Talk:Moufang loop

(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)

: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)

Thanks. (1)Unfortunately my efforts only give an associative table.

(2) True. The rest is silence.

Roger Beresford 14:29, 22 May 2006 (UTC)

: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)

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)

A Cayley table for M(S₃,2) can be seen at [http://www.theoremoftheday.org/Theorems.html#114 Theorem of the Day, no. 114] Charleswallingford (talk) 17:59, 24 June 2008 (UTC)

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 [http://mathworld.wolfram.com/MoufangLoop.html Mathworld]). I'm correcting this. 2A01:E34:EF75:CCE0:223:12FF:FE57:5ADD (talk) —Preceding undated comment added 11:06, 10 January 2017 (UTC)

: 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)

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 $(xyx)(x^\{-1\}z)=((xy)x)[x^\{-1\}z]=x(y\backslash left[x(x^\{-1\}z)\backslash right])=x(yz)$
• 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"):
  - $(yz)x=x(x^\{-1\}\backslash left[(yz)x\backslash right]x^\{-1\})x=x\backslash left[x^\{-1\}(yz)\backslash right]x$
  - $(yx^\{-1\})(xzx)\; =(x(x^\{-1\}yx^\{-1\}))((xz)x)\; =x\backslash left[(x^\{-1\}yx^\{-1\})(xz)\backslash right]x$

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)

Hi again,
A little correction, I just got a part of the "autotopism philosophy", with this, we can get M1=M2 :

From M1 : $z(x(yx)=\backslash left[((x^\{-1\}y^\{-1\})x^\{-1\})z^\{-1\}\backslash right]^\{-1\}=\backslash left[(x^\{-1\}(y^\{-1\}(x^\{-1\}z^\{-1\}))\backslash right]^\{-1\}=((zx)y)x$. 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)