If $G = H$ is a non-trivial group, is it possible for $G*H$ to be isomorphic to $\mathbb Z^2$

Question

If $G = H$ is a non-trivial group, is it possible for $G*H$ to be isomorphic to $\mathbb Z^2$

Answer 1

A non trivial free product is never commutative.

You could argue using reduced words and stuff but it would soon be tedious. Here's a proof using only the universal property : let $X = G\sqcup H$ modded out (as a set) by $e_G = e_H$, and have $G,H$ act on it in the following way : $G$ acts on the copy of itself, and acts trivially on $H\setminus \{e\}$, and symmetrically for $H$.

Then we get maps $f, k : G,H \to \mathfrak S X$, so by the universal property $G*H\to \mathfrak SX$.

It's easy to check that if $g\in G\setminus \{e\}, h\in H\setminus \{e\}$, $f(g)$ and $k(h)$ don't commute, e.g. you may look at how $f(g)k(h)$ and $k(h)f(g)$ act on $e$.