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 g∈X, there is a unique x∈X 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 x∈S there is a unique y∈S such that xyx=x.
Proof Suppose that S is a non-empty semigroup, and for every x∈S there is a unique y∈S such that xyx=x. For each x∈S, let x′ denote the unique element of S such that xx′x=x. Note x(x′xx′)x=(xx′x)x′x=xx′x=x, so, by uniqueness, x′xx′=x′, and therefore x=x″.
For any x∈S, the element xx′ is idempotent, by (xx′)2=(xx′x)x′=xx′. As S is nonempty, we infer that S has at least one idempotent. If i∈S 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 x∈S, hence S is a group.
Conversely, if S is a group then xyx=x clearly has a unique solution, namely y=x−1.