Lately I have been thinking about commutator formulas, sparked by rereading the following paragraph in Isaacs (p.125):

An amazing commutator formula is the Hall-Witt identity: [x,y−1,z]y[y,z−1,x]z[z,x−1,y]x=1, which holds for any three elements of every group. … One can think of the Hall-Witt formula as a kind of three-variable version of the much more elementary two-variable identity, [x,y][y,x]=1. This observation hints at the possibility that a corresponding four-variable formula might exist, but if there is such a four-variable identity, it has yet to be discovered.

To my knowledge a four-variable formula hasn’t been discovered since this was written. I was thinking about this and found myself unable to even decide whether or not I thought one could exist (i.e. whether one would, hypothetically, try to find an identity or disprove its existence). Thus, I am attempting to “write down the problem.”

How can one rigorously formulate the question, “Does a four-variable analog of the Hall-Witt identity exist?”

Let’s start by explicitly making the free group on 4 letters. Suppose we have a free generating set A={a,b,c,d}, so that S=A∪A−1 and A−1={a−1,b−1,c−1,d−1}. Let FA the group of all

reduced wordsin S. Words are reduced when they have been can no longer be simplified by cancelling adjacent x and x−1s (for x∈A).

Cyclically reducedwords are words where the first and last letters are not inverse to each other. Every word is conjugate to a cyclically reduced word, so we can consider ^FA, the quotient of FA by the equivalence relation of being cyclically reduced. (Note that this isnotthe equivalence relation of being conjugate.)Let Φ be the set of functions φ:A4→^FA defining words in ^FA that contain at least one instance of each free generator or its inverse. Now let Ψ:Φ→^FA be formally defined by Ψ(φ)(u,x,y,z)=φ(u,x,y,z)φ(x,y,z,u)φ(y,z,u,x)φ(z,u,x,y).

So, if a 4-variable Hall-Witt identity exists, it will be among the functions in the preimage of 0 (the function which just maps everything to the empty word) under Ψ.

Question:Assuming the above formulation is sound, does there exist a (nontrivial) four-variable analog to the Hall-Witt identity? Can this approach be used to further refine the question?My use of “nontrivial” above is somewhat ambiguous: what I mean is that the identity should be made with commutators and conjugations in a free set of four letters, and done so in a way that it does not reduce to the two- or three-variable commutator identities by a substitution.

## Progress.

Let W(a,b,c)≜ Let’s chop W(a,b,c) in half and name the two parts. Define w_1(a,b,c)\triangleq a^{-1}b^{-1}ac^{-1}a^{-1} \hspace{30pt} \text{and} \hspace{30pt} w_2(a,b,c)\triangleq bab^{-1}cb, so that W(a,b,c)=w_1(a,b,c)w_2(a,b,c). Now, the Hall-Witt Identity can be written as W(x,y,z)W(y,z,x)W(z,x,y)=1, that is, \overbrace{\underbrace{x^{-1}y^{-1}xz^{-1}x^{-1}}_{w_1(x,y,z)}\underbrace{yxy^{-1}zy}_{w_2(x,y,z)}}^{W(x,y,z)} \overbrace{\underbrace{y^{-1}z^{-1}yx^{-1}y^{-1}}_{w_1(y,z,x)}\underbrace{zyz^{-1}xz}_{w_2(y,z,x)}}^{W(y,z,x)} \overbrace{\underbrace{z^{-1}x^{-1}zy^{-1}z^{-1}}_{w_1(z,x,y)}\underbrace{xzx^{-1}yx}_{w_2(z,x,y)}}^{W(z,x,y)} =1.

It’s clear the cancellation works by w_1(b,c,a)=w_2(a,b,c)^{-1}. This makes sense: we should be able to cyclically permute the overall word and have it still work, since 1^a=1. So, what we should be looking for are four letter strings w_1 and w_2 such that w_1(a,b,c,d) is the inverse of w_2(b,c,d,a).

Update:

In fact, the above observation is not just sufficient, but necessary for the existence of a four-variable Hall-Witt identity. Consider for example the case where W(a,b,c,d) would be split into three subwords, rather than two: W(a,b,c,d)=w_1(a,b,c,d)w_2(a,b,c,d)w_3(a,b,c,d). We would in this case have to insist that w_3(a,b,c,d)=w_1(b,c,d,a)^{-1}\hspace{14pt}\text{ and }\hspace{14pt}w_2(a,b,c,d)=w_2(b,c,d,a)^{-1}. After the w_1‘s and w_3‘s cancelled out, we’d be left with w_2(x,y,z,u)w_2(y,z,u,x)w_2(z,u,x,y)w_2(u,x,y,z)=1. But of course asking whether that that can happen is the same as asking whether W can exist, so eventually we will see a separation into two subwords, or we have a contradiction by infinite descent. If we divided up W(a,b,c,d) into n>3 subwords, we would reduce in the case of odd n to the case of a self-inverse word analogously to the n=3 case, or we would have an even number of subwords, which is the same thing as the 2 subword case.So, it suffices to either find a word W(a,b,c,d)=w_1(a,b,c,d)w_2(a,b,c,d), nontrivial in each variable, such that w_1(a,b,c,d)=w_2(b,c,d,a)^{-1}, or to prove that such a word does not exist.

