a characterization of groups

idempotent element

Theorem.

A non-empty semigroup $S$ is a groupif and only iffor every $x\mathrm{\in }S$ there is a unique $y\mathrm{\in }S$ such that $x\mathit{}y\mathit{}x\mathrm{=}x$.

Proof.

Suppose that $S$ is a non-empty semigroup,and for every $x\in S$ there is a unique $y\in S$ such that $xyx=x$.For each $x\in S$,let $x^{\prime }$ denote the unique element of $S$ such that $x{x}^{\prime }x=x$.Note that $x(x^{\prime }x{x}^{\prime })x=(x{x}^{\prime }x)x^{\prime }x=x{x}^{\prime }x=x$,so, by uniqueness, $x^{\prime }x{x}^{\prime }=x^{\prime }$,and therefore $x^{\prime \prime }=x$.

For any $x\in S$, the element $x{x}^{\prime }$ is idempotent (http://planetmath.org/Idempotency),because ${(x{x}^{\prime })}^2=(x{x}^{\prime }x)x^{\prime }=x{x}^{\prime }$.As $S$ is nonempty, this means that $S$ has at least one idempotent element.If $i\in S$ is idempotent,then $ix=ix{(ix)}^{\prime }ix=ix{(ix)}^{\prime }iix$, and so ${(ix)}^{\prime }i={(ix)}^{\prime }$,and therefore ${(ix)}^{\prime }={(ix)}^{\prime }{(ix)}^{\prime \prime }{(ix)}^{\prime }={(ix)}^{\prime }ix{(ix)}^{\prime }={(ix)}^{\prime }x{(ix)}^{\prime }$,which means that $ix={(ix)}^{\prime \prime }=x$.So every idempotent is a left identity,and, by a symmetric argument, a right identity.Therefore, $S$ has at most one idempotent element.Combined with the previous result,this means that $S$ has exactly one idempotent element,which we will denote by $e$.We have shown that $e$ is an identity,and that $x{x}^{\prime }=e$ for each $x\in S$, so $S$ is a group.

Conversely, if $S$ is a groupthen $xyx=x$ clearly has a unique solution, namely $y=x^{-1}$.∎

Note. Note that inverse semigroups do not in generalsatisfy the hypothesis of this theorem:in an inverse semigroup there is for each $x$ a unique $y$ such that $xyx=x$ and $yxy=y$,but this $y$ need not be unique as a solution of $xyx=x$ alone.