# Anagram quotient

### From BluWiki

## Contents

## 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 **Z**^{26}, 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 *S _{n}* of

*A*such that each

*An(S*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

_{n})*An(S)*is always free Abelian for a singleton

*S = {α}*, we omit singletons from our trajectories. Note that, if

*S*is contained in

_{1}*S*, then

_{2}*An(S*surjects onto the subgroup of

_{1})*An(S*generated by

_{2})*S*. Thus any finite collection of trajectories has an "upper bound" in the obvious sense.

_{1}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

*S*_{1} = {a, d}

_{1}= {a, d}

Since add = dad, ad = da.

*S*_{2} = {a, d, i, s}

_{2}= {a, d, i, s}

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

### Trajectory 2

*S*_{1} = {n, o}

_{1}= {n, o}

on = no.

### Trajectory 3

*S*_{1} = {a, b, r}

_{1}= {a, b, r}

bar = bra, so ar = ra. **Incomplete**.

### Trajectory 4

*S*_{1} = {a, e, p, s, t}

_{1}= {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.

*S*_{2} = {a, e, p, r, s, t}

_{2}= {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.

*S*_{3} = {a, e, n, p, r, s, t}

_{3}= {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.