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During my research in Algebraic Geometry, I was led to the following problem in Combinatorial Group Theory, strictly related to finite quotients of pure surface braid groups.

Let $G$ be a finite group. I am looking for ordered $9$-tuples $$\mathfrak{S}=(\mathsf{r}_{11}, \, \mathsf{t}_{11}, \, \mathsf{r}_{12}, \, \mathsf{t}_{12}, \, \mathsf{r}_{21}, \,\mathsf{t}_{21}, \, \mathsf{r}_{22}, \, \mathsf{t}_{22}, \, \mathsf{z})$$ of non-trivial elements in $G$ such that the following relations hold: \begin{equation} \renewcommand{\t}{\mathsf{t}} \renewcommand{\r}{\mathsf{r}} \newcommand{\z}{\mathsf{z}} \begin{aligned} (R_1) & \, \, [\r_{11}, \, \r_{22}]=1 & (R_6) & \, \, [\r_{12}, \, \r_{22}]=1 \\ (R_2) & \, \, [\r_{11}, \, \r_{21}]=1 & (R_7) & \, \, [\r_{12}, \, \r_{21}]= \z^{-1}\,\r_{21}\,\r_{22}^{-1}\,\z\,\r_{22}\,\r_{21}^{-1} \\ (R_3) & \, \, [\r_{11}, \, \t_{22}]=1 & (R_8) & \, \, [\r_{12}, \, \t_{22}]=\z^{-1} \\ (R_4) & \, \, [\r_{11}, \, \t_{21}]=\z^{-1} & (R_9) & \, \, [\r_{12}, \, \t_{21}]=[\z^{-1}, \, \t_{21}] \\ (R_5) & \, \, [\r_{11}, \, \z]=[\r_{21}^{-1}, \, \z] & (R_{10}) & \, \, [\r_{12}, \, \z]=[\r_{22}^{-1}, \, \z] \\ & & & \\ (T_1) & \, \, [\t_{11}, \, \r_{22}]=1 & (T_6) & \, \, [\t_{12}, \, \r_{22}]= \t_{22}^{-1}\, \z \,\t_{22} \\ (T_2) & \, \, [\t_{11}, \, \r_{21}]= \t_{21}^{-1}\, \z \, \t_{21} & (T_7) & \, \, [\t_{12}, \, \r_{21}]=[\t_{22}^{-1}, \, \z] \\ (T_3) & \, \, [\t_{11}, \, \t_{22}]=1 & (T_8) & \, \, [\t_{12}, \, \t_{22}]=[\t_{22}^{-1}, \, \z] \\ (T_4) & \, \, [\t_{11}, \, \t_{21}]=[\t_{21}^{-1}, \, \z] & (T_9) & \, \, [\t_{12}, \, \t_{21}]=[\t_{22}^{-1}, \, \z]\,\t_{21}\, [\z, \, \t_{22}^{-1}]\,\t_{21}^{-1}\\ (T_5) & \, \, [\t_{11}, \, \z]=[\t_{21}^{-1}, \, \z] & (T_{10}) & \, \, [\t_{12}, \, \z]=[\t_{22}^{-1}, \, \z] \\ \end{aligned} \end{equation} Here I am using the convention $[a, \, b]:=aba^{-1}b^{-1}$. Note that I am not requiring that the elements of $\mathfrak{S}$ generate $G$. Let me call such $\mathfrak{S}$ an RT-structure on $G$.

With the help of GAP4 and some (non-trivial) work, I was able to classify all groups of order $|G| \leq 127$ admitting RT-structures. There are few of them, and the possible orders are $32$, $64$, and $96$. In every case, the element $\mathsf{z}$ belongs to the centre of $G$, so that (a posteriori) the relations $(R_i)$ and $(T_j)$ are greatly simplified.

I also have some infinite families of examples (for instance, it is not too difficult to see that some extra-special groups admit RT-structures); again, $\mathsf{z} \in Z(G)$ in every case.

I do not know a single example of a finite group $G$ admitting an RT-structure such that $\mathsf{z}$ is non-central. So let me ask the

Question. Let $G$ be a finite group admitting an RT-structure $$\mathfrak{S}=(\mathsf{r}_{11}, \, \mathsf{t}_{11}, \, \mathsf{r}_{12}, \, \mathsf{t}_{12}, \, \mathsf{r}_{21}, \,\mathsf{t}_{21}, \, \mathsf{r}_{22}, \, \mathsf{t}_{22}, \, \mathsf{z})$$ as above. Is it always true that $\mathsf{z} \in Z(G)$? If not, what is a counterexample?

Bonus question. What can we say if we additionally assume that the elements of $\mathfrak{S}$ generate $G$?

