# Prove whether $xyz =1$ implies that $yzx=1$ or $yxz=1$.

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.

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

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

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 $$yzx=1$$.

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.

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.

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}