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Math Help - Semigroups, Problems For Fun

  1. #1
    MHF Contributor Swlabr's Avatar
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    Semigroups, Problems For Fun

    A semigroup is a non-empty set S together with an associative binary operation *, *:S \times S \rightarrow S.

    A rectangular band is a semigroup S := A \times B, A, B non-empty sets, under the operation (a_1, b_1)*(a_2, b_2) := (a_1, b_2) for all a_i \in A, b_i \in B.

    1) Let S be a semigroup such that x^2 = x and xyz = xz for all x,y,z \in S. Prove that S is isomorphic to a rectangular band.

    Spoiler:
    You might wish to show that it is isomorphic to Sa \times aS for a fixed, Sa := \{sa : s \in S\} and aS := \{as : s \in S\}.


    2) Prove that a semigroup S is a rectangular band if and only if for all a,b \in S we have that (ab=ba \Rightarrow a=b).
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  2. #2
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    Quote Originally Posted by Swlabr View Post
    A semigroup is a non-empty set S together with an associative binary operation *, *:S \times S \rightarrow S.

    A rectangular band is a semigroup S := A \times B, A, B non-empty sets, under the operation (a_1, b_1)*(a_2, b_2) := (a_1, b_2) for all a_i \in A, b_i \in B.

    1) Let S be a semigroup such that x^2 = x and xyz = xz for all x,y,z \in S. Prove that S is isomorphic to a rectangular band.

    Spoiler:
    You might wish to show that it is isomorphic to Sa \times aS for a fixed, Sa := \{sa : s \in S\} and aS := \{as : s \in S\}.

    fix a \in S and define the map f: S \to Sa \times aS by f(x)=(xa,ax). clearly f is well-defined.

    1) f preserves multiplication: this is equivalent to xya=xa, \ axy=ay, for all x,y \in S, which is given in the problem.

    2) f is injective: suppose f(x)=f(y), i.e. xa=ya, \ ax=ay. then x=x^2=xax=xay=xy and y=y^2=yay=xay=xy. thus x=y.

    3) f is surjective: suppose (xa,ay) \in Sa \times aS. then f(xy)=(xya,axy)=(xa,ay).



    2) Prove that a semigroup S is a rectangular band if and only if for all a,b \in S we have that (ab=ba \Rightarrow a=b).
    the non-trivial side: suppose ab=ba \Longrightarrow a=b. we claim that x^2=x and xyz=xz for all x,y,z \in S and therefore we're done by the first part of the problem.

    1) x^2=x: by associativity we have x^2 \cdot x = x \cdot x^2 and thus x^2=x.

    2) xyz=xz: let a,b \in S. since a^2=a, we have aba=a \cdot aba=aba \cdot a and so aba=a for all a,b \in S. thus zxz=z, \ xzx=x. hence: xz \cdot xyz=xzx \cdot yz=xyz=xy \cdot zxz=xyz \cdot xz

    and therefore xyz=xz.
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