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Anagram quotient

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Definition of the anagram quotient

Let S be a subset of the English alphabet A = {a, b, c, ..., x, y, z}. Let An(S) be the quotient of the free group on S by the relation under which two words are equivalent if they are valid English words which are anagrams (for example, ape = pea in An(a, e, p)). The group An(A) is called the anagram quotient. Unlike the homonym quotient, the quotient of the free group on A by the relation under which two words are equivalent if they are valid English words which are homonymous (for example, led = lead in this group), this group is not trivial. In fact, each of the homomorphisms

word |-> (# of a's in word)
word |-> (# of z's in word)

into the integers, defined on the free group on 26 letters, is trivial on the kernel of the projection to An(A), so induces a map from An(A) to the integers. Clearly each such map is non-trivial. The question is whether these 26 maps provide an isomorphism to Z26, i.e., whether the anagram quotient is isomorphic to the free Abelian group on 26 letters. This page is dedicated to showing that many pairs of generators commute.

Notions of commutativity

This sort of definition introduces an interesting analogue of commutativity; namely, if σ is a permutation of {1, 2, ..., n} and α : {1, 2, ..., n} → G is a set map into a group G, then we may say that α σ-commutes if α(1)α(2)...α(n) = α(σ(1))α(σ(2))...α(σ(n)). Note that α σ-commutes if and only if the composition of α with σ (σ-1)-commutes.

It seems that the case n = 3 is worth some attention. Obviously the permutation σ = (1) is uninteresting, and any of (1 2), (2 3), (1 2 3) and (1 3 2) just express commutativity relations in the ordinary sense. This leaves σ = (1 3). To say that α σ-commutes (for this choice of σ) is to say that α(1) conjugates α(2)α(3) into α(3)α(2) (or similarly for the conjugation action of α(3) on α(2)α(1)). This kind of behaviour is constantly to be observed in the anagram quotient (where, for example, not = ton and bad = dab). Is this kind of "commutativity" of interest anywhere else in algebra?

Update: It appears that the so-called plactic algebra is defined using relations similar to this. However, a quick search (on Google and Mathscinet) hasn't turned up any useful references (elementary enough for me just to check the definition!).

Proofs of commutativity

Since there are many different ways to prove that a pair of generators commute, and since proofs become easier as the number of known commutativity relations increases, it seems worthwhile to have a place to accumulate such proofs. A first stab at sorting the proofs is organising them into "trajectories". A "trajectory" is a sequence of nested subsets Sn of A such that each An(Sn) is trivial. Ideally, a trajectory would consist of twenty-six sets, the first a singleton and each set a proper superset of the previous one; but in practice this seems unlikely to happen. Since An(S) is always free Abelian for a singleton S = {α}, we omit singletons from our trajectories. Note that, if S1 is contained in S2, then An(S1) surjects onto the subgroup of An(S2) generated by S1. Thus any finite collection of trajectories has an "upper bound" in the obvious sense.

A hope in this context is that, if An(S) (or even the subgroup of An(A) generated by S) can be proved to be commutative, eventually some group-theoretic facts might kick in and establish that the entire group is Abelian. However, as above, in practice this seems unlikely.

Trajectory 1

S1 = {a, d}

Since add = dad, ad = da.

S2 = {a, d, i, s}

Since aids = dais = adis, id = di. Since ads = sad, adis = aids = said = sadi = adsi, so is = si. Incomplete.

Trajectory 2

S1 = {n, o}

on = no.

Trajectory 3

S1 = {a, b, r}

bar = bra, so ar = ra. Incomplete.

Trajectory 4

S1 = {a, e, p, s, t}

In An({a, e, p, t}), five of the six possible pairs of generators commute. Since apt = pat, ap = pa. Since pea = ape = pae, ea = ae. Since ate = eat = aet, te = et. Since ate = tea = tae, at = ta. Since ptea = peat = tape = tpae, pt = tp. However, it seems difficult to prove that e and p commute in this group.

Since pets = pest, ts = st. Since sap = asp, sa = as. Since asps = pass = apss, sp = ps. Since seat = eats = esat, se = es. Since step = pest = stpe, ep = pe.

S2 = {a, e, p, r, s, t}

Since art = rat, ar = ra. Since art = tar = atr, rt = tr. Since are = ear = aer, re = er. Since prat = tarp = rpta, pr = rp. Since ears = sear = easr, rs = sr.

S3 = {a, e, n, p, r, s, t}

Since pans = span = pasn, ns = sn. Since ant = tan = atn, nt = tn. Since net = ten = ent, ne = en. Since apena = apnea = paean = apean, na = an. (It's probably less trouble to use the relation ante = neat to prove that a and n commute.) Since apn = pan = nap = anp, pn = np.

Miscellaneous observations

It's occurred to me (only a few years later) that, since producing trajectories is hard work not soon completed, but I do want to have a record of my efforts, it might be useful to record some commutativity observations even if they don't fill out any obvious trajectory. dab = bad and not = ton were observed above; also there is cork = rock (another example of the non-trivial kind of σ-commutativity discussed above). Others?

Other commutativity relations (some already proven above): west = wets, so st = ts. tries = tires, so ri = ir. quiet = quite, so et = te. tire = tier, so re = er.

Some relations in the homonym quotient: led = lead, so 1 = a. by = bye, so 1 = e. by = buy, so u = 1. mien = mean, so i = ie = ea = 1. die = dye (or, even better, I = eye), so 1 = i = y.

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This page was last modified on 16 September 2007, at 20:11.
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