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Suppose that $A$ and $B$ are subsets of a group or semigroup. We call $A$ left upper [resp. lower] $B$-periodic if $BA\subseteq A$ [resp. $A\subseteq BA$]. If $A$ is both left upper and lower $B$-periodic (i.e. $BA=A$ ), then we call it left $B$-periodic. It is a generalization of the conceptions ideals, sub-semigroups and subgroups (for more information, one can see http://www.worldscientific.com/doi/abs/10.1142/S1005386711000332?journalCode=ac) .

Now, we are looking for some (finite, infinite) groups/semigroups containing a non-singleton subset $B\neq \emptyset$ such that for every subset $A$ the following property holds $$ A\subseteq BA\Rightarrow A\subseteq bA\; \mbox{; for all } b\in B $$

(Note that in groups it is equivalent to "$A\subseteq BA\Rightarrow B^{-1}A\subseteq A$". Also, if $B^{-1}\subseteq B$ then $A=BA$ implies $A=bA$, for all $b\in B$)

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  • $\begingroup$ Let B be a right zero, so that ab=z for all a in S and b in B. Then S and B yield your property. $\endgroup$ Nov 14, 2014 at 8:54
  • $\begingroup$ You may also need z (and B) to be a left zero to guarantee associativity. and I think I reversed left and right above: try ba=z instead. $\endgroup$ Nov 14, 2014 at 8:59
  • $\begingroup$ If $B$ is a subset of left zeros with more than two elements, then it does not have the property. If $B$ is a subset of left identities, then it has the property. But, I'm looking for some other (non-trivial) subsets with the property (specially in groups). $\endgroup$ Nov 15, 2014 at 9:30
  • $\begingroup$ As I understand your property, (and if I get the direction right), the only time the condition A contained in BA holds is if A is contained in singleton z, in which case the consequent is obvious. So the condition does hold, but may be less interesting. Another case is when B consizts of a single elements and acts like a cyclic derangement. $\endgroup$ Nov 15, 2014 at 16:19

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