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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?

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    $\begingroup$ What does $\approx$ mean? $\endgroup$ – Gerry Myerson May 29 '18 at 22:50
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ Besides terminology, do you have a specific mathematical open problem to offer? $\endgroup$ – Wlod AA May 30 '18 at 4:00
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    $\begingroup$ 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
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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]).

References:

  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.

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    $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ 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
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    $\begingroup$ Center-by-metabelian is the the same as $[G'',G]=1$. $\endgroup$ – Derek Holt May 30 '18 at 17:32
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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.

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  • $\begingroup$ 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
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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.

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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.

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    $\begingroup$ 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

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