Proof. Since G is nonabelian, there exist elements a and b of G such that $ab\neq ba$ . The result follows by considering the pairwise $1$ -noncommuting set $\{\{a\}, \{b\}, \{ab\}\}$ and the pairwise $2$ -noncommuting set $\{\{a,b\}, \{ab,ba\}\}$ .

Question 2.4. Let G be a nonabelian group of order n. Are there rational numbers $f(r, n)$ and $g(r, n)$ such that $g(r,n)\leq w_G(r) \leq f(r,n)$ ?

Proof. By definition, there exists a pairwise r-noncommuting set of the form

$$ \begin{align*}\mathcal{Y}_r=\{X_1,X_2,\ldots,X_{w_G(r)}\}\end{align*} $$

in G, which implies that for every $i\neq j$ , we have $X_i \bigcap X_j=\varnothing $ . Assume that $C_G(a)$ is an abelian centraliser of G. Then, there exists a subset $\mathcal {R}$ with cardinality $w_G(r)-1$ of $\mathcal {Y}_r$ such that $X_i\bigcap C_G(a)=\varnothing $ for any $X_i\in \mathcal {R}$ , and also $Z(G)\bigcap X_i=\varnothing $ . Therefore,

$$ \begin{align*}\bigcup_{i=1}^{w_G(r)-1}X_i \subseteq G\backslash C_G(a).\end{align*} $$

It follows that $rw_G(r)\leq |G|-p|Z(G)|+r$ . Therefore,

$$ \begin{align*}w_G(r)\leq \frac{|G|-p|Z(G)|}{r}+1.\end{align*} $$

The second result follows by a similar argument.

Proof. (1) First, note that it follows from results of [Reference Pyber4] that if $t = |G : Z(G)|$ , then

$$ \begin{align*}c\log_2t \leq w(G)\leq t-1\end{align*} $$

for some positive constant c. Since $1 \leq r \leq w(G)-1$ , this gives

$$ \begin{align*} r \leq w(G)-1 \leq |G:Z(G)|-2.\end{align*} $$

Thus, $|Z(G)|r< |G|$ and $w_G(|Z(G)|r)$ is meaningful. Now assume that

$$ \begin{align*}w(G)=|\{x_1,x_2,\ldots,x_m\}|, \quad Z(G)=\{z_1=e,z_2,\ldots,z_t\} \quad\text{and}\quad 1 \leq r \leq w(G)-1.\end{align*} $$

Let $\lfloor w(G)/r\rfloor =n.$ Then, it is not hard to see that the following set is a pairwise $|Z(G)|r$ -noncommuting set:

$$ \begin{align*}\{\{x_1z_1,x_2z_1,\ldots,x_rz_1, \ldots, x_1z_t,x_2z_t,\ldots,x_rz_t\},\end{align*} $$

$$ \begin{align*}\{x_{r+1}z_1,x_{r+2}z_1,\ldots,x_{2r}z_1, \ldots, x_{r+1}z_t,x_{r+2}z_t,\ldots,x_{2r}z_t\},\end{align*} $$

$$ \begin{align*}\cdots\end{align*} $$

$$ \begin{align*}\{x_{w(G)-r+1}z_1,x_{w(G)-r+2}z_1,\ldots,x_{w(G)}z_1, \ldots, x_{w(G)+r-1}z_t,x_{w(G)-r+2}z_t,\ldots,x_{w(G)}z_t\}\}.\end{align*} $$

Hence,

$$ \begin{align*}w_G(|Z(G)|r)\geq \lfloor w(G)/r \rfloor.\end{align*} $$

Since $w_G(|Z(G)|r)\geq \lfloor w(G)/r \rfloor $ and $1 \leq r \leq w(G)-1$ , by setting $r=\lfloor w(G)/2\rfloor $ , we deduce that $w_G(|Z(G)|\lfloor w(G)/2\rfloor )\neq 1$ and so $\min (G)> |Z(G)|(w(G)/2)$ .

(2) If $Z(G)\neq 1$ , then $\min (G)> |Z(G)|(w(G)/2)\geq w(G)$ and so $w(G)+1\leq \min (G)$ .

(3) By considering $r=1$ in part (1), we see that $w(|Z(G)|)\geq [w(G)]$ and so

$$ \begin{align*}w(G)\geq w(|Z(G)|)\geq w(G).\end{align*} $$

Hence, $w_G(i)=w(G)$ for any i with $1\leq i\leq |Z(G)|$ . Therefore, $\min (G)\geq |Z(G)|+1$ .

