free-groupsfree-productgroup-theory
What's the free product of $\Bbb Z/3\Bbb Z *\Bbb Z/3\Bbb Z $?
Let $G=H=\Bbb Z/3\Bbb Z *\Bbb Z/3\Bbb Z =\langle g\in\{0,1,2\}\mid 3g\equiv0\pmod 3\rangle$
Then I have that the free product of groups $J$ is:
$J=G*H=\langle g,h\mid 3g\equiv3h\equiv0\pmod 3\rangle$
I'm seeing this as the set of ordered pairs $\{g,h\}$ which has $9$ elements, which I know is wrong but I don't understand why.
What do the elements of $J$ look like?
Yes, the free product of the free group of rank $n$ and the free group of rank $m$ is isomorphic to the free group of rank $n+m$. Note also that if $G$ is free of rank $2$, and two elements $a$ and $b$ generate $G$, then $a$ and $b$ must freely generate $G$. One way to see this is to invoke the fact that a free group of finite rank is Hopfian. Another is to assume there is a nontrivial reduced word in $a$ and $b$ equal to the identity, take a free generating set $x$ and $y$, express $a$ and $b$ in terms of $x$ and $y$, and replace them in the nontrivial word expressing the identity; this will yield a nontrivial word in $x$ and $y$ equal to the identity (some work needs to be done here, of course), contradicting the choice of $x$ and $y$ as a free generating set. And there are other ways, of course.
An easy way to get this is as a corollary to the fact that the free group functor is the left adjoint of the underlying set functor. That is, for every group $G$ and every set $X$, $$\mathcal{G}roup(\mathbf{F}(X),G) \longleftrightarrow \mathcal{S}et(X,\mathbf{U}(G)),$$ where $\mathbf{F}(X)$ is the free group on the set $X$ and $\mathbf{U}(G)$ is the underlying set of the group $G$.
Because the free group functor is a left adjoint, it sends coproducts to coproducts. That is, the coproduct of two free groups $F(X)$ and $F(Y)$ in the category of groups is the free group on the coproduct of $X$ and $Y$ in the category of sets. The coproduct in the category of groups is the free product, and the coproduct in sets is the disjoint union. Therefore, there is a natural isomorphism $$F(X\amalg Y) \cong F(X)*F(Y).$$
You can also prove it directly from the universal property: the universal property of the free group on $X\amalg Y$ (the disjoint union) is that for every set-theoretic map $f\colon X\amalg Y\to G$ to a group $G$, there is a unique group homomorphism from $F(X\amalg Y)\to G$ that extends $f$. On the other hand, the universal property of the free product $F(X)*F(Y)$ is that for every pair of group homomorphisms $\varphi\colon F(X)\to G$ and $\psi\colon F(Y)\to G$, there is a unique group homomorphism $\Psi\colon F(X)*F(Y)\to G$ such that $\Psi\circ i_{F(X)}=\varphi$ and $\Psi\circ i_{F(Y)}=\psi$, where $i_{F(X)}\colon F(X)\to F(X)*F(Y)$ and $i_{F(Y)}\colon F(Y)\to F(X)*F(Y)$ are the canonical inclusions.
A set-theoretic map $f\colon X\amalg Y\to G$ is equivalent to a pair of maps $g\colon X\to G$ and $h\colon Y\to G$; the map $g\colon X\to G$ induceds a map $\varphi\colon F(X)\to G$, while the map $h\colon Y\to G$ induces a map $\psi\colon F(Y)\to G$, which in turn induces a map $F(X)*F(Y)\to G$; it is now straightforward to verify that this map extends $f$, and that it is unique, so that $F(X)*F(Y)$ has the universal property of $F(X\amalg Y)$, and therefore they are isomorphic.
Or: the inclusions $X\to F(X)\to F(X)*F(Y)$ and $Y\to F(Y)\to F(X)*F(Y)$ induce an inclusion $X\amalg Y\to F(X)*F(Y)$, which in turn induces a morphism $F(X\amalg Y)\to F(X)*F(Y)$. Conversely, the map $X\to F(X\amalg Y)$ induces a map $F(X)\to F(X\amalg Y)$, and $Y\to F(X\amalg Y)$ induces a map $F(Y)\to F(X\amalg Y)$, and these two maps together induce a map $F(X)*F(Y)\to F(X\amalg Y)$. It is now easy to verify that the induces maps $F(X\amalg Y)\to F(X)*F(Y)$ and $F(X)*F(Y)\to F(X\amalg Y)$ are inverses of each, in the usual abstract nonsense argument about their compositions having the same universal property as the corresponding identity.
P.S. The equality $F(a,b)=\langle a\rangle*\langle b\rangle$ presumably just means that there is a unique isomorphism between the two objects that maps $\{a\}$ and $\{b\}$ to themselves as the identity; this is a consequence of their respective universal properties/adjointness of free group construction.
