Algebraic Hoop
Algebraic HoopThis object (now called a hoop) is a non-empty set S combined with a BinaryOperator, "*", where the following axioms are satisfied -
What is this property; is it already proved on this page? -- sh
The property is that for any finite hoop H=(S,*), S has order 3^n, where n is an integer. Note that n is called the size of the hoop.
Has the above sentence discouraged further contributions to this page?
The simplest example uses a set with just one element.
Naming ideas: words similar to 'group', such as 'hoop', 'soup, 'troop', and 'scoop'.
Has anyone noticed this?
1. g*(g*h) = h (self-inverse) 2. (g*g)*(g*h) = h (self-distributivity) 3. g*g = g (???)Is this valid? Or are there cases where a != b and a*c = b*c? -- TaralDragon
Try this reasoning
1. g*(g*h) = (g*g)*(g*h) (self-distributivity) 2. (g*(g*h))*(g*h) = ((g*g)*(g*h))*(g*h) (right * (g*h)) 3. (g*h)*((g*h)*g) = (g*h)*((g*h)*(g*g)) (repeated commutativity) 4.g = (g*g)(repeated self-inverse)So, hoops are idempotent. -- SunirShah
Hmm... so does a*c = b*c -> a = b? -- td [Yes (see below). This could be termed cancellation.]
a = c*(c*a) = c*(a*c) = c*(b*c) = c*(c*b) = b
The fact that a * a = a was already noted (see the line beginning 'first of all' lower down), but it's worth having proofs in full as it's all too easy to make a slip when trying to be brief. There are less trivial identities waiting to be found.
Hmmm, 4 is the killer. It's an exclusive-or style thingy until 4 sticks its nose in. Are all members "1" ? Perhaps you could give an example as I am JustAProgrammer ;). -- RichardHenderson
To avoid excessive use of parentheses, it is convenient to adopt the convention that an ambiguous expression, such as a * b * (c * d) is evaluated on a 'left to right' basis, i.e., as though written (a * b) * (c * d).
In appropriate contexts, expressions can be further shortened by omitting the '*' symbol entirely.
It might be useful to differentiate between a*b and ab, according to the precedence. So it would hold that ab*cd = (a*b)*(c*d) = (ab)(cd) = (a*b)(c*d). -- anon
Indeed, and combining this with the previous shortcut means one would write simply ab cd. An expression such as ab cde ef would then mean ((a * b) * ((c * d) * e)) * (e * f). Most use of parentheses can be avoided by these means.
I should think having different notations for the same operator be differently ranked is worse than having parentheses. What do we have against them, anyways?
For example, one can easily prove that (ab)(cd) = abcdc for any a,b,c,d in a hoop. In general, (x1...xn)(y1...ym) = x1...xnym...1...ym. I made some typos in previous versions, but this one should work: supposing (x[1]...x[n])(y[1]...y[m]) = x[1]...x[n]y[m]...y[1]...y[m], then (x[1]...x[n])(y[1]...y[m+1]) = (x[1]...x[n])((y1...y[m])y[m+1]) = x[1]...x[n]y[m+1](y[1]...y[m])y[m+1] = ((x[1]...x[n]y[m+1])(y[1]...y[m]))y[m+1] = (x[1]...x[n]y[m+1]y[m]...y[1]...y[m])y[m+1] = x[1]...x[n]y[m+1]...y[1]...y[m+1]. Happy?
Yes! It's a good example of where the suggested shortcut notation should make such things easier to follow, but any minor typo will frustrate that.
For some fun, see if you can simplify abcdbcdbcdbcdbcdbcdbc.
As an easy exercise, using such notation, prove that if a hoop has exactly N elements, N is 1 or a multiple of 3.
As a further exercise, construct a hoop for which abcd = adcb is not always true. Such a hoop will serve as a counterexample for some of the hypotheses suggested below.
Specifically, o(ab) multiplications are associative if and only if abcd = adcb always holds. Constructing a counter-example and proving it works don't seem trivial. I'd be interested to hear some details of your solution, if you're willing to give them out. -- JoshuaGrosse
[It's straightforward to construct (without proof) a counterexample. Given patience, one can find the smallest, which has 81 elements. (Consider what hoops can be generated by four elements.)]
