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In the abstract algebra class, we have proved the fact that right identity and right inverse imply a group, while right identity and left inverse do not.

My question: Are there any good examples of sets (with operations on) with right identity and left inverse, not being a group?

To be specific, suppose $(X,\cdot)$ is a set with a binary operation satisfies the following conditions:

(i) $(a\cdot b)\cdot c=a\cdot (b\cdot c)$ for any $a,b,c\in X$;

(ii) There exists $e\in X$ such that for every $a\in X$, $a\cdot e=a$;

(iii) For any $a\in X$, there exists $b\in X$ such that $b\cdot a=e$.

I want an example of $(X,\cdot)$ which is not a group.

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marked as duplicate by Martin Sleziak, graydad, Robert Cardona, Najib Idrissi, Lord_Farin Mar 9 at 17:15

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@ZhenLin: I think rhenskyyy means that she know that right identity and left inverse cannot ensure the conditions being a group, but I think she want a concrete example to show that there exists a non-group set with multiplication satisfying associative law, right identity and left inverse. – Yuchen Liu Sep 11 '12 at 12:16
@jerrysciencemath You don't need to put your name in the post if you edit it. I think your edit is good, so leave it all, but take out "Edit(by jerrysciencemath):" – Graphth Sep 11 '12 at 12:39

1 Answer 1

up vote 7 down vote accepted

$$\matrix{a&a&a\cr b&b&b\cr c&c&c\cr}$$ That is, $xy=x$ for all $x,y$.

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This is a nifty and easy to remember example! – rschwieb Sep 11 '12 at 12:51
You didn't tell which one is $e$, so I have difficulty finding the left inverse of (say) $b$. I know any one will do, but I think you need to be specific. – Marc van Leeuwen Sep 11 '12 at 14:16
OP didn't posit a unique left inverse, and I don't see why I need to be specific. Isn't it enough that, as you point out, any one will do? – Gerry Myerson Sep 11 '12 at 22:41
@GerryMyerson: Is there any example satisfying unique left inverse and unique right identity besides the conditions in the problem? – Yuchen Liu Sep 12 '12 at 15:06
@jerry, I don't know. – Gerry Myerson Sep 13 '12 at 2:53

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