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# a characterization of groups

###### 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, 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.

## Mathematics Subject Classification

20A05*no label found*

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