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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.

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  • $\begingroup$ Note that $g_1g_2 = g_2g_1$ always holds for $g_1 = g_2$. $\endgroup$ – TastyRomeo Oct 9 '18 at 9:43
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The statement $g_1g_2 = g_2g_1 \implies g_1 = g_2^{-1}$ for any $g_1$, $g_2$ is not correct.

There are a few trivial counterexamples:

  • 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 $g_1 = x^2 \neq y^{-1} = g_2^{-1}$.

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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 $[x^my^n,x^{m'}y^{n'}]$.

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