Let $G = \left\langle S | R \right\rangle$ be a finitely presented group where S is a set of generators and R is a set of relations. We say that the presentation is "locally commuting" if whenever two generators $a, b$ appear in a word in R, the word $aba^{-1}b^{-1}$ also belongs to R. Following Slofstra, I'll call a group which admits a locally commuting presentation a "solution group".

Problem: Which finite nonabelian groups are solution groups?

(Slofstra showed that every finitely-presented group embeds into a solution group. However, the methods used there seem too coarse for this problem. In particular, he gives a construction which takes finitely-presented G and outputs a solution group G' and embedding G -> G', but if we feed different presentations of the same group in for G we can get out different groups for G'.)

I know of one infinite family of nonabelian finite solution groups. Quantum computing theorists know these as the $n$-qubit Pauli groups for $n\geq 2$. They are also known as the Heisenberg groups over $\mathbb F_2$. These are the $(n+2)\times (n+2)$ upper-triangular matrices with the following structure: $$ \begin{pmatrix} 1 & a & c \\ 0 & I_n & b \\ 0 & 0 & 1 \end{pmatrix},\quad a,b, \in \mathbb F_2^n, c \in \mathbb F_2. $$ There are two cute constructions for $n=2$ and $n=3$. Larger $n$ can be formed by appropriate products.

Problem, restated: Are there any nonabelian finite solution groups other than the Pauli groups on $n$-qubits, $n \geq 2$?

Here I'll show the cute constructions. This paper with Andrea Coladangelo has picture-proofs that these presentations give the right groups. (The results there stated for general $d$ are false, but still hold when $d=2$.)

For $n=2$, we have the Mermin--Peres "Magic Square". $G = \left\langle S | R_0 \cup R_1 \cup R_2 \cup R_\text{comm} \right \rangle$, where

$S = \left\{J,e_1,\ldots,e_9\right\}$ $R_0 = \left\{[s, J] | s\in S \right\},$ $R_1 = \left\{s^2 | s \in S \right\},$ $R_2 = \left\{e_1e_2e_3, e_4e_5e_6, e_7e_8e_9, e_1e_4e_7, e_2e_5e_8, Je_3e_6e_9\right\},$ $R_\text{comm} = \left\{[e_i,e_j] | \text{$e_i$ and $e_j$ appear together in some relation of $R_2$} \right\}$.

(Here $[x,y] := xyx^{-1}y^{-1}$ denotes the group commutator.)

For $n=3$, we have the Mermin--Peres "Magic Pentagram". $G = \left\langle S | R_0 \cup R_1 \cup R_2 \cup R_\text{comm} \right \rangle$, where

$S = \left\{J,e_1,\ldots,e_{10}\right\}$ $R_0 = \left\{[s, J] | s \in S \right\},$ $R_1 = \left\{s^2 | s \in S \right\},$ $R_2 = \left\{e_1e_2e_8e_9, e_2e_3e_6e_7, e_3e_4e_9e_{10}, e_4e_5e_7e_8, Je_5e_6e_{10}e_1\right\},$ $R_\text{comm} = \left\{[e_i,e_j] | \text{$e_i$ and $e_j$ appear together in some relation of $R_2$} \right\}$

  • $\begingroup$ What are $J$ and $S$? $\endgroup$ – M. Farrokhi D. G. Dec 12 '18 at 9:51
  • $\begingroup$ Edited for clarity. S is the set of generators and J is a specific generator which is a central involution. $\endgroup$ – Jalex Stark Dec 12 '18 at 15:34
  • 1
    $\begingroup$ Are you saying that you have a paper on the arXiv with results that you know are wrong? If so, then you should probably update or withdraw it …. $\endgroup$ – LSpice Dec 15 '18 at 15:56
  • $\begingroup$ Yes; we were just made aware of the error recently and are working to put out a revision. $\endgroup$ – Jalex Stark Dec 16 '18 at 9:55

Base on the case $n=2$ with $J=\emptyset$ (the most simple case), we can construct non-abelian $p$-groups for every prime $p$. Let $$G_p=\langle x_1,\ldots,x_9:R\rangle,$$ where $R=R_1\cup R_2\cup R_3$ with $$R_1=\{x_1^p,\ldots,x_9^p\},$$ $$R_2=\{x_1x_2x_3,x_4x_5x_6,x_7x_8x_9,x_1x_4x_7,x_2x_5x_8\},$$ and $$R_3=\{[x_i,x_j]:x_i,x_j\ \text{ belong to a relation in }R_2\}.$$ Then $$G_p\cong(C_p \times ((C_p \times C_p) \rtimes C_p)) \rtimes C_p$$ is a nonabelian finite $p$-group. Of course, this is not a new group!

  • $\begingroup$ How did you deduce the isomorphism? It's not immediately obvious to me. $\endgroup$ – Jalex Stark Dec 12 '18 at 15:37
  • $\begingroup$ It seems to me that 1. $x_7$ is central. 2. If you quotient out $x_7$, the quotient group is a free product between the subgroup generated by $x_1, \ldots, x_6$ and the subgroup generated by $x_8,x_9$. Your $R_2$ has 5 relations. Is it meant to have more? $\endgroup$ – Jalex Stark Dec 12 '18 at 16:35
  • $\begingroup$ @Jalex Stark Sorry! There was a typo in the last relation. The group $G_p$ is the extra-special $p$-group of order $p^5$ with exponent $p$. Indeed, $G_p$ is the central product of the $p$-groups $\langle x_1,x_5\rangle$ and $\langle x_2,x_4\rangle$ of order $p^3$ and exponent $p$. The generators, $x_3,x_6,x_7,x_8,x_9$ are superfluous and can be omitted. $\endgroup$ – M. Farrokhi D. G. Dec 15 '18 at 7:24
  • $\begingroup$ The above groups $G_p$ can be defined not only for primes $p$ but also for any natural number $n$ in the same way. For instance, $G_4\cong (C_4 \times ((C_4 \times C_4) \rtimes C_4)) \rtimes C_4$ and $G_{15}\cong \big((C_3 \times ((C_3 \times C_3) \rtimes C_3)) \rtimes C_3\big) \times \big((C_5 \times ((C_5 \times C_5) \rtimes C_5)) \rtimes C_5\big)$. $\endgroup$ – M. Farrokhi D. G. Dec 15 '18 at 7:40

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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