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Let * be a binary operation on a set S. Assume the domain of definition of * is S^2, assume * is associative and n is the neutral element for *. Now, let s and t be two elements of S. We say that t is a left inverse of s under * iff t*s=n. We say that t is a right inverse of s iff s*t=n. Show that if an element of S has both a left and right inverse under * then the left inverse and the right inverse are equal.

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closed as off-topic by This is much healthier., Jyrki Lahtonen, RecklessReckoner, Claude Leibovici, glace Jul 14 at 7:56

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1 Answer 1

Hint: $l=l*n=l*(s*r)=...$ continue from here using the associativity of $*$.

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I see that l*(s*r)=(l*s)*r. Is this the last step of the proof? I don't know how to take that any further. –  Karl Feb 11 at 2:55
This is just next step. Now. What is the definition of $l$ being the left inverse of $s$? Use it. You should get $=r$ at the end, because this is what you need to show: $l=...=r$. –  Vadim Feb 11 at 3:00
so (s*r) = n and (l*s) = n I can write l*n=n*r and because n is the neutral element l=r? –  Karl Feb 11 at 3:15
Yes, right, because that's the definition of the neutral element. –  Vadim Feb 11 at 3:24
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