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Create a structure by using Cayley Table which has element with more than one inverse.

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I thought a cayley table was used to describe a finite group, where we have uniqueness of inverses so this wouldn't be possible? – hmmmm Apr 5 '12 at 9:30
Michael: it would be nice if you shared some of your own thoughts on this matter. It is not the purpose of this site to simply provide solutions to your homework. Please read this thread. @hmmmm: I suppose that "Cayley table" simply refers to the multiplication table of a finite set equipped with a binary operation with unit. – t.b. Apr 5 '12 at 9:39
I'm thinking about this structure. If I have set with, for example, three elements - {a, b, c} where a is neutral element, the position of a will give us inverses. If a is in the section of c row and b column, that means c is inverse of b. Also, I can put neutral element in the section of b row and b column and get b is inverse to itself. In that way I will get element with two inverses. Am I correct? – Michael Apr 5 '12 at 10:00
@Michael This is not enough. Let $a$ be a neutral element. Then an inverse of $b$ is such an element $x$ that $bx=xb=a.$ It's not enough that $xb=a.$ This would only be a left inverse of $b$. – user23211 Apr 5 '12 at 10:28

Note that you cannot achieve this if your operation is associative; and you cannot achieve this if there are fewer than three elements. So the smallest example will be with a set of four, and a product which is not associative.

Let $\{e,a,b,c\}$ be your three elements, with $e$ the identity. Define the product as: $ex=xe=x$ for all $x$, and $xy=e$ if neither $x$ nor $y$ are equal to $e$. That is: $$\begin{array}{c||cccc} &e&a&b&c\\ \hline e&e&a&b&c\\ a&a&e&e&e\\ b&b&e&e&e\\ c&c&e&e&e \end{array}$$ Then every element except $e$ has two two-sided inverses. Note, however, that the operation is not associative, since $a(bc) = ae = a$, but $(ab)c = ec=c$.

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