(4) Assume that $w(G)=|\{x_1,x_2,\ldots ,x_n\}|$ . We claim that the set

$$ \begin{align*}\{C_G(x_i)\setminus Z(G) \mid 1\leq i\leq n\}\end{align*} $$

is a pairwise noncommuting set. Assume not, then there exist $x_i$ and $x_j$ for $i\neq j$ such that $C_G(x_i)\setminus Z(G) $ and $C_G(x_j)\setminus Z(G) $ are not pairwise noncommuting sets. Therefore, there exist $a\in C_G(x_i)\setminus Z(G) $ and $b\in C_G(x_j)\setminus Z(G) $ such that $ab=ba$ . Thus, $x_i$ and b belong to $C_G(a)$ , and since $C_G(a)$ is abelian, $x_ib=bx_i$ . This implies that $x_i$ and $x_j$ belong to $C_G(b)$ , but since $C_G(b)$ is abelian, we have $x_ix_j=x_jx_i$ , which is a contradiction. By considering

$$ \begin{align*}t=\min\{| C_G(x_i)\setminus Z(G)| \mid 1\leq i\leq n\},\end{align*} $$

one can show that $w_G(t) \geq w(G)$ , as desired.

Proof. (1) Suppose that $N=\{m_1,m_2,\ldots ,m_r\}$ . Since $G/N$ is nonabelian, there exist two elements $aN$ and $bN$ of $G/N$ such that $[a,b]\not \in N$ . In this case, it is not difficult to see that

$$ \begin{align*}\{ \{am_1,am_2,\ldots,am_r\}, \quad \{bm_1,bm_2,\ldots,bm_r\}, \quad \{abm_1,abm_2,\ldots,abm_r\}\}\end{align*} $$

is a pairwise r-noncommuting set in G. This implies that $w_G(r)\geq 3 $ , as desired.

(2) The result follows from the fact that, for two positive integers $i, j$ , if $i\leq j$ , then $w_G(i)\geq w_G(j)$ .

Proof. Let $bc(G)$ be the size of the largest conjugacy class in G and $\mathrm {Rad}(G)$ be the largest solvable normal subgroup of G. Then, $G/\mathrm {Rad}(G)$ is nonabelian. Let $k(G)$ be the number of conjugacy classes of G. Since ${|G|}/{k(G)} \leq bc(G)$ , it follows that $|G/\mathrm {Rad}(G) |\leq bc(G)^2$ by [Reference Guralnick and Robinson3, Theorem 9]. According to [Reference Pyber4], $bc(G)\leq 4w(G)^2$ . Now by Lemma 3.1(1), $w(G)< 2\min (G)$ and also by Lemma 3.2, $|\mathrm {Rad}(G)| < \min (G)$ as desired.

If $Z(G)\neq 1$ , then Lemma 3.1(2) gives the result.

Proof. (1) Assume that there exist elements a and b in G such that $ab\neq ba$ . Set $q-1=n$ . Since $C_G(x^i)=C_G(x)$ for $(|x|,i)=1$ , one can see that the set

$$ \begin{align*}\{\{a,a^2,\ldots,a^n\}, \quad \{b,b^2,\ldots,b^n\}, \quad \{ab,(ab)^2,\ldots,(ab)^n\}, \quad \{ab^2,\ldots,ab^{n-1},a^2b\} \}\end{align*} $$

is a pairwise n-noncommuting set which gives the result.

(2) Assume that $w(G)=|\{x_1,x_2,\ldots ,x_m\}|$ . Again, since $C_G(y^i)=C_G(y)$ for ${(|y|,i)=1}$ , one can see that

$$ \begin{align*}\{\{x_1,x_1^2,\ldots,x_1^{q-1}\}, \quad \{x_2,x_2^2,\ldots,x_2^{q-1}\}, \quad \ldots, \quad \{x_m, x_m^2,\ldots,x_m^{q-1}\} \}\end{align*} $$

is a pairwise $(q-1)$ -noncommuting set. Therefore, $w_G(q-1)\geq w(G)$ and the result follows.

(3) Since G is a nilpotent group, we have

$$ \begin{align*}\prod_{p\in \pi (|G|)}p \leq |Z(G)|\end{align*} $$

and Lemma 3.1(3) gives the result.

Proof. By Lemma 3.5(2), $|G|$ is even. Then, Lemma 3.1(3) and (4) give the result.