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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\}$

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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!

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  • $\begingroup$ 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

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