# Semigroups, Problems For Fun

• June 16th 2009, 02:19 AM
Swlabr
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 $(ab=ba \Rightarrow a=b)$.
• June 16th 2009, 04:02 AM
NonCommAlg
Quote:

Originally Posted by Swlabr
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 $f(xy)=(xya,axy)=(xa,ay).$

Quote:

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 $xyz=xz.$