**Answer**

\renewcommand{\mod}[1]{~(\text{mod $#1$})}

\newcommand{congr}{\equiv}

The last statement in your question (which is not a mere question)

also suggests a solution of the more general problem,

namely the n-variable analog of the Hall-Witt formula for any n\geq 2.

Let x_1, \ldots, x_n be the variables.

If w=w(x_1,x_2,\ldots,x_n) is any word in x_i‘s and their inverses,

define the word \gamma w by

(\gamma w)(x_1,x_2,\ldots,x_n):=w(x_2,\ldots,x_n,x_1),

and set W:=w(\gamma w)^{-1}. Then

W(\gamma W)\cdots(\gamma^{n-1}W)=1~.

The requirement that W is nontrivial in each variable is easily satisfied.

In this way you can produce n-variable analogs of the Hall-Witt formula by truckload.

To make things more interesting, you may require, say,

that W must be representable by an expression

built from the variables and their inverses (the ‘tokens’)

by making commutators [u,v],

where u and v are already built expressions,

and by conjugations u^y,

where u is an already built expression, but not a token, and y is a token;

then the requirement is that W=u

for an expression built in this way that is not a single token.

*Now* the things are no longer so simple.

Am I far wrong in supposing that in fact you had some such additional conditions in mind

when you formulated your question?

I have no idea how to go about finding an example for n=4

or proving that it does not exist.

The only (rather weak) restriction on the word w I have found so far

is l_1=l_2=\cdots=l_n,

where l_i is the sum of the exponents

of the appearances of x_i^{\pm1} in w.

If we relax the condition imposed on W

and require only that W\in[G,G],

where G is the free group generated by the variables x_1, \ldots, x_n,

then we can give the full solution of the problem,

since for every w\in G we have w(\gamma w)^{-1}\in[G,G]

if and only if l_1(w)=l_2(w)=\cdots=l_n(w).

Here \gamma is the automorphism of G

that sends x_i to x_{i+1} for 1\leq i<n, and sends x_n to x_1.

We must also tell how the functions l_i: G\to\mathbb{Z} are defined.

Let A be the free abelian (additive) group generated by x_1, \ldots, x_n,

and let h\colon G\to A be the homomorphism

sending x_i\in G to x_i\in A for 1\leq i\leq n;

note that \ker h=[G,G].

For every w\in G we have h(w)=\sum_{i=1}^n l_i(w)x_i: this defines the l_i's.

For example, if w=x_1^{-1}x_2^{-1}\cdots x_{n-1}^{-1}x_n^{-1},

then W=w(\gamma w)^{-1}=[x_1,x_nx_{n-1}\cdots x_2].

When n=3 we obtain the identity

[x,zy][y,xz][z,yx]=1~,

which is a humble cousin of the Hall-Witt formula,

and is probably quite useless (nice as it is).

If you wish you can rewrite W=[x,zy]=x^{-1}y^{-1}z^{-1}xzy

as a product of iterated commutators, W=[x,y][x,z][[x,z],y],

or perhaps in the well-known form W=[x,y][x,z]^y.

(Mark that here [x,y]=x^{-1}y^{-1}xy;

some authors define the commutator as [x,y]:=xyx^{-1}y^{-1}.)

Note that the example w=w_1(a,b,c,d)=a^{-1}bc^{-1}b^{-1}dad^{-1}

provided by weux082690

has l_a(w)=l_b(w)=l_d(w)=0 but l_c(w)=-1,

thus w(\gamma w)^{-1}\notin[G,G].

By a sort of a retrograde progress, let us return to the beginning.

We are going to prove the following:

Let G be the free group with the free generators x_1, \ldots, x_n, n\geq 2,

and let \gamma be the automorphism

of G that rotates the generators one place to the right,

that is, \gamma x_i=x_{i+1} for 1\leq i<n and \gamma x_n=x_1.

Then g\in G has the property that g(\gamma g)\cdots(\gamma^{n-1}g)=1

iff there exists h\in G such that g=h(\gamma h)^{-1}.

**Proof.** ~The sufficiency is clear.

Necessity.

Let T be the set of 'tokens' \{x_1,x_1^{-1},\ldots,x_n,x_n^{-1}\},

and let T^* denote the free monoid (of 'words') generated by T;

the neutral element of T^* is the empty word \varepsilon.

