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If we have a relation $\sim$ on $\mathbb{Z}/6\mathbb{Z}\times (\mathbb{Z}/6\mathbb{Z}\setminus\{0\})$ so that $(w,x)\sim(y,z)$ if $wz=xy$, how is $\sim$ not an equivalence relation?

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    To give a countere$x$ample proving that the relation is not transitive2011-11-17

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In your comment (which you could preferably have added to the question), you cancel $z$ on both sides of an equation. This is not valid in $\mathbb Z/6\mathbb Z$, since it contains zero divisors. Thus, if you have $wb=2$ and $xa=4$, then with $z=3$ you get $wzb=xza$ despite $wb\ne xa$. Thus for instance $(1,1)\sim(3,3)$ and $(3,3)\sim(2,4)$, but $(1,1)\nsim(2,4)$.