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Possible Duplicate:
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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    @Derek Bingo-that was the point o$f$ my proof below and correspondin$g$ 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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    @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