Then G=T^*/{\sim}, where \sim is the congruence on the monoid T^*

generated by x_i^\alpha x_i^{-\alpha}\sim\varepsilon, 1\leq i\leq n, \alpha=\pm1.

The equivalence class of the empty word

is the multiplicative identity of the free group: \varepsilon/{\sim}=1_G=1.

Every equivalence class g\in G contains a unique reduced word \varrho(g)

which does not contain any pair of consecutive tokens that are inverses of each other.

Given a word w\in T^*, we obtain the reduced word \varrho(w/{\sim})

by repeatedly applying the reductions ux_i^{\alpha}x_i^{-\alpha}v\to uv,

1\leq i\leq n, \alpha=\pm1, u,v\in T^*;

it is easy to verify that this system of reductions is locally confluent,

thus it has the Church-Rosser property,

and so there is in fact a unique reduced word in each equivalence class.

The rotation \gamma of generators induces the double rotation of tokens

(the rotation of the tokens with the exponent 1

and also the rotation of the tokens with the exponent -1);

the automorphism of the free monoid T^* determined by this double rotation

we still denote by \gamma.

\quadLet w=x_{i_1}^{\alpha_1}x_{i_2}^{\alpha_2}\cdots x_{i_m}^{\alpha_m}\in T^*.

We denote by |w| the length m of the word w.

For each i, 1\leq i\leq n, we denote by l_i(w) the sum of the exponents

of all tokens x_i^{\pm1} appearing in w,

and by l(w) we denote the sum of all exponents

\alpha_1+\alpha_2+\cdots+\alpha_m=l_1(w)+\cdots+l_n(w).

Always |w|\congr l(w)\mod{2},

because every term in |w|-l(w)=(1-\alpha_1)+(1-\alpha_2)+\cdots+(1-\alpha_m)

is ether 0 or 2.

For each i, 1\leq i\leq n, we have l_i(uv)=l_i(u)+l_i(v) for all u,v\in T^*,

and l_i(w) is constant on every equivalence class g\in G.

\quadNow suppose that g\in G

has the property g(\gamma g)\cdots(\gamma^{n-1} g)=1,

and let w:=\varrho(g).

Then l_n(w)+l_n(\gamma w)+\cdots+l_n(\gamma^{n-1} w)=l_n(\varepsilon)=0;

since l_n(\gamma w)=l_{n-1}(w), \ldots, l_n(\gamma^{n-1} w)=l_1(w)

we have l(w)=l_1(w)+\cdots+l_n(w)=0,

therefore the word w is of even length, |w|=2k.

We split the word w as w=w_1w_2, where |w_1|=|w_2|=k.

We are assuming that k>0, since the case k=0 is trivial.

The reduction process must reduce the word

W := w_1w_2(\gamma w_1)(\gamma w_2)(\gamma^2w_1)(\gamma^2w_2)\cdots

(\gamma^{n-1}w_1)(\gamma^{n-1}w_2)

to the empty word.

Since the words w_1w_2, (\gamma w_1)(\gamma w_2), \ldots,

(\gamma^{n-1}w_1)(\gamma^{n-1}w_2) are reduced,

the only places in the word W where the reductions can be applied

are at the points of contact between subwords w_2 and \gamma w_1,

\gamma w_2 and \gamma^2 w_1, \ldots\,

Consider the effect of the reduction process on the subword w_2(\gamma w_1)

(mark that we can carry out the reductions in any order we choose,

the result will be always the same).

The first reduction eliminates some product t\,t^{-1}

from the center of w_2(\gamma w_1),

where t is the last token in w_2 and t^{-1} is the first token in \gamma w_1.

We are left with w_2'(\gamma w_1'), where |w_1'|=|w_2'|=k-1.

If k>1, there may be another reduction applicable

at the center of w_2'(\gamma w_1') (and nowhere else in this word), so we apply it, and so on.

In fact the reductions must proceed to the bitter end,

reducing the initial word w_2(\gamma w_1) to the final empty word.

For suppose that the reduction process stops after r<k reductions;

then the reduction process, applied to each of the subwords (\gamma w_2)(\gamma^2 w_1),

\ldots, (\gamma^{n-2}w_2)(\gamma^{n-1}w_1), will likewise stop after r reductions,

and we will have a nonempty reduced word on our hands, which cannot be,

because the word W must reduce to the empty word.

Let h:=w_1/{\sim}.

Since w_2(\gamma w_1)\sim\varepsilon,

it follows that w_2/{\sim}=(\gamma h)^{-1},

whence g=(w_1/{\sim})(w_2/{\sim})=h(\gamma h)^{-1}.~ **Done.**

**Attribution***Source : Link , Question Author : Alexander Gruber , Answer Author : chizhek*