Free product not abelian

Question

If $(G_i)$, $i \in I$ with $|I| > 1$ and $|G_i| > 1$ is the free product $G$ of $(G_i)$,$i \in I$ always not-commutative? The free product of a family of groups $(G_i)$, $i \in I$ is defined by a tuple $(G, (\phi_i)_{i \in I})$ where $G$ is a group and $\phi_i : G_i \rightarrow G$ are homomorphisms, with the following property:
For every tuple $(U, (g_i)_{i \in I})$ where $U$ is a group and $g_i : G_i \to U$ are homomorphisms, there exists a unique homomorphism $f : G \to U$ with $g_i = f \circ \phi_i$ for all $i \in I$. I thought that since $|G_i| > 1$ and $|I| > 1$ I can choose $a_i \in G_i$ and $a_j \in G_j$ with $i \neq j$, $a_i \neq e_i $ with $a_j \neq e_j$. I wanted to show this by contraposition assuming $\phi_i(a_i) \phi_j(a_j) = \phi_j(a_j) \phi_i(a_i)$ for all $i \neq j$ and wanted to show that $a_i = e_i$ or $a_j = e_j$ but I can't really come to this conclusion since all Groups $G_i$ can be commutative and the only think that comes in my mind is that $G$ is generated by \begin{equation} \bigcup_{i \in I}{\phi_i(G_i)} \end{equation} which doesn't really help.
I would be thankful for every hint I can get.

Answer 1

One construction of a free product of groups is that it is the set of all words

$$ \phi_i(a_i) \cdots \phi_j(a_j) $$

where

• Every pair of consecutive letters are from different $G_i$ and not identity,
• Each $\phi_i$ is an embedding of $G_i$, and
• The pairwise images of $\phi_i$ and $\phi_j$ intersect only at the empty word when $i \neq j$.

These words are considered reduced, and the operation is concatenation followed by reduction until the word satisfies the first condition.

If $a$ and $b$ are two nonidentity elements from $G_i$ and $G_j$ where $i \neq j,$ then $\phi_i(a) \phi_j(b) \phi_i(a)^{-1} \phi_j(b)^{-1}$ is a reduced word (it satisfies the first condition), but it is not the empty word.

Thus,

$$ \phi_i(a) \phi_j(b) \neq \phi_j(b) \phi_i(a), $$

and the free product is not abelian. Of course, this depends on the fact that there are at least two nontrivial groups which are factors in the free product.

Answer 2

An argument directly from the universal property:

Let $A$ be an "almost-disjoint union" of the $G_i$s, sharing a single identity element for all of them: $$ A = \{0\} ~\cup~ \{ (i,x) \mid i\in I, x\in G_i, x\ne 0_{G_i}\} $$

Let $U$ be the group of permutations of $A$. Each $G_i$ acts on itself by left multiplication, which transfers in the natural way to an action on $A$ (namely, one that always fixes the elements that come from the other groups). This can be represented as homomorphisms $\phi_i: G_i \to U$.

Now, whenever we have nontrivial $x\in G_i$ and $y\in G_j$ with $i\ne j$, the two permutations $\phi_i(x)$ and $\phi_j(y)$ do not commute: $\phi_i(x)\phi_j(y)(0) = (j,y)$ whereas $\phi_j(y)\phi_i(x)(0) = (i,x)$.

Therefore when we apply the universal property to our $\phi_i$s, we get an $f$ with $$ f(g_i(x)g_j(y)) = \phi_i(x)\phi_j(y) \ne \phi_j(y)\phi_i(x) = f(g_j(y)g_i(x)) $$ so $g_i(x)g_j(y)$ and $g_j(y)g_i(x)$ must be different elements of the free product.