A semigroup XX is a group iff for every g∈Xg\in X, ∃!x∈X\exists! x\in X such that gxg=ggxg = g

The following could have shown up as an exercise in a basic Abstract Algebra text, and if anyone can give me a reference, I will be most grateful.

Consider a set X with an associative law of composition, not known to have an identity or inverses. Suppose that for every gX, there is a unique xX with gxg=g. Show that X is a group.

Note that I’m not asking for a proof (though a really short one would please me!), just some place where this has been published.


I believe this is well-known to those working in regular/inverse semigroups. So well-known in fact that one can even find proofs on the web, e.g. see this proof on PlanetMath. Alas, I cannot recall the history of this result, though I vaguely recall reading something about such decades ago.

For completeness, I reproduce below the proof presented on PlanetMath (slightly edited).

Theorem   A non-empty semigroup S is a group if and only if for every xS there is a unique yS such that xyx=x.

Proof   Suppose that S is a non-empty semigroup, and for every xS there is a unique yS such that xyx=x. For each xS, let x denote the unique element of S such that xxx=x.  Note x(xxx)x=(xxx)xx=xxx=x, so, by uniqueness, xxx=x, and therefore x=x.

For any xS, the element xx is idempotent, by (xx)2=(xxx)x=xx. As S is nonempty, we infer that S has at least one idempotent. If iS is idempotent, then ix=ix(ix)ix=ix(ix)iix, so, by uniqueness, (ix)i=(ix), hence (ix)=(ix)(ix)(ix)=(ix)ix(ix)=(ix)x(ix), so, by uniqueness, x=(ix)=ix. So every idempotent i is a left identity, and, by a symmetry, 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, denoted e . We have shown that e is an identity, and that xx=e for each xS, hence S is a group.

Conversely, if S is a group then xyx=x clearly has a unique solution, namely y=x1.

Source : Link , Question Author : Lubin , Answer Author : Bill Dubuque

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