Does the commutator of $\langle x ,y : xyx^{-1}y = e \rangle$ is trivial?

I don't know whether it's correct but I wish to show that $$g_1g_2=g_2g_1 \Leftrightarrow g_1=g_2^{-1}$$ for any $$g_1\neq g_2$$ in $$\langle x,y : xyx^{-1}y=e\rangle$$

I was trying to work be definition, taking $$x^{n_1}y^{m_1}...x^{n_k}y^{m_k}$$ , $$x^{s_1}y^{t_1}...x^{s_j}y^{t_k}$$ and getting a contradiction when assuming:

$$x^{n_1}y^{m_1}...x^{n_k}y^{m_k} \cdot x^{s_1}y^{t_1}...x^{s_j}y^{t_j} \cdot y^{-m_1}x^{-n_k}...y^{-m_1}x^{-n_1} \cdot y^{-t_j}x^{-s_j}...y^{-t_1}x^{-s_1}$$ is the unit element, by using the properties: $$xy=y^{-1}x , xyx^{-1} = y^{-1}$$ but I didn't succeed to advance from that point.

• Note that $g_1g_2 = g_2g_1$ always holds for $g_1 = g_2$. – TastyRomeo Oct 9 '18 at 9:43

The statement $$g_1g_2 = g_2g_1 \implies g_1 = g_2^{-1}$$ for any $$g_1$$, $$g_2$$ is not correct.
• Setting $$g_1 = g_2 = x$$, then $$g_1g_2 = x^2 = g_2g_1$$, but $$g_1 = x \neq x^{-1} = g_2^{-1}$$, since $$x$$ is an element of infinite order.
• Set $$g_1 = e$$ and $$g_2 \neq e$$.
For a more non-trivial example: set $$g_1 = x^2$$ and $$g_2 = y$$. Since $$xy = y^{-1}x$$ and $$xy^{-1} = yx$$, we have that $$g_1g_2 = x^2y = x(xy) = x(y^{-1}x) = (xy^{-1})x = (yx)x = yx^2 = g_2g_1,$$ but $$g_1 = x^2 \neq y^{-1} = g_2^{-1}$$.
If $$xyx^{-1}y=e$$ then $$y^{-1}x=xy$$ (and also $$yx=xy^{-1}$$), so every element of $$G=\langle x,y\mid xyx^{-1}y\rangle$$ can be represented by $$x^my^n$$ for some $$m,n\in\mathbb{Z}$$. Clearly $$G'\leq\langle y\rangle$$ (the number of $$x$$ appearing is invariant), and $$y^2=[x,y]\neq e$$ so $$G'\neq\langle e\rangle$$.
In fact it isn't hard to show $$G'=\langle y^2\rangle$$ by calculating the commutator $$[x^my^n,x^{m'}y^{n'}]$$.