Talk:Inverse in Monoid is Unique

The paragraph from "Similarly:" seems unnecessary... -- TheSpleen (talk) 16:03, 8 September 2013 (UTC)


 * There is a reason, I just can't think what it is at the moment. --prime mover (talk) 16:24, 8 September 2013 (UTC)


 * It's just that if you read the second equation system backwards and switch $y$ and $z$, you get exeactly the same. -- TheSpleen (talk) 17:09, 8 September 2013 (UTC)


 * Yes all right it's not necessary, but it's informative for a beginner. --prime mover (talk) 06:07, 9 September 2013 (UTC)


 * I think the original rationale for including both was because $\circ$ isn't commutative (under the monoid axioms); however, I think that since $\circ$ commutes when one argument is the identity element (by the definition of the identity element, $x\circ 1 = 1 \circ x = x$), this clarification isn't necessary. Luolimao (talk) 05:45, 12 February 2014 (UTC)

merge
I put the mergeto the corollary because this theorem is about magmas that are associative and have an identity element. The corollary is about monoids. They are exactly the same: a monoid is an associative magma with identity. Hence the merge. --barto (talk) (contribs) 15:24, 4 November 2017 (EDT)


 * Good call, my bad. &mdash; Lord_Farin (talk) 16:46, 4 November 2017 (EDT)


 * This page was deliberately written so as to refer to a general algebraic structure which may or may not be closed. A monoid is specifically closed. Hence the explicit difference between one and the other.


 * So can we go back to how this page was originally?


 * No, let me rephrase that, we are going back to how this page was originally. --prime mover (talk) 17:58, 4 November 2017 (EDT)


 * Oh right, thanks. --barto (talk) (contribs) 03:42, 5 November 2017 (EST)

A subtle distinction worth preserving indeed. I already wondered how it got to be set up like this. &mdash; Lord_Farin (talk) 11:42, 5 November 2017 (EST)


 * Wait... In order for $\circ$ to be associative, $S$ has to be closed: $(x\circ y)\circ z$ only makes sense if $x\circ y\in S$. See Definition:Associative/Algebraic Structure. --barto (talk) (contribs) 12:12, 5 November 2017 (EST)

I'll move the first line of the proof to the theorem statement, because the theorem really doesn't make any sense without requiring that $S$ has identity. --barto (talk) (contribs) 12:09, 5 November 2017 (EST)


 * No, please leave it as it is. There exists a definition of inverse which does not depend on there being an identity. I want to preserve that. --prime mover (talk) 12:14, 5 November 2017 (EST)


 * Ehm okay.. I'm curious which definition, something not yet at ProofWiki? --barto (talk) (contribs) 12:18, 5 November 2017 (EST)


 * Definition:Inverse Semigroup -- I think I got it off Twikipedia, have no references to it in any of my own hardcopies. --prime mover (talk)


 * Well, in an inverse semigroup the inverse is unique by definition/assumption. It is unrelated. --barto (talk) (contribs) 12:40, 5 November 2017 (EST)


 * This whole category of mathematics was set up by someone who lied about his capabilities. Delete the lot and start all over again, it's complete shit. --prime mover (talk) 12:25, 5 November 2017 (EST)