disjoint union of the underlying sets of $G$ and $H$...) –  Thomas Andrews Dec 5 '12 at 18:42

1 Answer 1

up vote 5 down vote accepted

Assume that we have a group $T$ and morphisms $\phi_G:G\to T$ and $\phi_H:H\to T$. Denote by $\iota_G:G\hookrightarrow G\ast H$ the inclusion, same with $\iota_H$. For a letter $g\in G\mathop{\dot\cup} H$, let $$D(g) = \left\{ \begin{array}{lcl} G &;& g \in G \\ H &;& g \in H \end{array}\right. $$ and define the map $\phi$ given by $$\begin{align*} G\ast H &\longrightarrow T \\ g_1\ldots g_r &\longmapsto \phi_{D(g_1)}(g_1)\cdots\phi_{D(g_r)}(g_r). \end{align*}$$ The relations of $G\ast H$ make this well-defined and a morphism of groups with the property $\phi\circ\iota_G=\phi_G$ and also $\phi\circ\iota_H=\phi_H$.

On the other hand, given any morphism $\psi:G\ast H \to T$ with $\psi\circ\iota_G=\phi_G$ and $\psi\circ\iota_H=\phi_H$, it must map a letter $g\in G\mathop{\dot\cup} H$ to $\phi_{D(g)}(g)$, forcing $\psi=\phi$. This should prove existence and uniqueness.

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