Lately, I have been constructing finite involution monoids that generate varieties with $2^{\aleph_0}$ subvarieties. One construction requires groups that violate the identity ${ [x,y]^2 \approx 1 }$, where ${ [x,y] = x^{-1} y^{-1} xy }$.

Is there a name for groups satisfying the identity ${ [x,y]^2 \approx 1 }$? Has there been any work done on these groups?

  • 7
    $\begingroup$ What does $\approx$ mean? $\endgroup$ – Gerry Myerson May 29 '18 at 22:50
  • 2
    $\begingroup$ I guess it means "identically equal to", i.e. "$w(x,y)\approx 1$" means "$w(x,y)=1$ for all $x,y$" $\endgroup$ – YCor May 29 '18 at 22:56
  • 2
    $\begingroup$ Yes, for example, a group $G$ that satisfies $xy \approx yx$ means that $ab=ba$ for all $a,b \in G$. $\endgroup$ – E W H Lee May 29 '18 at 23:59
  • 2
    $\begingroup$ Besides terminology, do you have a specific mathematical open problem to offer? $\endgroup$ – Wlod AA May 30 '18 at 4:00
  • 6
    $\begingroup$ Asking about existence of a non-metabelian example is a reasonable question, more interesting than the terminology one (which probably has no answer). $\endgroup$ – YCor May 30 '18 at 7:33

This variety of groups has indeed been considered in the literature. It is known that the following conditions hold for every group $G$ satisfying the identity $[x,y]^2=1$:

  1. $[[x,y_1,\ldots,y_m],[x,z_1,\ldots,z_n]]=1$ for all $x,y_1,\ldots,y_m,z_1,\ldots,z_n\in G$ (see [1]).
  2. $[[x_1,x_2],[x_3,x_4]]=[[x_{\pi(1)},x_{\pi(2)}],[x_{\pi(3)},x_{\pi(4)}]]$ for all $x_1,\ldots,x_4\in G$ and $\pi\in S_4$ (see [2]).
  3. $[[x,y],[z,w]]=[x,y,z,w][x,y,w,z]$ (see [1]).
  4. $[\gamma_2(G),\gamma_3(G)]=1$ (see [2]).
  5. $[G'',G]=1$ (see [2]).
  6. $G'^4=1$ (see [2]).
  7. The group $G$ need not be metabelian (see [3]).
  8. If $G$ is finite, then $G=P\rtimes H$, where $P$ is a normal Sylow $2$-subgroup of $G$, $H$ is abelian of odd order, and $[P,H]$ is an elementary abelian $2$-group (see [1]).


  1. M. Farrokhi D. G., On groups satisfying a symmetric Engel word, Ric. Mat. 65 (2016), 15–20.

  2. I. D. Macdonald, On certain varieties of groups, Math. Z. 76, (1961) 270–282.

  3. B. H. Neumann, On a conjecture of Hanna Neumann, Proc. Glasgow Math. Assoc. 3 (1956), 13–17.

  • 2
    $\begingroup$ Could you be more explicit about (7)? in [3] there are two examples of finite groups, of order $2^7$ and $2^{14}$ respectively, that are not metabelian and have every 2-generator subgroup being metabelian. But I don't see the claim that they satisfy the identity $[x,y]^2$. Is it the case? $\endgroup$ – YCor May 30 '18 at 12:24
  • 1
    $\begingroup$ @YCor: I have checked that the first example in [3] does not satisfy $[x,y]^2=1$ in general. I have not understood the second example. $\endgroup$ – Neil Strickland May 30 '18 at 12:29
  • $\begingroup$ @YCor: In [2] (on page 279) Macdonald says that "it is not difficult to verify that" the second example in [3] satisfies $[x,y]^2=1$ $\endgroup$ – Neil Strickland May 30 '18 at 12:41
  • 4
    $\begingroup$ Since the question is "Is there a name for groups satisfying …?", what 'problem' do you mean when you say "The problem is settled before"? $\endgroup$ – LSpice May 30 '18 at 13:42
  • 2
    $\begingroup$ Center-by-metabelian is the the same as $[G'',G]=1$. $\endgroup$ – Derek Holt May 30 '18 at 17:32

Not every group satisfying the law $[x,y]^2=1$ is metabelian. But by Theorem 4 of McDonald, I. D. "On certain varieties of groups" Math. Z. 76 1961 270–282. every group satisfying this law has second derived subgroup in the center so it is center-by-metabelian, and the derived subgroup of exponent 4.

Note. I had a wrong answer before and did not notice Farrokhi's answer when I made corrections - one hour after his answer. Farrokhi's answer is more complete.

