Free Product on 3 Groups confusion

Question

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?

Answer 1

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:

1. If $i_j = i_{j + 1}$ for some position $j$, then replace $w$ by $$ w' = ( (g_1, i_1), \dotsc, (g_{j-1}, i_{j-1}), \, (g_j g_{j + 1}, i_j), \, (g_{j+2}, i_{j + 2}), \dotsc, (g_ℓ, i_ℓ) ) \,. $$
2. 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 $b c^3$.

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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 $W_{\mathrm{red}}$.