Goal: be able to undo Cayley's theorem; be able to recover (R,+,0) from (Tr,∘,id)
Introduction
The modern notion of a group does not come with any ‘semantics’ attached; a group need be no more than a collection of things (having significance or not) satisfying the proper axioms.
However, many groups do admit some canonical interpretation:
One may wonder whether there are mechanical ways to go between groups and their interpretations: from (R,+,0), can we recover translations? From the group of translations, can we recover (R,+,0)?
Cayley's theorem provides a partial answer for one direction. In particular, for each groupG, Cayley's theorem produces an isomorphism between G and a symmetric group over G. This can be seen as giving an interpretation to G. In some cases Cayley's theorem produces an interpretation that matches our intuition; it interprets (R,+,0) and (Z,+,0) as translation and (C,⋅,1) as combined dilation/rotation.
Here we provide a partial answer in the opposite direction. In particular, given a permutationgroup over a setA, and assuming the groupsatisfies some preconditions, we are able to construct an isomorphicgroup whose underlying set is A. In some cases the group produced is the one we want; for instance, from translations over the reals we recover (R,+,0) (up to a particular stringent form of isomorphism).
Concept
The general idea of the contruction can be seen by looking at the particular example of translations over R, which we will denote as (Tr, ; ,id)⊆S(R). This group is isomorphic with (R,+,0).
If one thinks of every point r∈R as specifying the (unique!) translation taking 0 to r, you construct the groupTr out of R. (This is exactly what Cayley's theorem on R produces.)
To construct R out of Tr, then, it seems we would want to identify each translation t∈Tr with the image of 0 under that translation.
Indeed this would work, but it “pulls the element 0∈R out of thin air”, seeming to use knowledge that we aim to construct, in particular, (R,+,0). Morally speaking we want a construction that uses no such “extra knowledge”. Hence we need to identify “from the outside” what the identity of our target group is.
This is likely impossible. After all, instead of thinking of any r∈R as specifying the translation taking 0 to r, we could think of it as specifying the translation taking, say, 12 to r. Addition over R would then be given by
a+′b=12+(a−12)+(b−12)
with identity 12. This feels disconcerting because we are used to the particular group(R,+,0), but in fact there’s nothing wrong with it. If I had an infinitely-long ruler, I could choose for “zero” to be at any point along it, and all choices will work just as well as each other—one just need fixsome origin.
Point being, 0 is not actually special in order to create an “addition” operator that acts like translation; for every r0∈R we can construct a group(R,+′,r0)acting like translation.
Hence, we should not expect to be able to recover (R,+,0) on the nose from Tr, and in the construction presented we do not do so. Instead we recover all “translated” groups(R,+′,r0), which by the above discussion is likely the best one can hope for.
〈🌷 6〉 If z is an identifier for G, then all x∈orb(z) are also identifiers.
Let x∈orb(z), and assume xf=x. Then z(z→x)f(z→x)−1=xf(z→x)−1=x(z→x)−1=z and hence (z→x)f(z→x)−1=id and hence f=id. This satisfies definition 〈🌷 2〉.
As a consequence of uniqueness in 〈🌷 4〉, we have for any identifier z that the set of arrows {z→a:a∈orb(z)} is just G.
Let x,y be identifiers in the same orbit. Then x→y=(y→x)−1.
Proof: both functions take x to y, and x is an identifier.
Remarks
The equivalence〈🌷 2〉⇔〈🌷 3〉 is trivial, but we include both definitions in light of their differing qualities
Definition 〈🌷 2〉 is more attuned to proof-writing. Definition 〈🌷 3〉 makes it easy to quickly spot the identifiers of a known G: think of all the actionsf∈G, think of their pertubation setspb(f), and think whether any element lies in all of them. (This quickly reveals, for instance, that the permutationgroup of linear transformations over R2 has many identifiers, but the permutationgroup of Rubik’s Cube movements has none.)
As discussed in the section “Concept”, taking a permutationgroupG⊆S(X) and trying to construct an isomorphicgroup(G⊆X,⋆,e) runs into the problem of not being able to pick out the proper identity e∈X.
Instead of picking one identity, we pick a set of values from X that can work equally well as identities; this is exactly the set of identifiers. In other words, the point of the ‘identifier’ condition is to pick out elements that can act as identities in our construction.
Consider the group over R2 consisting of rotations of the unit ball. Every element of the unit ball is an identifier, and every element outside of it is not.
I found this particular group to be a very useful example to have in mind as I did all this math. Usually when I wrote G I was thinking about this group.
Let (G, ; ,id)⊆S(X) and let z be an identifier for G.