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  • $\begingroup$ You need to specify your convention for defining the commutator $[a,b]$ ($aba^{-1}b^{-1}$, $bab^{-1}a^{-1}$, $a^{-1}b^{-1}ab$ all exist). $\endgroup$ Commented Oct 7 at 15:18
  • $\begingroup$ Are you familiar with the nilpotent product of groups? I think that if you take a suitable nilpotent product of $G$ with a suitable relatively free finite nilpotent group will give you a group that contains $G$ as a subgroup (so that all relations still hold); but if $z$ lies in the $k$th but not the $(k+1)$st term of the lower central series of $G$, then taking the $(k+2)$-nil product should do it. $\endgroup$ Commented Oct 7 at 15:18
  • $\begingroup$ There is an approach consisting in considering $R_1\dots T_{10}$ as a presentation of a group $G_0$ (9 generators, 20 relators) and then the specific question here is whether $z$ is in the center inside the profinite completion. $\endgroup$ Commented Oct 7 at 15:22
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    $\begingroup$ The original definition is due to O. N. Golovin, Nilpotent products of groups, Amer. Math. Soc. Transl. (2) 2 (1956), 89--115; MR0075947; it is a special case of the verbal product, given by S. Moran, Associative operations on groups. I, Proc. London Math. Soc. (3) 6 (1956), 581--596; MR0097439. I am on my way to a class, but I'll try to write up the basics later today. $\endgroup$ Commented Oct 7 at 15:29
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    $\begingroup$ @FrancescoPolizzi The groups involved don't need to be nilpotent themselves. But for example, if you take the $2$-nilpotent product $P$ of two (not necessarily nilpotent) groups $G_1$ and $G_2$, then $P$ contains copies of $G_1$ and $G_2$, and $Z(P)\cap G_i = Z(G_i)\cap[G_i,G_i]$. The $2$-nilpotent product won't work here because your $z$ lies in $[G,G]$, but I expect that taking a larger nilpotency class for the product will work. $\endgroup$ Commented Oct 7 at 15:33

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We can build examples where $z$ is not central out of your examples of order $32$ and $64$ with the nilpotent product.

My idea is to use the $k$-nilpotent product of groups. This was introduced by Golovin, and then generalized to "verbal products" by Moran. If $H$ and $K$ are groups, then we define their $r$-nilpotent product as $$H\amalg^{\mathfrak{N}_r}K = \frac{H*K}{N\cap(H*K)_{r+1}},$$ where $H*K$ is the free product of $H$ and $K$, $N$ is the cartesian, which is the kernel of the canonical map $H*K\to H\times K$ (it is the normal subgroup of $H*K$ generated by elements of the form $[k,h]$ with $k\in K$ and $h\in H$), and $G_m$ is the $m$th term of the lower central series of $G$, defined by $G_1=G$ and $G_{n+1}=[G_n,G]$.

The "$1$-nilpotent product" is just the direct product; the $2$-nilpotent product consists of elements of the form $xyu$ with $x\in H$, $u\in K$, and $u$ in the image of the cartesian, which is isomorphic to $K^{\rm ab}\otimes H^{\rm ab}$ via the map $\overline{[k,h]}\mapsto \overline{k}\otimes \overline{h}$, and multiplication $$ (xyu)(hkv) = (xh)(yk)(\overline{[y,h]}vk).$$ For higher $r$, the expressions become a little more complicated. But in any case, $H\amalg^{\mathfrak{N}_r}K$ contains a copy of $H$ and a copy of $K$, and if $h\in H$ and $k\in K$, then $h$ commutes with $k$ if $h\in H_m$, $K\in K_n$, and $m+n\geq r+1$, or if certain order conditions are met (for example, if we you can express $h$ as $h=x^u$ for some suitably high $u$ that is a multiple of the order of $k$). If $H$ and $K$ are finite, then $H\amalg^{\mathfrak{N}_k}K$ is also finite.

In the groups of order $32$ and $64$, there exists $r$ such that $z\in G_r\setminus G_{r+1}$, since $z\neq e$. We can then consider $G\amalg^{\mathfrak{N}_{2r+1}}G$. This is finite, and since it contains two copies of $G$,any $\mathsf{RT}$-structure that was present in $G$ will remain present in this group, but $z$ will not be central since it does not commute with elements in the "other" copy. You could replace the second copy of $G$ with a suitable finite cyclic group as well, if I'm not mistaken, to simplify and so that your resulting group is generated by $G$ plus an additional element. But this would not answer the bonus question.

If you wanted to build a similar overgroup for an arbitrary finite group $G$ that admits an $\mathsf{RT}$-structure, you could try to use a verbal subgroup of $G$ that does not contain $z$ and replace $(H*K)_{r+1}$ with $\mathfrak{V}(H*K)$, where $\mathfrak{V}$ is the corresponding variety; for instance if $|G|=n$ should could take $\mathfrak{B}_n(H*K)$, the subgroup of $H*K$ generated by all $n$th powers of elements. But in those general cases it may be that the resulting group is not finite.

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  • $\begingroup$ Thank you very much for your answer. I will need some time to digest it and check all the details. In what case does your construction answer the bonus question? Perhaps if the RT-structure in $G$ generate? And why are you not considering the groups of order $96$? $\endgroup$ Commented Oct 7 at 18:45
  • $\begingroup$ @FrancescoPolizzi In no case does my construction answer the bonus question, since I am taking an example you already have and constructing a larger group; even if the original set generated the group you start with, it will not generate the group I am constructing. At best, I can give you a group that is generated by the set of nine elements plus one additional element. $\endgroup$ Commented Oct 7 at 19:52
  • $\begingroup$ @FrancescoPolizzi A group of order $96$ need not be nilpotent, so my argument that $z$ is not central in the resulting group would fail. For example, if you take $S_3\amalg^{\mathfrak{N}_r} K$, then the element of order $3$ in $S_3$ would commute with every element of $K$, because the element of order $3$ lies in $(S_3)_r$ for all $r$, so the commutator with any element of $K$ would lie in $N\cap(S_3*K)_{r+1}$. If in the group of order $96$ you know that $z$, though central, does not lie in some term of the lower central series, you can use the construction. But otherwise, it may not work. $\endgroup$ Commented Oct 7 at 19:55

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