Finding an equivalence relation that isn't a congruence.

Question

Let $B=S \times T$ be a rectangular band such that $|S|=|T|=3$.

I've got to find an equivalence relation which is not a congruence in order to prove that at least one exists.

I've tried many different equivalence relations, but they have all turned out to be a congruence. Any help on how I would find one would be great.

Thanks.

EDIT: The operation on $B$ is: $(s,t)(s',t')=(s,t')$

Congruence is defined as : for $((s,t),(s',t')),((x,y),(x',y'))$ contained in the the equivalence relation $R$, $((s,t)(x,y),(s',t')(x',y'))$ is also contained in $R$.

Answer 1

Hint. Try to find a counterexample when $|S| = |T| = 2$, this is easier. An equivalence relation on a set defines a partition of the set. Just find a partition of $S \times T$ into two sets which does not define a congruence on $S \times T$.