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Thread: 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 $\displaystyle S$ together with an associative binary operation $\displaystyle *$, $\displaystyle *:S \times S \rightarrow S$.

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

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

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


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

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

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

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

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

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

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

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



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

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

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

    and therefore $\displaystyle xyz=xz.$
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