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.


3 Answers 3


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 $6$.

  • $\begingroup$ 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

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 $\le s+1$).

  • 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

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

  • $\begingroup$ 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

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