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Prove if an element of a monoid has an inverse, that inverse is unique

How to show that the left inverse x' is also a right inverse, i.e, x * x' = e?

Also, how can we show that the left identity element e is a right identity element also?

Thanks

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Double checking the title for typos is usually a great idea! –  Mariano Suárez-Alvarez Feb 2 '12 at 3:00
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(Presumably you are in a group or something? Add details to the body of the question so that it makes sense :) ) –  Mariano Suárez-Alvarez Feb 2 '12 at 3:01
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Same ground covered: math.stackexchange.com/questions/102882/… –  anon Feb 2 '12 at 3:04
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It depends on the definition of a group that you are using. If you define a group to be a set with associative binary operation such that there exists a left identity $e$ such that all elements have left inverses with respect to $e$ then showing that left identity/inverses are unique and also right identity/inverses can be a challenging exercise. –  Derek Holt Feb 2 '12 at 5:11
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@Derek Bingo-that was the point of my proof below and corresponding response to Dylan. When I first learned algebra, my professor DID in fact use those very weak axioms and go through this very tedious-but enlightening-process. –  Mathemagician1234 Feb 2 '12 at 7:21
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marked as duplicate by anon, Jonas Meyer, Arturo Magidin, t.b., Asaf Karagila Feb 2 '12 at 21:22

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2 Answers

The idea for these uniqueness arguments is often this: take your identities and try to get them mixed up with each other. Assuming that you are working with groups, suppose that we have $x, y, z$ in a group such that $yx = xz = e$. The products $(yx)z$ and $y(xz)$ are equal, because the group operation is associative. Evaluate these as written and see what happens. The story for left/right identities is even simpler: if I have two elements in a group, what's the obvious thing to do with them?

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The argument for identities is very simple: Assume we have a group G with a left identity g and a right identity h.Then strictly by definition of the identity:
g = gh = h.
So g=h. Q.E.D.

The argument for inverses is a little more involved,but the basic idea is given for inverses below by Dylan. Here's a straightforward version of the proof that relies on the facts that every left identity is also a right and that associativity holds in G. Assume x' is a left inverse for a group element x and assume x'' is a right inverse. Let h a 2 sided identity in G (note we did NOT assume it's unique!It in fact is,but we haven't proven that yet! Be careful!) Then:
x' = x'h = x'(xx'') = (x'x) x'' = hx''= x''. So x'=x'' and every left inverse of an element x is also a right. Q.E.D.

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The part of Dylan's answer that provides details is answering a different question than you answer. –  Jonas Meyer Feb 2 '12 at 3:34
    
@Jonus Yes,he's answering a similar question for inverses as for identities,which involves a little more care in the proof. I fixed my answer in light of this carelessness.Note my answer depends on the identity being 2 sided,so it's important in my version to prove that first. But either way works. –  Mathemagician1234 Feb 2 '12 at 5:16
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Where you wrote "we haven't proven that yet!", I thought that you did prove that in your first paragraph. –  Jonas Meyer Feb 2 '12 at 5:19
    
@Jonus Nope-all we proved was that every left identity was also a right. That does not imply uniqueness-suppose there's more then one left identity? If we specify in the axioms that there is a UNIQUE left identity,prove there's a unique right identity and then go from there,then YES,it does. But I guess it depends on how general your starting axioms are. In any event,there's nothing in the proof that every left is also a right identity BY ITSELF that shows that there's a unique 2 sided identity. I tend to be anal about such matters.In any event,we don't need the uniqueness in this case. –  Mathemagician1234 Feb 2 '12 at 7:17
    
You showed that if $g$ is a left identity and $h$ is a right identity, then $g=h$. This means that $g$ is a 2-sided identity, and that it is unique, because if $k$ is another 2-sided identity, it is also a right identity, so $g=k$ by what was already shown. –  Jonas Meyer Feb 2 '12 at 7:26
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