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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!2012-02-02
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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 :) )2012-02-02
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    Same ground covered: http://math.stackexchange.com/questions/102882/prove-if-an-element-of-a-monoid-has-an-inverse-that-inverse-is-unique/102883#1028832012-02-02
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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.2012-02-02
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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.2012-02-02

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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.2012-02-02
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    @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.2012-02-02
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    Where you wrote "we haven't proven that yet!", I thought that you did prove that in your first paragraph.2012-02-02
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    @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.2012-02-02
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    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.2012-02-02
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    @Dylan Again, it really depends on the strength of your assumptions and the order in which they're taken. Suppose we instead presume k is a right identity i.e. for every x in G, xk= x but it is not assumed kx = x. It does not automatically follow that if every left identity is a right,the converse holds. Unless G is Abelian, then your conclusion does not necessarily follow. It gets even more problematic if our definition of a group assumes there is AT LEAST ONE left or right identity. Then we need to show the left(right) identity is unique before we even go as far as we did in our answer.2012-02-02
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    I think you intended that comment for me, not Dylan. I agree that it depends on the strength of the assumptions. In this case the assumptions were sufficiently strong, as indicated in my previous comment. In any case, we agree that being precise and wary of hidden assumptions is a good idea, and I'm happy to leave it at that.2012-02-02
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    @Jonas Ok and apologize for the earlier misdirect.I don't agree the axioms begun with were strong enough-I think if you begin with the weakest possible axioms for a group,my point of view on this will become clearer.I agree we should leave it at that or move this discussion elsewhere-the moderators are getting restless.2012-02-02