Note that it is not demanded that associativity of * holds. It is easy to see that any non-trivial "hoop" must have a non-associative *. I will show the following results:
1. every hoop with an identity consists only of that identity 2. every associative hoop has an identityAd 1: Suppose we have an identity, e.g. an e with e*g=g*e = g for all g. Let f be an element of the hoop. Then:
f = f*e = f*(e*e) = (f*e)*(f*e) = f*(f*e) = eAd 2: Let the hoop be associative, e.g. f*(g*h) = (f*g)*h. Define e = f*f, for some f in the hoop. We will show that e is an identity. Let g be some element in the hoop. Then:
g*e = e*g = (f*f)*g = f*(f*g) = gThus the only associative hoop is {e}.
So that would be a yes then? One element and it is the identity element? It did say empty sets were invalid at the start. -- RIH.
Sorry, missed the non-emptiness condition. It's not a yes: there might be non-trivial non-associative hoops.
I believe I have an example: a hoop with 3 elements a, b and c:
*| a b c ------------ a| a c b b| c b a c| b a c-- StephanHouben
[Congratulations. There are a few more things just as easy as this but not yet discovered on this page.]
I wrote this as you wrote that. I suppose we can always come up with a binary operator that satisfies the conditions for any known set. We would merely need to define rules for each possible combination of two elements such that they result in another element. This would allow us to quantify the number of possible distinct binary operators in terms of a factorial combination. Except for rule 4, which will limit us. Is that the point? -- RIH
There are non-trivial hoops. Let G be an abelian group in which every element has order 1 or 3. (So, if G is finitely generated, I suppose we must have G = C3^n, which is rather appropriate in this forum. Anyone got another problem that puts C2 everywhere?)
Define a*b = (ab)^2 for all a, b. Then (G,*) is a hoop.
I haven't been able (after maybe about 30 minutes of scribbling) to prove that all hoops arise in this way, nor to find any exceptions. Here's the line I took:
First of all, it's easy to show that a*a = a for any a.
Now choose any element o of the hoop, and define ab = o*(a*b). Then axiom 3 immediately gives ox=x for any x, and x(o*x)=o.
So we've made ourselves a commutative operation with identity. So, can we prove it's associative? I got all tangled when I tried.
(By the way, in the case G=C3, this example is the same as (1) Stephan's above and (2) one of the pile of examples I proposed before axiom #4 was added to the definition of a hoop.)
In reply to Richard: I don't think it's possible to come up with an operation for every ground set. For instance, every hoop has an odd number of elements.
(Proof: take any o as before. Note that o*a=a if and only if a=o. (Because a*(o*a) = o and a*a=a.) So we can pair off all elements other than o itself, pairing a with o*a. So we have a bunch of pairs, and o on its own: an odd number of elements.
Someone anonymous, whom I conjecture to be the original author of the page, said: "I'll have to check this offline. I suspect a minor slip somewhere. Otherwise, it's quite clever. How did you think of it? BTW, please explain the notation C3^n."
I'll answer the questions just as soon as I know what this person is querying. C3^n means the direct product of n copies of the cyclic group of order 3.
[The algebraic manipulation part was fine; I was merely unsure whether the 'I suppose we must have' assertion was correct.] No question about that. A finitely generated abelian group of exponent 3 is the product of a number of copies of C3. This follows immediately, e.g., from the structure theorem for finitely generated abelian groups, although you don't need nearly so much machinery to prove it. -- gjm
[OK - I retract my original doubt.]
After playing somewhat more with hoops, I came to roughly the same conclusions as Gareth. The trick to turn a hoop into an Abelian group is clever; I didn't think of that.
For every pair of hoops H1, H2 one can define the product hoop H1 x H2 with multiplication
(a1,a2)*(b1,b2) = (a1*b1, a2*b2)Since we have the C3 hoop, we can create an infinite number of (finite) hoops by taking the C3^N hoop for some natural number N. For N = 0, this gives us the original trivial hoop. I have the following conjecture (reformulation of what Gareth said)
Every finite hoop is isomorphic with C3^N for some N.
Outline of proof:
If this is true, then the number N could be called the dimension of the hoop. Of course, we can now also construct infinite hoops, e.g. the hoop consisting of functions from the natural numbers to C3, with pointwise multiplication.
