# a characterization of groups

idempotent element

###### Theorem.

A non-empty semigroup  $S$ is a group if and only if for every $x\in S$ there is a unique $y\in S$ such that $xyx=x$.

###### 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 $xx^{\prime}x=x$. Note that $x(x^{\prime}xx^{\prime})x=(xx^{\prime}x)x^{\prime}x=xx^{\prime}x=x$, so, by uniqueness, $x^{\prime}xx^{\prime}=x^{\prime}$, and therefore $x^{\prime\prime}=x$.

For any $x\in S$, the element $xx^{\prime}$ is idempotent  (http://planetmath.org/Idempotency), because $(xx^{\prime})^{2}=(xx^{\prime}x)x^{\prime}=xx^{\prime}$. 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 $xx^{\prime}=e$ for each $x\in S$, so $S$ is a group.

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

Note. Note that inverse semigroups do not in general satisfy 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.

Title a characterization of groups ACharacterizationOfGroups 2013-03-22 14:45:08 2013-03-22 14:45:08 yark (2760) yark (2760) 10 yark (2760) Theorem msc 20A05 Group RegularSemigroup AlternativeDefinitionOfGroup