16
$\begingroup$

Let $G$ be a (discrete) group. Define $k^*(G)$ as the minimal cardinality of a set $S \subset G$ such that $C_G(S) = Z(G)$. Define $k(G) = k^*(G)$ if $G$ has trivial center (i.e. $|Z(G)| = 1$), and $k(G) = \bot$ otherwise. If $k(G) = \bot$, then the convention is that neither $k(G) \leq n$ nor $k(G) \geq n$ holds, for any cardinal $n$.

Question: Does there exist a finite group $G$ such that $k(G) \geq 3$, and more generally does every natural number $k(G) \geq 3$ occur for some finite group $G$?

I do not even know such examples for $G$ infinite, and would also be interested in such, though I do not have an immediate application for this. I am not an expert on group theory (especially finite group theory), so I do not know very effective search terms for this, and would also be interested in pointers to the literature.

What I have tried so far (though don't take my word on these):

  1. No abelian group or a p-group or a nilpotent group is an example, since they have nontrivial centers (in the finite case), thus $k(G) = \bot$.

  2. No finite simple group is an example, since they are all 2-generated (by CFSG), thus satisfy $k(G) \leq 2$ or $k(G) = \bot$.

  3. $k(G \times H) = \max(k(G), k(H))$ for any groups $G, H$ (by a simple proof).

  4. I did a quick search in GAP and seems that there are no finite groups of size up to $1151$ with this property (this is the first time I used GAP, so not sure how much proof value this has).

  5. $k(G) = 0$ for precisely the trivial group, and $k(G) = 1$ is impossible (since any $g$ commutes with itself).

  6. For infinite cardinal $\kappa$, $k(G) = \kappa$ where $G$ is the group of finite-support permutations on a set of cardinality $\kappa$, but of course $k(G)$ is finite (or $\bot$) for finite groups.

  7. Arbitrarily large $k^*(G)$ are provided by wreath products $\mathbb{Z}_2 \wr \mathbb{Z}_2^d$, where $\mathbb{Z}_2 = \mathbb{Z}/2\mathbb{Z}$, but I wrote a quick proof sketch that $k(G \wr H) = 2$ whenever $|H| \geq 2$ and $k(G) = 2$ so it seems one cannot use wreath products to get examples.

  8. For a single permutation $\pi$ on a set of size $n$, The centralizer of $\pi$ in $S_n$ never has $k(G) \geq 3$ (I prove this as Proposition 7.9 in this paper of mine, by a rather ad hoc case analysis).

  9. In terms of the commuting graph $\Gamma(G)$ with vertices $G \setminus Z(G)$ and edges $\{(g, h) \;|\; gh = hg\}$, the question of whether $k(G) \geq 3$ is possible is equivalent to whether there exists a finite group $G$ with trivial center such that $\mathrm{diam}(\Gamma(G)) = 2$, where $\mathrm{diam}$ is the diameter, i.e. maximal minimal distance between a pair of vertices. For a finite minimal nonsolvable group the diameter is always at least $3$ according to this paper which implies $k(G) = 2$ for a minimal nonsolvable group (a different definition is used in that paper, but it should be equivalent to mine for minimal nonsolvable groups, as they have trivial center). Most literature I know about this graph and its diameter are about finding upper bounds, but I do not know if that has a relation to $k(G)$. Larger values of $k(G)$ also correspond to statements about this graph, but not about its diameter.

For context, the question arose from the study of automorphism groups of one-dimensional subshifts. I am interested in quantitative versions (or lack thereof) of the so-called Ryan's theorem, which states that the center of the automorphism group of a mixing subshift of finite type consists of only the shift maps. I ask the above question about finite groups after Lemma 7.7 here and Lemma 7.7 is my application for it. The paper of Boyle, Lind and Rudolph is a standard reference for these groups.

$\endgroup$
1

3 Answers 3

22
$\begingroup$

I believe that you can construct examples with arbitrary $k = k(G) > 1$ as follows.

Let $G$ be a semidirect product of an elementary abelian group $N$ of order $3^{2^k-1}$ with an elementary abelian $2$-group $H$ of order $2^k$, with action defined as follows.

There are $2^k-1$ subgroups of $H$ of order $2^{k-1}$, and each of the $2^k-1$ direct factors of $N$ is normalized by $H$, and centralized by one of these subgroups of $H$ of order $2^{k-1}$, where each factor has a different centralizer in $H$.

