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 $n$.



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> 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 $G$?
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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](https://www.cambridge.org/core/journals/ergodic-theory-and-dynamical-systems/article/transitive-action-on-finite-points-of-a-full-shift-and-a-finitary-ryans-theorem/FB39F6A639A28DBACE0569E9ADD13011), 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](https://link.springer.com/article/10.1023%2FA%3A1013180005982) 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](https://www.cambridge.org/core/journals/ergodic-theory-and-dynamical-systems/article/transitive-action-on-finite-points-of-a-full-shift-and-a-finitary-ryans-theorem/FB39F6A639A28DBACE0569E9ADD13011) and Lemma 7.7 is my application for it. [The paper of Boyle, Lind and Rudolph](http://www.ams.org/tran/1988-306-01/S0002-9947-1988-0927684-2/) is a standard reference for these groups.



If the center of a finite group is trivial, are there two elements whose centralizers intersect trivially?
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 $6$.