Then the orbitorb(z) admits the following groupstructurea⊕b:=a(z→b)⊖a:=z(a→z)e:=z
The idea is to think of z as a choice for “zero” and then identify each a∈orb(z) with the arrow z→a. The definition for a⊕b is interpreted as “starting from a, travel the length of b aka z→b". The definition for ⊖ is interpreted as “travel the length of b, backwards, starting at z". (Consider what this looks like on translations over R)
We now present a structure-preserving bijection(φ,ψ):(orb(z),⊕,⊖,z)≅(G, ; ,(−)−1,id). This suffices to prove both that (1) (orb(z),⊕,z) is a group (with inverses given by ⊖); and (2) it’s isomorphic with (G, ; ,id).
Define
φ:orb(z)gf(z)⇄G:ψ↦(z→g)↤f
This is a bijection, as gφψ=z(z→g)=g and (fψφ)(z)=(z→f(z))(z)=f(z) which by 〈🌷 1〉 means fψφ=f.
Also, φ preserves all desired structure:
⊕ is preserved
φ(g⊕h)(z)=(z→g(z→h))(z)=g(z→h)=(z→g ; z→h)(z)=(gφ ; hφ)(z)
and hence φ(g⊕h)=gφ ; hφ since z is an identifier
⊖ is preserved
φ(⊖a)(z)=(z→z(a→z))(z)=z(a→z)=z(z→a)−1=(aφ)−1(z)
and hence φ(⊖a)=(aφ)−1 since z is an identifier.
e is preserved
(eφ)(z)=(z→z)(z)=z=id(z)
and hence eφ=id since z is an identifier.
This construction is called the orbit groupOz(G).
Properties
Due to 〈🌷 6〉, each y∈orb(z) is also an identifier, so it admits its own groupOy(G).
These groups will be isomorphic and have the same underlying setorb(y)=orb(z) and hence differ by automorphism only.
〈🌷 7〉 If G is commutative, then the automorphism is specifically (z→y)=(−⊕(y⊖z))=(−⊕y)∈S(orb(z)). We say that the groups differ by translation.
Proof. Let z,y be identifiers in the same orbit, and induce Oz(G) and Oy(G).
By construction, these groups will differ exactly by the composite isomorphismOz(G)⇄φz,ψzG⇄φy,ψyOy(G)
Which is a permutation of orb(z)=orb(y) and is given by
(φx ; ψy)(g)=y(z→g)
If G is commutative then so is Oz(G) and we have
y(z→g)=y⊕g=g⊕y=g(z→y)meaning that
φx ; ψy=z→y
〈🌷 8〉 If G⊆S(X) is an image under Cayley's theorem of a group(G,⋆,e), then e will be an identifier for G and Oe(G) will be exactly G. By 〈🌷 7〉, the worst that will happen if we choose z=e is automorphing G by translation.
Hence, the orbitgroup construction is a limited inverse of Cayley's theorem.
Remarks
Result 〈🌷 7〉 applied to the permutationgroup(Tr(R), ; ,id) of translations over R says that all orbitgroups on Tr(R) are constructed by taking an r0∈R and defining a group(R,+′,r0) by
a+′b=r0+(a−r0)+(b−r0)
This matches what was teased in the section “Concept”.
Conjecture — there should be some kind of result that states that arrows are “invariant under translation” in some sense, like (a⊕b)(q→r)=a(q→r)⊕b=b(q→r)⊕a=z(q→r)⊕(a⊕b)
Conjecture — this construction is the only natural (in the sense of CT) way to undo Cayley's theorem
Conjecture — z is an identifier if and only if∣orb(z)∣=∣G∣
Conjecture — there’s some relationship between identifiers of permutationgroups and the significance of an identity element in a monoid when viewed as right-actions (see Monoids as “self-actions” in monoids-n-ways). The identity element is useful because it means that a monoid element can be recovered from its right-action, just like how an identifier is useful because it means that a function can be recovered by the value it takes on at the identifier.
We should be able to extend this construction to work on anypermutationgroup as follows. Let G⊆S(X). For a cardinalα, define an α-identifier to be an α-tuple (z1…zα) such that (z1f…zαf)=(z1…zα)⟹f=id. Now construct the “orbitgroup” whose underlying set is the “tuple orbit” {(z1f…zαf):f∈G} and where an element (a1…aα) is interpreted as the “tuple arrow” (z1…zα)→(a1…aα); ie, the uniquef taking (z1…zα) to (a1…aα). We expect this group to be isomorphic to G. Its underlying set will be a subset of Xα. This construction works on all G because every G will at least admit the ∣X∣-identifier which is the tuple of all elements of X; we expect that it will perform better on most groups, though. Also, letting dim(G) be the leastα such that G admits an α-identifier, I conjecture that dim(G⊗H)=dim(G)+dim(H) (nb. for E the trivial group, dim(E)=0).