Talk:Moufang loop

smallest

(1) Re - M(S3,2)is the smallest nonassociative Moufang loop, which has order 12. Can anyone provide a Cayley table for this? I can only find the three associative groups, and a query to sci.math (etc) elicited no response.

(2) Vector division and conservation. In Mathsource/4894 & /6189 I (empirically) establish that if Moufang Loops are used as multiplication tables for vectors, every vector has a multiplicative inverse, and the factors of the determinant of the symbolic loop table are conserved on multiplication and division; they are denominators for a partial fraction formulation of this inverse. This leads to the formulation of many named algebras. Would a note on this be acceptable? I ask on the discussion page because an entry on ArcTanh was rejected as "original research". Roger Beresford 08:14, 20 May 2006 (UTC)[reply]

You should be able to reconstruct the Cayley table for M(S_3,2) given the Cayley table for S_3 and the information given in the article. Regarding point 2: it sounds like what you are describing constitutes original research. See the offcial policy if you are in doubt. -- Fropuff 04:01, 22 May 2006 (UTC)[reply]

Thanks. (1)Unfortunately my efforts only give an associative table. (2) True. The rest is silence. Roger Beresford 14:29, 22 May 2006 (UTC)[reply]

I can at least explain why M(S₃,2) is nonassociative. Let g and h be two noncommuting elements of S₃ (such as (1 2 3) and (1 2)). Then

g(hu) = (hg)u ≠ (gh)u.

So g, h and u form a nonassociative triple. -- Fropuff 20:18, 22 May 2006 (UTC)[reply]

If you really want a Cayley table, download the LOOPS package for GAP mentioned in the references. Alternatively, see the reference to Goodaire et al. --Michael Kinyon 15:14, 3 August 2006 (UTC)[reply]

A Cayley table for M(S₃,2) can be seen at Theorem of the Day, no. 114 Charleswallingford (talk) 17:59, 24 June 2008 (UTC)[reply]

Equivalent identities?

Unless I'm deeply mistaken, the 4 defining Moufang identities are not equivalent in general, and the four of them have to be verified. (see Mathworld). I'm correcting this. 2A01:E34:EF75:CCE0:223:12FF:FE57:5ADD (talk) —Preceding undated comment added 11:06, 10 January 2017 (UTC)[reply]

Almost all sources list three identities, and call them all equivalent for loops. MathWorld makes no statement as to whether they are equivalent or not. Most of the on-line sources just mention that they are equivalent and refer to published sources. I will try to track down a precise reference. -- Walt Pohl (talk) 10:52, 2 September 2017 (UTC)[reply]

Hi,
I'm (definitely) not a specialist but this week, I've try to work on Moufang loops, here is what I know about the question:

For quasigroup, identities M3 ( (zx)(yz) = (z(xy))z ) and M4 ( (zx)(yz) = z((xy)z) ) are equivalent. You can easily get the identity element from both, then get flexibility, and with flexibility z((xy)z) = (z(xy))z. I guess that's why specialists don't care about one of them. We'll call them both M3.
From either M1 ( z(x(zy)) = ((zx)z)y ), M2 ( x(z(yz)) = ((xz)y)z ) or M3 (=M4) quasigroup, you can get : - identity element (see Kunen for M1 or M2) - flexibility and alternativity - existence of two side inverse - "simple" inverse property ( $a^{-1}(ab)=b$ and equivalents) - "inverse flexibility" $(ab)a^{-1}=a(ba^{-1})$ (you just need flexibility and "simple" inverse property to get it).

We'll call "Inverse identities" the identities:

II-1 :  $(xyx)(x^{-1}z)=x(yz)$ 
II-2 :  $(yx^{-1})(xzx)=(yz)x$

It's easy to get:

II-1 from M1 $(xyx)(x^{-1}z)=((xy)x)[x^{-1}z]=x(y\left[x(x^{-1}z)\right])=x(yz)$
II-2 from M2 (by symmetry)
With M3, I can't get any, but there is the equivalence (II-1 <=> II-2) like this (use "inverse flexibility"):
- $(yz)x=x(x^{-1}\left[(yz)x\right]x^{-1})x=x\left[x^{-1}(yz)\right]x$
- $(yx^{-1})(xzx)=(x(x^{-1}yx^{-1}))((xz)x)=x\left[(x^{-1}yx^{-1})(xz)\right]x$

I tried to get equivalence between each by making variables change. I didn't manage, for the following reasons:

From M1 : I need II-2 to get M2 or M3. (I don't have II-2 from M1)
From M2 : I need II-1 to get M1 or M3. (same)
From M3 : I need any of II-1 or II-2 to get M1 or M3. (I don't have any)

But we have then: (M1 and M2) = (M1 and M3) = (M2 and M3) = (M1 and M2 and M3), here "and" means "the quasigroup respects each ... and ...".
As I didn't get the equivalence, I tried to get from references, from Kunen, you get this reference:

Bruck (book): A survey of binary system Springer 1971

page 115: link to two articles:

Bruck (huge article): Contributions to the theory of loops, Transactions of the American Mathematical Society, Vol. 60, No. 2 (Sep., 1946), pp. 245-354
Bol (article) Gewebe und Gruppen ,Mathematische Annalen (December 1937, Volume 114, Issue 1, pp 414–431)

The second one (Bol) is in germain, Bruck pretends it prove M3 from M1. All I manage to understand was (p.4) M3 from M1 and M2 (and I already had). From Bruck, The theorem 7a page 58 (noted 301) is Moufang's theorem, there is a reference to lemma 4a (page 52, noted 295) which should prove the equivalence M1 - M3. The lemma make use of "autotopism group", a notion defined at page 42 (285) and which I don't understand (it sounds like any equational symmetry of a magma could define an automorphism on the magma and those automorphisms generate a group).
That's all I got. Hope it could help

--Un autre type (talk) 13:33, 26 April 2018 (UTC)[reply]

Hi again,
A little correction, I just got a part of the "autotopism philosophy", with this, we can get M1=M2 :
From M1 : $z(x(yx)=\left[((x^{-1}y^{-1})x^{-1})z^{-1}\right]^{-1}=\left[(x^{-1}(y^{-1}(x^{-1}z^{-1}))\right]^{-1}=((zx)y)x$ . Then M1 -> M2 same way, we get M2 -> M1. So we get M1=M2 (-> M3, because II-1 and II-2 are then true). Still don't know how to get M1 from M3, and I'll stop working on it for the moment.
Bye

--Un autre type (talk) 14:20, 26 April 2018 (UTC)[reply]