  • $\begingroup$ In the dihedral group $\langle e,f\rangle$ of order 8, $[e,f]$ has order 2. $\endgroup$ – YCor May 30 '18 at 7:01
  • $\begingroup$ That is correct. $\endgroup$ – Mark Sapir May 30 '18 at 7:02
  • $\begingroup$ I guess you mean some particular central extension. Which one? $\endgroup$ – YCor May 30 '18 at 13:27
  • $\begingroup$ Yes, I have described it now. $\endgroup$ – Mark Sapir May 30 '18 at 13:35
  • $\begingroup$ Is that not metabelian? It looks to me like the commutator subgroup is generated by elements $x_ix_j$ and $y_my_n$ and $c$, and those all commute. $\endgroup$ – Neil Strickland May 30 '18 at 13:49

Here are some additional details for the answer of M. Farrokhi. In the paper " On a conjecture of Hanna Neumann", B.H.Neumann constructs a certain group $G$. There are generators $a_1,\dotsc,a_4$, and additional elements defined in terms of these as follows: \begin{align*} b_{12} &= [a_1,a_2] = [a_2,a_1] \\ b_{13} &= [a_1,a_3] = [a_3,a_1] \\ b_{14} &= [a_1,a_4] = [a_4,a_1] \\ b_{23} &= [a_2,a_3] = [a_3,a_2] \\ b_{24} &= [a_2,a_4] = [a_4,a_2] \\ b_{34} &= [a_3,a_4] = [a_4,a_3] \\ c_1 &= [a_2,b_{34}] = [a_4,b_{23}] \\ c_2 &= [a_3,b_{14}] = [a_4,b_{13}] \\ c_3 &= [a_4,b_{12}] = [a_1,b_{24}] \\ d &= [a_1,[a_2,[a_3,a_4]] = [a_2,[a_3,[a_4,a_1]] = [a_3,[a_4,[a_1,a_2]] \end{align*}

There are some relations implicit in the above equations. There are also additional relations as follows:

  • $a_i^2=b_{jk}^2=c_l^2=d^2=1$
  • All commutators $[a_i,b_{jk}]$ that have not already been listed, are trivial.
  • $[a_i,c_j]=1$ whenever $i\neq j$, and $[a_i,d]=1$

We can define a map $\phi\colon\{0,1\}^{14}\to G$ by $$ \phi(u) = a_1^{u_1}\dotsb a_4^{u_4} b_{12}^{u_5} \dotsb b_{34}^{u_{10}} c_1^{u_{11}}c_2^{u_{12}}c_3^{u_{13}}d^{u_{14}} $$ One can check that this is bijective, and one can write formulae for the permutations of $\{0,1\}^{14}$ corresponding to right multiplication by the elements $a_i$, $b_{jk}$, $c_l$ and $d$. In particular, this proves that $|G|=2^{14}$. One can also check that $$ [b_{12},b_{34}] = [b_{13},b_{24}] = [b_{14},b_{23}] = d \neq 1, $$ so $G$ is not metabelian.

In the paper "On certain varieties of groups", Macdonald states that it is easy to verify that the above group has $[x,y]^2=1$ for all $x,y\in G$. I don't see how to prove this myself. However, I have checked it by computer for 10000 randomly chosen pairs $(x,y)$, so it must be true.


Here are some preliminary results. Let $F$ be freely generated by $x$ and $y$, let $N\leq F$ be generated by all squares of commutators, and put $G=F/N$. We are interested in the structure of $G$ and $G'$. For $i,j\in\mathbb{Z}$ put $z_{ij}=[x^i,y^j]$ (which might be interepreted as an element of $F$ or $G$). I believe that $F'$ is generated by the elements $z_{ij}$ subject only to $z_{0j}=z_{i0}=1$, so $G'$ is also generated by the $z_{ij}$. In $G$ we have $z_{ij}^2=1$, but there are extra relations as well.

First, as $z_{ij}^2=z_{kl}^2=1$ we see that $[z_{ij},z_{kl}]=(z_{ij}z_{kl})^2$, and the square of this must be trivial, so $(z_{ij}z_{kl})^4=1$. If we could improve this to $(z_{ij}z_{kl})^2=1$ then we would see that $G'$ is abelian.

In $F$ one can check that $z_{ip}z_{jp}^{-1}=[x^{i-j},x^jy^p]$, so in $G$ we have $(z_{ip}z_{jp}^{-1})^2=1$ as well as $z_{ip}^2=z_{jp}^2=1$. From this it follows easily that $z_{ip}$ commutes with $z_{jp}$. Similarly, we see that $z_{mi}$ commutes with $z_{mj}$.

Along the same lines, one can check that $$ [x^iy^j,x^ky^l] = z_{ij}\;z_{i+k,j}^{-1}\;z_{i+k,l}\;z_{k,l}^{-1} $$ so the square of the right hand side is the identity. Adjacent terms on the right hand side commute in $G$ by the two rules that we have already established.

This feels like it is getting close, but I am not sure what to do next.

  • 1
    $\begingroup$ I believe the group G generated by two elements of order three with all commutators having order 2 has order 288, with commutator subgroup G' having order 32, and is non-abelian, G'' being the group of order 2. I have only checked a few of the orders of the commutators though, so this could be erroneous. $\endgroup$ – Thomas May 30 '18 at 10:18
  • $\begingroup$ @Thomas: It seems that your group is this one: people.maths.bris.ac.uk/~matyd/GroupNames/288/Omega+(4,3).html. Randomly chosen examples show that some commutators have order $4$, so this is not a counterexample. $\endgroup$ – Neil Strickland May 30 '18 at 10:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.