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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$$\mathfrak{S}=(\mathsf{r}_{11}, \, \mathsf{t}_{11}, \, \mathsf{r}_{12}, \, \mathsf{t}_{12}, \, \mathsf{r}_{21}, \,\mathsf{t}_{21}, \, \mathsf{r}_{22}, \, \mathsf{t}_{22}, \, \mathsf{z})$$ ofnon-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 amnot requiring that the elements of$\mathfrak{S}$ generate$G$. Let me call such$\mathfrak{S}$ anRT-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$G$?

askedOct 7 at 14:34
Francesco Polizzi's user avatar
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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).$\endgroup$CommentedOct 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.$\endgroup$CommentedOct 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.$\endgroup$CommentedOct 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.$\endgroup$CommentedOct 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.$\endgroup$CommentedOct 7 at 15:33

1 Answer1

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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.

answeredOct 7 at 18:36
Arturo Magidin's user avatar
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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 $96$?$\endgroup$CommentedOct 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.$\endgroup$CommentedOct 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.$\endgroup$CommentedOct 7 at 19:55

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