Answer to Question #350788 in Abstract Algebra for Eliz

Question

Assume that the equation xyz = 1 holds in a group G. Does

it follow that yzx = 1? That yxz = 1? Justify your answer.

Answer 1

$xyz=1$ implies that $x(yz)=1$. Let $yz=a$. Then we have $xa=1$ and so $ax=1$ since $a$ is invertible and $a^{-1}=x$. It follows that $(yz)x=1$. Hence $yzx=1$.

On the other hand, if $xyz=1$ it is not always true that $yxz=1$. To see that, let $G$ be the group of $2\times2$ real matrices and let $x=\left(\begin{smallmatrix}1&2\\0&2\end{smallmatrix}\right)$, $y=\left(\begin{smallmatrix}0&1\\2&1\end{smallmatrix}\right)$ and $z=\left(\begin{smallmatrix}-\frac{1}{2}&\frac34\\1&-1\end{smallmatrix}\right)$. Then $xyz=\left(\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right)=1$ in $G$. But $yxz=\left(\begin{smallmatrix}2&-2\\5&-\frac92\end{smallmatrix}\right)\ne1$.