Since you mentioned in the comments you have, and are familiar with Topology by Munkres, and I think Munkres takes a similar approach to what your question is discussing, I will use it as a guide, and explain some of what is going on there.
So first off, to actually construct an example of a free product of groups, typically you go through this reduced word definition, in which case you need disjoint groups, otherwise there is interaction between subgroups. This is purely convenient for construction, and not necessary for definition of free products (although this is actually what Munkres takes as the definition of free product of groups).
Munkres then defines what it means to be an external free product of $G_\alpha$ relative to monomorphisms $i_\alpha:G_\alpha \to G$ to be when $G$ is a free product of $i_\alpha(G_\alpha)$. Note that he does not require $G_\alpha$ to be disjoint, but by his definition of free product the $i_\alpha(G_\alpha)$ must be disjoint. To show that given $\{G_\alpha\}$ that an external free product always exist, what Munkres does is move to assuming that they are disjoint, which can be done by passing to $\{G_\alpha \times \{\alpha\} \}$ and doing the word construction. Taking this route there are natural monomorphism $j_\alpha:G_\alpha \times \{\alpha \} \to G$, so $G$ is an external free product of this collection, and there are natural isomporphism $\iota_\alpha:G_\alpha \to G_\alpha \times \{\alpha \}$, so $i_\alpha= j_\alpha \iota_\alpha$ is a collection of monomorphisms, and $G$ is in fact an external free product of the $G_\alpha$, and you don't really have to assume the $G_\alpha$ are disjoint, it is just helpful for the actual construction. This "little trick" is frequently used in other places (basically whenever you are constructing things from maps satisfying certain properties), and I think it is one of those things that Munkres, and others would probably expect you to fill or "just know" how to do/it doesn't effect anything.
In a couple of the other answers here, free products is defined in terms of universal property (Munkres uses the term extension condition, or something similar), and note that this does not assume things are disjoint either, for essentially the same reasons as why you don't require the external free product to not be disjoint.
In a little bit more concrete example, if we want to think about $\mathbb{Z} * \mathbb{Z}$ (which is just a name for a group satisfying some universal property) in terms of external free product, we get that there are monomorphisms $i_1,i_2: \mathbb{Z} \to \mathbb{Z} * \mathbb{Z}$ such that $i_1(\mathbb{Z}) \cap i_2(\mathbb{Z})= \{id\}$, and it is a free product $i_1(\mathbb{Z})*i_2(\mathbb{Z})$ in the reduced word sense.
If you look in Munkres I don't think he ever writes anything like $\mathbb{Z} * \mathbb{Z}$, basically because it disagrees with his approach, but from the above example it is pretty easy to see what that would mean, no matter what approach you take to understanding free products.
Best Answer
Its not the set of ordered pairs. Its more natural in this setting to write our group operation multiplicatively, so $g^3=1$ etc., and then we can view this group $\mathbb{Z}/3\mathbb{Z}\ast\mathbb{Z}/3\mathbb{Z}$ as the set of all words of the form: $$ g^{\alpha_1}h^{\beta_1}\cdots g^{\alpha_n}h^{\beta_n} $$ where $\alpha_2, \ldots, \alpha_n, \beta_1, \ldots, \beta_{n-1}\in\{1, 2\}$ and $\alpha_1, \beta_n\in\{0, 1, 2\}$ (this last condition just allows for the first and end letters of the word to change).
Multiplication is concatenation-then-reduction, for example multiplying $hgh^2$ and $hg^2h$ gives: $$ \begin{align*} hgh^2\cdot hg^2h&=hgh^3g^2h\\ &=hg1g^2h\\ &=hg^3g\\ &=h1g\\ &=h^2. \end{align*} $$ Using this view, its clear that this group is infinite. (Note that people often use $\epsilon$ for the identity element in this setting, where $\epsilon$ denotes the empty word, rather than $1$.)
Being slightly more formal, what is going on here is that the group is equivalence classes of words over the symbols $g$ and $h$, where $U$ and $V$ are equivalent if one can be obtained from the other by adding and removing words of the form $g^3$ and $h^3$. All I've done above is pick a single representative from each equivalence class. This is the kind of view taken in, for example, the book Combinatorial group theory by Magnus, Karrass and Solitar, or the book Combinatorial group theory by Lyndon and Schupp*.
For an added geometric twist, consider the infinite tree where each vertex has valency $3$ (the "trivalent tree"). Colour the vertices blue and red such that no adjacent vertices have the same colour. Then the free product $\mathbb{Z}/3\mathbb{Z}\ast\mathbb{Z}/3\mathbb{Z}$ acts by rotations on this tree (you can think of the "left" group as the "blue" group and the "right" group as the "red" group). For more details, see the very readable book "Groups, graphs and trees" by John Meier.
*Why yes, these books do have the same name.