Exercise 1.1.18 ($xy = yx \iff x^{-1} y^{-1} xy = 1$)

Question

Let $x$ and $y$ be elements of $G$. Prove that $xy = yx$ if and only if $y^{-1} x y =x$ if and only if $x^{-1} y^{-1} xy = 1$.

Richard Ganaye · Accepted Answer

\begin{proof} 
If $xy = yx$, then multiplying by $y^{-1}$ on the left, we obtain
$$y^{-1} xy = y^{-1}(yx) = (y^{-1}y )x = 1 x = x,$$
and multiplying by $x^{-1}$ on the left, we obtain
$$x^{-1} y^{-1} x y = x^{-1} x = 1.$$
Conversely, if $x^{-1} y^{-1} x y = x^{-1} x = 1,$ we obtain by multiplying  by $x$ on the left, $x (x^{-1} y^{-1} xy) = x 1$, so
$$y^{-1} x y = x,$$
and multiplying by $y$ on the left
$$xy = yx.$$
For all $x, y \in G$,
$$xy = yx \iff y^{-1} xy = x \iff x^{-1} y^{-1} xy = 1.$$
\end{proof}

Exercise 1.1.18 ($xy = yx \iff x^{-1} y^{-1} xy = 1$)

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Exercise 1.1.18 ($xy = yx \iff x^{-1} y^{-1} xy = 1$)

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