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 $g\in X$, there is a unique $x\in X$ 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.

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    $\begingroup$ I think Prof. Lubin already knows that this is a true statement (he is not asking us if this is true or not). So people should rethink if they believe they have a counterexample. $\endgroup$
    – Rankeya
    Dec 8, 2012 at 3:28
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    $\begingroup$ @Clayton $(0,1)(0,1)(0,1)=(0,1)=(0,1)(1,1)(0,1)$, failing the uniqueness criterion. $\endgroup$ Dec 8, 2012 at 3:37
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    $\begingroup$ No, Clayton. Reread the question. For each $g\in X$, there is a unique $x\in X$ (depending on $g$) such that $gxg=g$. This is not the same as there being a unique $x\in X$ such that for each $g\in X$, $gxg=g$. $\endgroup$
    – anon
    Dec 8, 2012 at 3:43
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    $\begingroup$ Perhaps this is a record: so far, 5 deleted answers out of 7 tries to this question. $\endgroup$
    – DonAntonio
    Dec 8, 2012 at 4:00
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    $\begingroup$ @DonAntonio Take a look at this gem :) $\endgroup$
    – anon
    Dec 8, 2012 at 5:21

2 Answers 2


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 $x\in S$ 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 $x'$ denote the unique element of $S$ such that $\,xx'x=x.\ $ Note $\,x(\color{blue}{x'xx'})x=(xx'x)x'x=x\color{#C00}{x'}x=x,\,$ so, by uniqueness, $\color{blue}{x'xx'}=\color{#C00}{x}',$ and therefore $\color{blue}x = \color{#C00}{x}''.$

For any $x\in S,$ 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 $i\in S$ is idempotent, then $ix =ix\color{#0A0}{(ix)'}ix=ix\color{#C00}{(ix)'i}ix,$ so, by uniqueness, $\color{#C00}{(ix)'i}=\color{#0A0}{(ix)'},$ hence $(ix)'=(ix)'(ix)''(ix)'=\color{#C00}{(ix)'i}x(ix)'=\color{#0A0}{(ix)'}x(ix)',$ 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 $x\in S,$ hence $S$ is a group.

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

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    $\begingroup$ Thanks! This proof is shorter and more direct than the one I found, too. $\endgroup$
    – Lubin
    Dec 8, 2012 at 23:23
  • $\begingroup$ How does one get the idea to consider the inverse of an arbitrary idempotent times x? $\endgroup$
    – user51427
    Dec 9, 2012 at 3:15
  • $\begingroup$ This is the kind of complete correct answer with effort that I love. $\endgroup$ Jan 21, 2017 at 13:10
  • $\begingroup$ The argument is beautiful and quite natural. $\endgroup$ Oct 25, 2017 at 1:14

I am just putting some sources here. Bill Dubuque already did all the work:

This book may be of some interest: Mark V. Lawson, Inverse Semigroups: The Theory of Partial Symmetries, at Google Books

I came across this book in this Wikipedia article, which gives some of the history of Inverse Semigroups.

Also, there is this shorter PDF by the author of the book whose link I gave above, and what you state is probably Proposition 2.4 in the notes (although it is stated in a different way) after some other things have already been proved.

  • $\begingroup$ The third link is broken. Could you please update it. $\endgroup$
    – Bumblebee
    Oct 17, 2020 at 5:51

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