On Wed, Jun 4, 2014 at 10:34 AM, Omari Norman
It's not quite idempotence, because more than one function is involved.
It's a restricted version of the quasigroup law. For non-commutative operators there are actually two laws: left-division: forall a, b. exists (a\b). a*(a\b) == b N.B., it follows that: forall a, b. a \ (a*b) == b right-division: forall a, b. exists (b/a). (b/a)*a == b N.B., it follows that: forall a, b. (a*b) / b == a The "division" in the name just comes from assuming (*) is a "multiplication", as is usually assumed in group theory. We call (*) a left- or right-quasigroup depending on which law holds, or call it a quasigroup if both laws hold. If the quasigroup has an identity element, then it's called a loop, and we can define left- and right-inverse operators by (x\1) and (1/x). If we have various weak forms of associativity then we get left Bol loops, right Bol loops, and Moufang loops depending on what sort of weak associativity we have. If we have full associativity then the loop is a monoid. If the left- and right-inverses coincide, then this monoid is in fact a group. cf., http://winterkoninkje.dreamwidth.org/79868.html also http://en.wikipedia.org/wiki/Quasigroup Alternatively, if you want to view the law as being associated the other way — i.e., (x*y)/y == x — then, as Alexander Vieth said, the way to think about it is in terms of the endomorphism group. That is, given any monoid (G,(*)) we can construct a monoid (Endo(G),(.)) where forall x:G we have (_*x) : Endo(G), and where (.) is function composition. When G happens to be a group every element has a unique multiplicative inverse, therefore every endomorphism has a unique compositional inverse, hence Endo(G) is a group. Thus, we'd just call (_*x) and (_/x) inverses since they're inverse elements in Endo(G). If you want to get fancy, whenever r . s == id we say that s is a "section" of r, and that r is a "retraction" of s. So you could use that terminology, though it's more general and it loses the fact that we actually have both (_*x).(_/x) == id and also (_/x).(_*x) == id. -- Live well, ~wren