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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.

Willie Wong
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Lubin
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  • I keep thinking that $gxg=g$ automatically implies $x=g^{-1}$ . . . I just have a feeling this is far too simple. Is it? The more I think about this, the more it confuses me. Isn't that the only possible way that equality holds for this situation? I can't see any other value for $x$ such that this holds for all $g\in X$ and $x\in X$. – 000 Dec 08 '12 at 02:36
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    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. – Rankeya Dec 08 '12 at 03:28
  • I'm still unsure what is wrong with the example $\mathbb{Z}_2\times \mathbb{Z}_2$ with multiplication. There is a unique $x$ such that $gxg=g$ for all $g\in X$, namely $x=(1,1)$, but it isn't a group. Will someone explain if anything is wrong with the counterexample? I would just like to see it if there is something wrong. – Clayton Dec 08 '12 at 03:32
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    The question is not that $\exists$ ! $x$ such that for all $g$ ..., but that $\forall$ $g$ there exists a unique $x$ such that ... I think you got your order of quantifiers wrong. – Rankeya Dec 08 '12 at 03:35
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    @Clayton Are you confusing "there is an $x$ such that for all $g$" with "for all $g$ there is an $x$"? – anon Dec 08 '12 at 03:35
  • I don't think so. In the example, $x=(1,1)$ is unique, so it works for all $g$. If you think about it for just a little while, you'll see what I mean (I hope). No other element will have the property, since, when multiplied by (1,1), at least one of these coordinates would go to $0$. @leo thanks! I hope so – Clayton Dec 08 '12 at 03:37
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    @Clayton $(0,1)(0,1)(0,1)=(0,1)=(0,1)(1,1)(0,1)$, failing the uniqueness criterion. – yearning4pi Dec 08 '12 at 03:37
  • But $(0,1)(1,1)(0,1)=(0,1)$, so your example doesn't work, I don't think. – Clayton Dec 08 '12 at 03:38
  • @peoplepower, I don't think the $x$ has to be unique in the sense that no other element works for some, but not others. I think the particular $x$ in question has to work for all $g\in X$, and it's okay if there is an element that fixes itself. – Clayton Dec 08 '12 at 03:41
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    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 $g\in X$, $gxg=g$. – anon Dec 08 '12 at 03:43
  • I think I see, never mind; I apologize. Thanks for helping my understanding along... :) – Clayton Dec 08 '12 at 03:44
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    Perhaps this is a record: so far, 5 deleted answers out of 7 tries to this question. – DonAntonio Dec 08 '12 at 04:00
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    @DonAntonio Take a look at this gem :) – anon Dec 08 '12 at 05:21

2 Answers2

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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 said common proof (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 $\ (\Leftarrow)\,$ For each $\,x\in S\,$ let $\,x'$ be the unique element of $\,S\,$ such that $\,xx'x=x.\ $ Note $\,x(\color{#0a0}{x'xx'})x=(xx'x)x'x=x\color{#C00}{x'}x=x,\,$ so, by uniqueness, $\,\color{#0a0}{x'xx'}=\color{#C00}{x}',\,$ hence $\,\color{#0a0}x = \color{#C00}{x}''.$

For all $\,x\in S\,$ the element $\,xx'$ is idempotent $\,(xx')^2=(xx'x)x'=xx'.$ Thus, being nonempty, $S\,$ 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.

$(\Rightarrow)\,$ if $\,S\,$ is a group then $\,xyx=x\iff xy=1\iff y = x^{-1},\,$ so $\,y\,$ is unique. $\ \bf\small QED$

Bill Dubuque
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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.

Brian M. Scott
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Rankeya
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