Talk:Group structure and the axiom of choice

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References needed[edit]

I wrote this article with most inspiration coming from here: http://mathoverflow.net/questions/12973/does-every-non-empty-set-admit-a-group-structure-in-zf/12988#12988

A reference for the ZF counterexample is needed. (A text outlining production of a model with the advertised properties.) Done
A reference for the AC -> group structure is needed too, but is less urgent since the proof is quite easy. Done
How to change title from "Group Structure and the Axiom of Choice" to the less screaming "Group structure and the axiom of choice"? Done

YohanN7 (talk) 04:04, 20 December 2013 (UTC)[reply]

A ZF set with no group structure[edit]

Rewritten with a new very easy example. Most of the old material is retained. Emphasis on the fact that a non-wellorderable set not necessarily is without a group structure. YohanN7 (talk) 21:59, 7 February 2014 (UTC)[reply]

The section as written is riddled with mistakes. Maybe someone knowledgeable can rewrite it. And if not, it should be removed. — Preceding unsigned comment added by 80.110.127.154 (talk) 14:13, 8 December 2017 (UTC)[reply]

Let me add on my previous comment, with some examples: it is certainly false that "Non-wellorderable sets have the properties of being infinite and Dedekind-finite.", and the next sentences also contain many mistakes that seem to stem from a misinterpretation of Randall Dougherty's post on sci.math which is linked in the MathOverflow answer by Justin Palumbo. — Preceding unsigned comment added by 80.110.126.105 (talk) 20:49, 10 December 2017 (UTC)[reply]

"Cancellative quasigroup"[edit]

Isn't every quasigroup cancellative? It claims so in the first statement of Quasigroup#Properties. —Quondum 19:55, 8 February 2014 (UTC)[reply]

Guess you're right. However, "cancellative" is the key word in our context. In the corresponding place in Axiom of choice it says "Cancellative binary operation". Most concise would perhaps be to simply say "Quasigroup" which sounds better than "Cancellative binary operation"? No, better to have both. Will fix. YohanN7 (talk) 20:49, 8 February 2014 (UTC)[reply]

The intention and meaning seems to be clearer now. But I think mentioning a quasigroup just muddies the waters here, because it is merely an "e.g.", not an "i.e.", that is to say, it seems that there must be (non-finite) cancellative structures that are not quasigroups. In this context, such a distinction can be significant. Evidently, division implies but is not implied by cancellation. —Quondum 00:53, 9 February 2014 (UTC)[reply]

Hmmm... If you look at the first section in the article, do you think a cancellative magma will do? Key sentence from article: But by elementary group theory, the y • α are all different as α ranges over ℵ(X). Something similarly sounding should hold with quasigroups because of the Latin square property. Perhaps it is the Latin square property that is the crucial one. YohanN7 (talk) 02:08, 9 February 2014 (UTC)[reply]

I may have misunderstood what you've said, what is given and what is inferred. The "key sentence" you mention above is not an argument though: that seems to be merely an argument that is based on the assumption that the operation is a group operation. What is not clear is what is meant by the add-on statement in Axiom of choice about a cancellative operation being sufficient – it actually seems to be saying that the binary operation being cancellative ensures that the structure is a group, which is clearly false. There is nothing that really points at quasigroups though. It sounds as though this will have to be obtained directly from a reference (maybe Hajnal?). We shouldn't be adding something based on undecidable wording in another article. If we can't figure out what the case is, the note should just be scrapped. —Quondum 02:55, 9 February 2014 (UTC)[reply]

The result from that "key sentence", namely that the y • α are all different is the important thing for the proof and very much an argument. This works with a quasigroup, but nothing less than a quasigroup. Arthur Rubin mentioned "(cancellative quasi-)group" in a context when I asked about something for the third section. See close to the end in his 2013 archive. He knows about this stuff. I have seen quasigroup elsewhere too in this context, can't remember exactly where. I'll put quasigroup back for now, but skip the "cancellative" which apparently is redundant. YohanN7 (talk) 03:40, 9 February 2014 (UTC)[reply]

Forget about me saying that "cancellative" is the keyword. That thought was taken from the AC article + my temporary belief that it actually meant quasigroup. I believe now that the AC article is wrong. YohanN7 (talk) 03:57, 9 February 2014 (UTC)[reply]

I removed the popup for now. From review of Hajnal: the axiom of choice is equivalent to the statement: on every nonempty set there exists a cancellative groupoid (groupoid = magma). Thus the AC article is right. YohanN7 (talk) 04:36, 9 February 2014 (UTC)[reply]

Wow. Then it should be mentioned as an alternative, I expect, not even as a popup. —Quondum 05:04, 9 February 2014 (UTC)[reply]

Mystery resolved! It all makes sense, ~~because the multiplication table of a cancellative magma is a (possibly infinite) Latin square. One can see a cancellative magma as a quasigroup stripped off of its left and right division.~~ Reverted to your last version. YohanN7 (talk) 05:08, 9 February 2014 (UTC)[reply]

Resolved, yes, thanks to you looking up that reference (could you add it as reference for that specific thing?). As per my talk page, the quasigroup–cancellative magma correspondence does not hold, though.

—Quondum 06:09, 9 February 2014 (UTC)[reply]

[Edit conflict]Hmm. I was perhaps too quick with some striked out conclusions. Nevertheless, If

y, α

are elements of a cancellative magma with

y

fixed, then y • α (and α • y) will all be different as α varies. This is what matters for the proof in the first section. YohanN7 (talk) 06:15, 9 February 2014 (UTC)[reply]

Interestingly, the left and right cancellative properties individually do not guarantee that a divisor, if it exists, is unique. However, for two-sided cancellation it does (quite a subtle point due to right cancellation ensuring the uniqueness of the left divisor and vice versa). —Quondum 06:34, 9 February 2014 (UTC)[reply]