Order of $xy$ in various quotients of the free product $G_1 * G_2$, where $x \in G_1, y \in G_2$ are nontrivial Say we have two arbitrary nontrivial groups $G_1$ and $G_2$, and some arbitrary nontrivial elements $x \in G_1, y \in G_2$. Then it is known that the order of $xy$ in the free product $G_1 * G_2$ is infinite. However, quotienting $G_1 * G_2$ by the normal subgroup generated by $(xy)^n$ should create a group where the image of $xy$ has order dividing $n$.
The question I have is about the word "dividing" there: does this process always set the order of $xy$ to exactly $n$ (for any $n \in \mathbb{N}$)? Or is there some counterexample where, say, quotienting the free product by the normal subgroup generated by $(xy)^4$ makes $xy$ only have order $2$, or something along those lines?
(If this isn't true for $xy$ but it is true for some other word like $xyx^{-1}y^{-1}$ then I would be interested in knowing a word that it is true for.)
 A: Rewritten
The intersection of the normal closure of $(xy)^n$ with $\langle xy\rangle$ will be exactly $\langle (xy)^n\rangle$; the order of the product in the named quotient will be exactly $n$.
To see this, let us first prove it when $G_1=\langle x\rangle$ and $G_2=\langle y\rangle$ are cyclic groups.
It is enough to produce any group $G$ with an element $r$ of order $|x|$, an element $s$ of order $|y|$, and where $rs$ is of order $n$. To see this, note that the universal property of $G_1*G_2$ will then yield a group homomorphism $\phi\colon G_1*G_2\to G$ with $\phi(x)=r$ and $\phi(y)=s$. The kernel of this map will contain $(xy)^n$, since $(rs)^n=1$; in particular, the kernel contains the normal closure of $\langle (xy)^n\rangle$. But since the image of $xy$ is of order exactly $n$, we have that $\ker(\phi)\cap\langle xy\rangle = \langle (xy)^n\rangle$, and hence the same holds for the normal closure of $(xy)^n$. That is, the normal closure of $(xy)^n$ in $G_1*G_2$ does not contain any element of the form $(xy)^k$ in which $n$ does not divide $k$.
If $x$ and $y$ have finite order, then we have the following:
For any positive integers $a,b,c\gt 1$, there exists a finite group $G$ and elements $r,s\in G$ such that the order of $r$ is $a$, the order of $s$ is $b$, and the order of $rs$ is $c$ (see e.g. this Mathoverflow question, and the answer which references Milne's course notes; or this one which includes a proof). This gives the desired result.
If $x$ and $y$ are both of infinite order, then we can find an abelian group with the desired property: considering $\mathbb{Z}\times(\mathbb{Z}_n)$, and taking $r=(1,1)$ and $s=(-1,0)$.
If $x$ has infinite order and $y$ has order $b\gt 1$, then pick any $a\gt 1$; we know there is a group as above with elements $r$ and $s$ of orders $a$ and $b$, and $rs$ of order $n$. The universal property of $G_1*G_2$ gives a morphism mapping $x$ to $r$ and $y$ to $s$ in which $rs$ has order $n$, and so by the same argument as above we conclude that the normal closure of $(xy)^n$ contains $(xy)^k$ if and only if $n$ divides $k$.
This proves the case where $G_1$ and $G_2$ are cyclic.
Now assume that $G_1$ and $G_2$ are arbitrary, $x\in G_1$, and $y\in G_2$.
By the above, there exists a group $G$ and elements $r$ and $s$ in $G$ such that $|r|=|x|$, $|s|=|y|$, and $|rs|=n$ (for example, a suitable quotient of $\langle x\rangle * \langle y\rangle$). Now consider the amalgam obtained by identifying $x\in G_1$ with $r$ and $y\in G_2$ with $s$,
$$ (G_1*_{x=r} G)*_{y=s}G_2.$$
This group contains copies of $G_1$ and $G_2$, so we obtain a morphism $\phi$ from $G_1*G_2$ into it is one-to-one on $G_1$ and on $G_2$. The image of $xy$ under $\phi$ has order $n$, so again we see that the normal closure of $(xy)^n$ in $G_1*G_2$ contains $(xy)^k$ if and only if $n$ divides $k$.
So we conclude that in all cases, in $G_1*G_2/N$ where $N$ is the normal closure of $(xy)^n$, the element $xyN$ has order exactly $n$.
