Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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. $\ldots$ 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 $S=A\cup A^{-1}$ and $A^{-1}=\{a^{-1},b^{-1},c^{-1},d^{-1}\}$. Let $F_A$ the group of all reduced words in $S$. Words are reduced when they have been can no longer be simplified by cancelling adjacent $x$ and $x^{-1}$s (for $x\in A$).

Cyclically reduced words 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 $\hat{F_A}$, the quotient of $F_A$ by the equivalence relation of being cyclically reduced. (Note that this is not the equivalence relation of being conjugate.)

Let $\Phi$ be the set of functions $\varphi:A^4\rightarrow \hat{F_A}$ defining words in $\hat{F_A}$ that contain at least one instance of each free generator or its inverse. Now let $\Psi:\Phi\rightarrow \hat{F_A}$ be formally defined by $$\Psi(\varphi)(u,x,y,z)=\varphi(u,x,y,z)\varphi(x,y,z,u)\varphi(y,z,u,x)\varphi(z,u,x,y).$$ 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 $\Psi$.

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.


Let $$\begin{eqnarray*} W(a,b,c) & \triangleq & [a,b^{-1},c]^b \\ &=& b^{-1}[a,b^{-1}]^{-1}c^{-1}[a,b^{-1}]cb\\ &=&a^{-1}b^{-1}ac^{-1}a^{-1}bab^{-1}cb.\end{eqnarray*}$$ 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 is are strings $w_1$ and $w_2$ in four letters such that $w_1(a,b,c,d)$ that is the inverse of $w_2(b,c,d,a)$.

Update: In fact, the above observation is not just sufficient, but necessary to the existence of a 4D Hall-Witt. Consider for example the case where $W(a,b,c,d)$ was split up 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 $w_3(a,b,c,d)=w_1(b,c,d,a)^{-1}$ and $w_2(a,b,c,d)=w_2(b,c,d,a)^{-1}$. 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.

share|improve this question
This is most certainly not an answer, but I remember thinking about this myself a few years ago. I didn't go very far at all, but I vaguely remember reading a result that all the commutator identities in free groups can be derived from the basic "level 1" identities (e.g., $[xy,z]=[x,z]^y[y,z]$) and the Hall-Witt identity. I do not recall where I read this, but a good bet would be in something Dan Segal wrote (probably his book Words). –  user641 Dec 6 '12 at 1:05
You might have a look at terrytao.wordpress.com/2012/05/11/… which offers up a derivation of Hall-Witt and pointers to some extensions of it... –  Steven Stadnicki Dec 6 '12 at 1:14
@StevenStadnicki I saw his Cayley graph proof sometime ago and liked it a lot. Makes me wonder whether the question would be well posed in terms of Cayley graphs. –  Alexander Gruber Dec 6 '12 at 1:17
Related: negative answer @ MO –  Grigory M Dec 24 '13 at 6:29
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.