Free Product on 3 Groups confusion This is a very basic question: How can the free product on three groups, say $G:=\mathbb{Z}*\mathbb{Z}*\mathbb{Z}$ be a group? For example, let $a,b,c$ be the generators for the three copies of $\mathbb{Z}$ respectively, so $g:=abc\in G$.
Now, as I understand it, the group operation is concatenation followed by reduction, where we are allowed to delete the neutral element and multiply elements in the same group.
But now, consider $g^2=(abc)\cdot (abc)=abcabc$
Is this an element of $G$? It's my understanding that elements of $G$ have the form $g_1g_2g_3$ where $g_1\in \mathbb{Z}_a$, $g_2\in\mathbb{Z}_b$ and $g_3\in\mathbb{Z}_c$ (I'm denoting $\mathbb{Z}_\alpha$ as the copy of $\mathbb{Z}$ generated by $\alpha$).
Can someone clear this up?
 A: 
It's my understanding that elements of $G$ have the form $g_1 g_2 g_3$ where $g_1 ∈ ℤ_a$, $g_2 ∈ ℤ_b$ and $g_3 ∈ ℤ_c$ […].

No, this is wrong.
Let $G_1, \dotsc, G_n$ be groups.

*

*By a word in $G_1, \dotsc, G_n$ we mean a tuple $( (g_1, i_1), \dotsc, (g_ℓ, i_ℓ))$ with $ℓ ≥ 0$ and $(g_j, i_j)$ contained in $G_{i_j}$ for every position $j = 1, \dotsc, ℓ$.
The number $ℓ$ is the length of this word.

Let us denote the set of all such words by $W$.

*

*We say that a word $( (g_1, i_1), \dotsc, (g_ℓ, i_ℓ) )$ is reduced if

*

*$i_j ≠ i_{j + 1}$ for every $j = 1, \dotsc, ℓ - 1$, and

*$g_{i_j}$ is not the neutral element of $G_{i_j}$ for every $j = 1, \dotsc, n$.



Let us denote the set of reduced words by $W_{\mathrm{red}}$.
As you already mentioned in your question, every words $w = ( (g_1, i_1), \dotsc, (g_ℓ, i_ℓ) )$ can be transformed into a reduced word by repeatedly applying the following two reduction rules:

*

*If $i_j = i_{j + 1}$ for some position $j$, then replace $w$ by
$$
  w' = ( (g_1, i_1), \dotsc, (g_{j-1}, i_{j-1}), \, (g_j g_{j + 1}, i_j), \, (g_{j+2}, i_{j + 2}), \dotsc, (g_ℓ, i_ℓ) ) \,.
$$

*If at some position $j$ the element $g_j$ is the neutral element of $G_{i_j}$, then replace $w$ by
$$
  w' = ( (g_1, i_1), \dotsc, (g_{j-1}, i_{j-1}), (g_{j+2}, i_{j + 2}), \dotsc, (g_ℓ, i_ℓ) ) \,.
$$
Starting with any word $w$, we get in this way a sequence of words
$$
  w \to w' \to w'' \to \dotsb
$$
until we arrive at a reduced word.
(This sequence must terminate, since each of the two reduction rules reduces the length by $1$.)
We have thus a reduction map
$$
  W \to W_{\mathrm{red}} \,.
$$
As you already explained in your post, we can now form the group $G_1 * \dotsb * G_n$ as follows:

*

*The underlying set of $G_1 * \dotsb * G_n$ is the set of reduced words $W_{\mathrm{red}}$.

*The multiplication of two reduced words is given by first concatenating them, and then reducing the resulting word.

The elements of $G_1 * \dotsb * G_n$ are typically not written as words $( (g_1, i_l), \dotsc, (g_ℓ, i_ℓ) )$, but as products $g_1 \dotsm g_ℓ$.
However, we must keep in the back of your minds that each factor $g_j$ in this product ‘remembers’ to which index $i_j$ it belongs.
In your example $G_1 = ℤ_a$, $G_2 = ℤ_b$, $G_3 =  ℤ_c$ we have, for example, the word
$$
  b^2 a c^{-1} a^2 b c^3
  =
  ((b^2, 2), (a, 1), (c^{-1}, 3), (a^2, 1), (b, 2), (c^3, 3))
$$
of length $6$.
This word is reduced, and therefore an element of $ℤ_a * ℤ_b * ℤ_c$.
It is the product of the two elements $b^2 a c^{-1}$ and $a^2 b c^3$, as well as the product of the three elements $b^2 a$, $c^{-1} a^2$ and $b c^3$.

It should be pointed out that the construction of $G_1 * \dotsb * G_n$ via reduced words is somewhat bad. We can instead do the following:

*

*We can turn $W$, the set of all words, into a monoid via concatenation.
The neutral element of $W$ is given by the empty word.

The monoid $W$ is very much not a group:
no element of $W$ except for the neutral element has an inverse, since the concatenation of two words is only empty if both words were empty to begin with.

*

*Let $∼$ be the equivalence relation on $W$ generated by $w ∼ w'$ whenever $w'$ results from $w$ by one of the two reduction rules.
This equivalence relation is compatible with the monoid structure of $W$:
if $w_1 ∼ w'_1$ and $w_2 ∼ w'_2$, then also $w_1 w_2 ∼ w'_1 w'_2$.
(In other words, the equivalence relation $∼$ is a congruence relation on $W$.)
It follows that the monoid structure on $W$ descends to a monoid structure on the quotient set $W / {∼}$.


*The quotient $W / {∼}$ is not only a monoid, but a group:
the inverse of an equivalence class $[((g_1, i_1), \dotsc, (g_ℓ, i_ℓ))]$ is given by $[((g_ℓ^{-1}, i_ℓ), \dotsc, (g_1^{-1}, i_1))]$.
We can now construct the free product $G_1 * \dotsb * G_n$ as the quotient $W / {∼}$.
The construction of the free product via reduced words uses the fact that $W_{\mathrm{red}}$ is a set of representatives for the equivalence relation $∼$.
This allows one to identity $W / {∼}$ with $W_{\mathrm{red}}$.
