BC MATHS
Saved by @Mollyycolllinss

Loops and their multiplication groups

A thread in 15 parts

(0/15)

 Recall that a quasigroup (Q,*) is a set Q with a binary operation * such that for each a,b in Q, the equations a*x=b and y*a=b have unique solutions x,y. Groups are quasigroups and this property is usually one of the first things proved in elementary group theory.

(1/15)
Note that we don't assume associativity of *!

A loop is a quasigroup with an identity element. The story of why they are called loops is an interesting one and may even be true, but I will save it for another day. I am going to focus on loops in this thread.

(2/15)
Natural examples of nonassociative loops:
- The nonzero octonions under multiplication
- The sphere S^7 under octonion multiplication
- I have discussed other examples previously:
https://t.co/q5LjmxHEIF
https://t.co/UPHSMwQo75
(3/15)
For each x in a loop Q, define the left & right translations L_x, R_x : Q->Q by L_x(y)=xy and R_x(y)=yx. These mappings are permutations of Q. The composition L_x L_y of two left translations is not necessarily a left translation because Q is not necessarily associative.

(4/15)
The Cayley table of a loop is a (possibly infinite) latin square with the first row and column corresponding to the identity element. You can visualize the L_x's as being the permutations corresponding to rows and the R_x's as being permutations corresponding to columns.

(5/15)
Let Mlt(Q) ("mult Q") be the group generated by all L_x's & R_x's. This is called the *multiplication group* of Q. In other words, it's the group generated by all rows and columns of the latin square of Q.

(6/15)
To help intuition, consider the case where Q is a group. Mlt(Q) puts together the left & right regular representations of Q into one group. QxQ acts on Q by (g,h)x = gxh^{-1} = R_h^{-1} L_g(x). This gives a homomorphism from QxQ to Mlt(Q).

(7/15)
Its kernel is K={(a,a)|a in Z(Q)} is a copy of the center of Q. Thus Mlt(Q) is isomorphic to QxQ/K, a complete description in this, the associative case.

Now back to the general case where Q is any loop.

(8/15)
Let Inn(Q) be the stabilizer in Mlt(Q) of the identity element, that is, the subgroup of all permutations in Mlt(Q) that fix the identity element. This is called the *inner mapping group* of Q. When Q is a group, Inn(Q) is precisely the inner automorphism group of Q.

(9/15)
Inn(Q) contains the familiar conjugations R_x^{-1} L_x. But it also contains permutations like L_{x*y}^{-1} L_x L_y, which measure nonassociativity.

Loops can be studied via their multiplication and inner mapping groups. I will give one example of how this works.

(10/15)
Some general permutation group theory: A block B of a permutation group G acting on a set X is a nonempty subset of X such that for all g in G, either gB = B (g fixes B) or gB\cap B is empty (g moves B entirely). The set {gB | g in G} forms a G-invariant partition of X.

(11/15)
For a subloop N of a loop Q, the following turn out to be equivalent:
1. N is the kernel of a homomorphism,
2. N is invariant under Inn(Q),
3. N is a block of Mlt(Q).
Such an N is called a normal subloop of Q. Q is simple if it has no nontrivial normal subloops.

(12/15)
A permutation group is *primitive* if it is transitive and has no nontrivial invariant partitions. (The trivial ones are the partition into singletons and the one part partition.)

Theorem (Albert 1941): A loop Q is simple if and only if Mlt(Q) is primitive.

(13/15)
This means that, in principle, to study simple loops, we should determine which primitive groups can occur as their multiplication groups. Much is still unknown about this!

(14/15)
Here is the takeaway for the main philosophical difference between semigroup theory and quasigroup theory:

Semigroups act.
Quasigroups are acted upon.

That's enough for one thread. As always, thanks for reading!

(15/15)

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It's also just kind of a cool combinatorial structure! I've been talking to @CreeepyJoe about this lately, as well as @grassmannian.

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Mathematically speaking.

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And why would the Black community be the first to get their own anthem over the American Indians? The Indians used to have all the land, and they still do have some of it.

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I chose to stick my neck out almost 4mo ago here:
https://t.co/b0rT5Lq9HI
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Please examine them, on the right, under the pie chart:


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