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Let $x,y,z$ be elements of a group $G$ and $xyz=1$. I am trying to prove whether this implies $yzx=1$ or $yxz=1$.

My proof goes as follows: Let $x^{-1}$ denote the inverse of $x$, then $xx^{-1} = 1 = x(yz)$. By the cancellation property of groups, $x^{-1}= yz$. This implies $(yz)x = x^{-1}x = 1$. Therefore $xyz = 1 \implies yzx = 1$.

By applying the same argument to $y(zx) = 1$ one can prove that $xyz = 1 \implies yxz =1$.

However while trying to my proof online I stumbled upon the following counterexample for the proposition $xyz = 1 \implies yxz =1$. If we take $G$ to be the group of $2\times 2$ matrices and let $x = \left( \begin{array} { c c } { 1 } & { 2 } \\ { 0 } & { 2 } \end{array} \right)$, $y = \left( \begin{array} { l l } { 0 } & { 1 } \\ { 2 } & { 1 } \end{array} \right)$ and $z = \left( \begin{array} { c c } { - 1 / 2 } & { 3 / 4 } \\ { 1 } & { - 1 } \end{array} \right)$. Then $x y z = \left( \begin{array} { c c } { 1 } & { 0 } \\ { 0 } & { 1 } \end{array} \right) = 1$ but $y x z = \left( \begin{array} { c c } { 2 } & { - 2 } \\ { 5 } & { - 9 / 2 } \end{array} \right) \neq 1$.

I don't understand where my proof went wrong.

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    $\begingroup$ I think you need to double-check your "By applying the same argument..." line. Are you sure it works out? $\endgroup$ Jan 11, 2019 at 14:57
  • $\begingroup$ Intuitively, this argument shows you can "peel off" a variable on one side, and slap it on the other. $\endgroup$ Jan 11, 2019 at 15:00

5 Answers 5

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You just mixed up some letters. $xyz=1$ implies $yzx=1$ and $zxy=1,$ not $yxz=1.$

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Well, if $xyz=1$, then by multiplying with the inverse of $x$ from the left, $yz=x^{-1}$. Now multiply the equation with $x$ from the right. Then $yzx=1$.

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For any group we have $aa^{-1}=a^{-1}a=1$. From $xyz=1$ we know that $(xy)z=1$, from where the first claim follows.

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Since $G$ is a group then if $x\in G$ , there exists $x^{-1}\in G$ such that $$x\cdot x^{-1}=x^{-1}\cdot x=e\in G$$having this result and exploiting the other properties of group we obtain $$x^{-1}\cdot (xyz)=(x^{-1}\cdot x)\cdot yz=e\cdot yz=yz=x^{-1}\cdot e=x^{-1}$$therefore $$yz\cdot x=yzx=x^{-1}\cdot x=1$$and the statement has been proved.

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Here $xyz=1$ gives

$$\begin{align} x^{-1}xyz=x^{-1}1 & \iff yz=x^{-1} \\ &\iff yzx=x^{-1}x \\ & \iff yzx=1, \end{align}$$

which gives

$$\begin{align} y^{-1}yzx=y^{-1}1 & \iff zx=y^{-1} \\ & \iff zxy=y^{-1}y \\ &\iff zxy=1. \end{align}$$

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