Then any subset of $G$ of order less than $k$ centralizes at least one of these direct factors of $N$.

To see that $k(G)=k$, let $H = \langle x_1,x_2,\ldots,x_k \rangle$, choose $y_1,y_2,\ldots,y_k \in N$ such that $[y_i,x_i] \ne 1$, but $[y_i,x_j]=1$ whenever $i \ne j$, and let $S = \langle x_1y_2,x_2y_3,\ldots x_ky_1 \rangle$. So the generators of $S$ all have order $6$, and since $x_1$ and $y_2$ are both powers of $x_1y_2$, etc, we have $H < S$ and also $y_i \in S$ for $1 \le i \le k$. Now $C_G(H) = H$ and $C_H(\langle y_1,y_2,\ldots,y_k \rangle) = \cap_{i=1}^k C_H(y_i) =1$, so we have $C_G(S)=1$.

Note that $G$ is a subgroup of the direct product of $3^{2^k-1}$ copies of the dihedral group of order $6$.

$\endgroup$
3
  • $\begingroup$ By "arbitrary", did you mean "arbitrarily large"? I see that indeed $k(G) > k-1$ for this group, and I think $k(G) \leq k + 1$ by taking shifts and a single translation on the side of $3$s, but I do not immediately see how to combine these to get $k(G) = k$. $\endgroup$
    – Ville Salo
    Apr 24, 2018 at 16:37
  • $\begingroup$ Yes sorry. I have added a proof that $k(G)=k$. $\endgroup$
    – Derek Holt
    Apr 24, 2018 at 17:18
  • 1
    $\begingroup$ I see, very nice. So my question is fully answered: the values of $k(G)$ are precisely the cardinals except 1. I will accept this answer. $\endgroup$
    – Ville Salo
    Apr 24, 2018 at 17:25
14
$\begingroup$

Here's another family of examples with arbitrary large $k$, more based on linear algebra. Fix any odd prime $p$. Consider the group $G_{p,s}$ of square matrices of size $s+2$ over the field $F=\mathbf{Z}/p\mathbf{Z}$ of the form $$m^\pm(u,v,z)=\begin{pmatrix}\pm 1 & ^tu & z\\ 0 & I_s & v\\ 0 & 0 & 1\end{pmatrix};$$ with $u,v\in F^s$ and $z\in F$. Its order is $2p^{2s+1}$. Its center is trivial (the center of the subgroup of index 2 is reduced to the cyclic group of elements $m^+(0,0,*)$.

For any $s-1$ elements in $G_{p,s}$. Then they are contained, for some hyperplane $H$ of $F^s$, in the subgroup $\Gamma$ consisting of those $m^\pm(u,v,z)$ with $u\in H$. Then there exists $w\in F^p\smallsetminus\{0\}$ such that $^tuw=0$ for all $u\in H$. Then $m^+(0,w,0)$ belongs to the centralizer of $\Gamma$. Hence $k(G_{p,s})\ge s$ (actually $\le s+1$).

$\endgroup$
2
  • 1
    $\begingroup$ This without the $\pm$ (generalized Heisenberg groups) was in fact the first thing I tried, but I concluded that nothing of this sort can work due to nilpotency. Apparently not much was needed to fix this. In your example, I agree that $k(G) \in \{k, k+1\}$; here also I don't see whether $k(G) = k$. As far as I can tell, both here and in Holt's, the large $k(G)$ arises from dimension considerations for a commutative group, and the $+1$ is needed because once you have enough dimension, you still need to kill the center which you can do with one extra generator. $\endgroup$
    – Ville Salo
    Apr 24, 2018 at 17:11
  • $\begingroup$ Fixed typo (I initially changed $k$ to $s$ to avoid confusion between function $k$ and parameter $s$, but forgot some $k$). $\endgroup$
    – YCor
    Apr 24, 2018 at 20:41
11
$\begingroup$

The finite group $G=$ SmallGroup(486,176) in GAP or Magma notation has $k(G)=3$.

$\endgroup$
2
  • $\begingroup$ Seems that $k(G) = 3$ for this group, so that answers my first question. Thanks! I must have had some problem in my GAP code (which does not surprise me, but certainly does embarrass me). $\endgroup$
    – Ville Salo
    Apr 24, 2018 at 12:36
  • 3
    $\begingroup$ Its order is $486=2.3^5$; it seems it's the case $(p,s)=(3,2)$ of my family of examples. $\endgroup$
    – YCor
    Apr 25, 2018 at 0:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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