Hmm. Nice approach. Seems to work, too; well done. The theorem you've proved isn't quite a reformulation of my conjecture, though, because it only covers finite hoops. However, all is well, because we can make a hybrid of the two approaches. Define the "multiplication" operation as in my proposed strategy. To prove it's associative, choose any x,y,z and argue that the hoop generated by o,x,y,z is isomorphic to C3^n for n <= 3, using your strategy. We've got ourselves an abelian group of exponent 3, and we're done.
[Beware of counter-examples!] As I say below, I am no longer convinced that this works. -- gjm
Good! Let's work out this page into a paper on Hoop Theory ;-). One question I have left is if there is actually any practical application for hoops. Another one is more an observation: 3 is an essential number in hoop theory. It's quite rare in mathematics to see 3 play such an important role; it is more common to see 2 play such a role. (To answer your question about C2: that would be Boole algebra's.) Perhaps the Martians from WarOfTheWorlds? would have invented hoops before groups, since they seem to do everything in threesomes.
I'm rather fond of places where the number 5 is magical (usually because of the sqrt(5) in the roots of x^2=x+1, one way or another). Another place where 3s are prominent is in the theory of elliptic curves; the group-making trick above is an adaptation of the way one makes an elliptic curve into an abelian group (not necessarily of exponent 3, though). I'd be mildly surprised if hoops had any practical, or impractical, application, and I'd love to know why the original author of this page was considering them in the first place.
Some more results came up during dinner. An interesting property is that if x*y=z holds, then we also have x*z=y and y*z=x. In a sense, x*y=z forms a trinary relation R(x,y,z) which is symmetric on all pair of slots. This observation allows us to consider a hoop as a trigraph, i.e. a graph with the edges replaced by triangles. Every triangle corresponds to a C3 sub-hoop. As an exercise, try to make a drawing of the C3^2 hoop.
We can also give a formula for the number of C3 sub-hoops in a hoop: the number of C3 sub-hoops is (3^n*(3^n-1))/6, where n is the dimension of the hoop. Some numbers:
nnum. of C3 sub-hoops 11 212 3117 41080 59801Finally, it is worth noting that for every a, b in a hoop, there exists an automorphism phi that maps a to b. This automorphism is given by
phi(x) = a*(b*x)Proof that this is an automorphism: let p, q be elements of the hoop. Then:
phi(p)*phi(q) = (a*(b*p))*(a*(b*q)) = a*((b*p)*(b*q)) = a*(b*(p*q)) = phi(p*q)This means in essence that the hoop looks the same as seen from every element; a hoop has no hair.
The symmetric trinary thing was one of the features that made me think of elliptic curves. I like the automorphism thing very much.
However, I am no longer convinced by Phase II of the Houben Plan for proving that all hoops are induced by groups. Am I being dim, or is there some more work still to do? -- GarethMcCaughan
Did you actually find a proof for the associativity of the o*(a*b) multiplication? I tried to, but I didn't succeed. Anyway, the "every finite hoop is isomorphic with C3^N" theorem, doesn't depend on that. Does anyone need a more detailed proof than the outline given above?
And now about the "mystery theorem" of finite hoops. The C3^N theorem tells us that every finite hoop has 3^N elements, (3^N*(3^N-1))/6 C3 sub-hoops, and is generated by N+1 suitably chosen elements of the hoop (a basis). Are we getting warm?
[Hot. See note inserted at top of page, but counter-examples do exist for your proposed isomorphism 'theorem'.]
See AlgebraicHoopConstruction for the various threads regarding the 9-hoop, and how to construct it.
The whole thing needs to be refactored out of ThreadMode anyways, but I don't have time to try right now.
Today I attended a geometry seminar, talking about KnotTheory?, and these axioms came up! An algebra satisfying the above axioms, except perhaps commutativity, is called an involutory quandle. You can make them by taking the order 2 elements of a group and defining the operation a*b = ba(b^-1). For instance, the 3-hoop given above comes from S3, the symmetry group of the triangle, and it's not hard to prove that any group which gives the 3-hoop will contain some copy of S3. I haven't looked at any of the other hoops, and a cursory glance suggests it would be difficult to determine which groups give hoops, but I don't have the time to think about it in any detail at the moment. -- JoshuaGrosse
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