Band (algebra)

In mathematics, a band (also called idempotent semigroup) is a semigroup in which every element is idempotent (in other words equal to its own square). Bands were first studied and named by . The lattice of varieties of bands was described independently in the early 1970s by Biryukov, Fennemore and Gerhard. Semilattices, left-zero bands, right-zero bands, rectangular bands, normal bands, left-regular bands, right-regular bands and regular bands are specific subclasses of bands that lie near the bottom of this lattice and which are of particular interest; they are briefly described below. A class of bands forms a variety if it is closed under formation of subsemigroups, homomorphic images and direct product. Each variety of bands can be defined by a single defining identity. Semilattices are exactly commutative bands; that is, they are the bands satisfying the equation xy = yx for all x and y. Bands induce a preorder that may be defined as if . Requiring commutativity implies that this preorder becomes a (semilattice) partial order. A left-zero band is a band satisfying the equation xy = x, whence its Cayley table has constant rows. Symmetrically, a right-zero band is one satisfying xy = y, so that the Cayley table has constant columns. A rectangular band is a band S that satisfies xyx = x for all x, y ∈ S, or equivalently, xyz = xz for all x, y, z ∈ S, In any semigroup the first identity is sufficient to characterize a Nowhere commutative semigroup. Nowhere commutative semigroup implies the first identity. In any flexible magma so every element commutes with its square. So in any Nowhere commutative semigroup every element is idempotent thus any Nowhere commutative semigroup is in fact a Nowhere commutative band. Thus in any Nowhere commutative semigroup So commutes with and thus - the first characteristic identity. In a any semigroup the first identity implies idempotence since so so idempotent (a band). Then nowhere commutative since a band So in a band In any semigroup the first identity also implies the second because xyz = xy(zxz) = (x(yz)x)z = xz.