Commutative free products

Question

Do there exist any non-trivial groups such that their free product is commutative?

That is, if $G, H$ are non-trivial groups is $G*H$ ever commutative? My thinking is no but I can't really formulate a reason as to why not.

Answer 1

Assume $g$ is an element in $G$ and $h$ is an element in $H$. If $gh=hg$, they should have the same impact on the other elements of the free group. Now $(gh)(h'g')=e$ where $h'$ and $g'$ are the inverses of $h$ and $g$ respectively and $e$ is the identity element. However $(hg)(h'g')$ is not equal to the identity, because it is a reduced word. So, our assumption ($gh=hg$) is not true.

Answer 2

Another way of doing this, which avoids talking about irreducible words and such, is using the universal property:

Universal Property of free products: For any group $K$ and any group homomorphisms $\phi_G:G\to K$ and $\phi_H:H\to K$, there exists a unique homomorphism $\phi:G*H\to K$ extending $\phi_G$ and $\phi_H$.

Thus, to prove that nontrivial elements of $G$ do not commute with nontrivial elements of $H$ in the free product, it suffices to find a group $K$ and homomorphisms $\phi_G:G\to K$ and $\phi_H:H\to K$ such that $\phi(g)$ and $\phi(h)$ do note commute for nontrivial $g\in G$ and $h\in H$.

The usual way to construct such things is with semidirect products. However, $G$ does not necessarily act on $H$ (and vice versa), so we need to employ a small trick.

Let $G$ act freely on itself by left multiplication, and consider the induced action of $G$ on the group $H^G$: $g\cdot(h_{g'})_{g'\in G}=(h_{g^{-1}g'})_{g'\in G}$.

Now consider a copy of $H$ in the first entry of $H^G$, i.e., consider the embedding $H\to H^G$, $h\mapsto (h_{g'})_{g'\in G}$, where $h_{g'}=h$ if $g'=1_G$ and $h_{g'}=1_H$ otherwise.

Then we have the crossed product $H^G\rtimes G$, which contains copies of $G$ and $H^G\supseteq H$. For any nontrivial $g\in G$ and $h\in H$, we have, inside this crossed product, with $h=(h_{g'})_{g'}$, $$ghg^{-1}=g\cdot h=(h_{g^{-1}g'})_{g'\in G}$$. So $ghg^{-1}$ has $h$ in the $g$-th entry. On the other hand, $h$ has $1_H$ in the $g$-th entry, because $g\neq 1_G$. Since $h\neq 1_H$, these threads are different, i.e., $ghg^{-1}\neq h$ in $H^G\rtimes G$, which means that $g$ and $h$ do not commute in $H^G\rtimes G$.