The conjugacy diameters of non-abelian finite $p$-groups with cyclic maximal subgroups

1.   Introduction

1.1.   Background on norms and boundedness

Let $G$ be a group. A norm on $G$ is a function $\nu:G\longrightarrow [0, \infty)$ which satisfies the following axioms:

(ⅰ) $\nu(g) = 0 \iff g = 1;$

(ⅱ) $\nu(g^{-1}) = \nu(g)$ for all $g\in G;$

(ⅲ) $\nu(gh)\le \nu(g)+\nu(h)$ for all $g, h \in G;$

We call $\nu$ conjugation-invariant if we also have

(ⅳ) $\nu(g^{-1}hg) = \nu(h)\, \, \mbox{for all}\, \, g, h\in G.$

The diameter of a group $G$ with respect to a conjugation-invariant norm $\nu$ on $G$ is defined as

A group $G$ is said to be bounded if every conjugation-invariant norm on $G$ has a finite diameter. The concepts of conjugation-invariant norms and boundedness were introduced by Burago et al. ^[5], and they provided a number of applications to geometric group theory, Hamiltonian dynamics, and finite groups. Since then, there has been a large amount of interest in providing further applications, examples, and analogues of these notions in geometry, group theory, and topology; see ^{[2,3,4,11,12,15]} for examples.

An important source of conjugation-invariant norm is a normally generating subset of a group. Let $G$ be a group and $S \subseteq G.$ The normal closure of $S$ in $G,$ denoted by $\langle\langle S \rangle\rangle,$ is the subgroup of $G$ that is generated by all conjugates of elements of $S$. In other words, it is the smallest normal subgroup of $G$ containing $S.$ We say that $S$ normally generates $G$ if $G = \langle\langle S \rangle\rangle.$ In this case, any element of $G$ can be written as a product of elements of

If $S$ normally generates $G,$ then the length $\|g\|_{S}\in \mathbb{N}$ of $g\in G$ with respect to $S$ is defined to be the length of a shortest word in $\mbox{Conj}_{G}(S^{\pm 1})$ that is needed to express $g.$ In other words,

It is important to stress that $\| 1\|_S$, i.e., the length of the identity element $1$ of $G$ with respect to $S$, is $0$. The word norm $\parallel.\parallel_S\colon G\rightarrow [0, \infty),$ $g\mapsto \| g\|_S$ is a conjugation-invariant norm on $G.$ The diameter of a group $G$ with respect to the word norm $\parallel.\parallel_S$ is given by

For other examples of conjugation-invariant norms, see ^[3,5].

If $G$ is a group and $S$ is a finite normally generating subset of $G,$ then $G$ is bounded if and only if $\|G\|_{S}$ is finite (see [12, Corollary $2.5$]). Therefore, word norms for finite normally generating subsets and their diameters are an important tool to study the boundedness of groups.

Moreover, word norms are used in the study of several refinements of the concept of bounded groups. To describe these refinements, we introduce some notation. For any group $G$ and any $n\geq 1,$ let

Set

For any group $G,$ $\Delta(G)$ is said to be the conjugacy diameter of $G.$ The group $G$ is called strongly bounded if $\Delta_{n}(G) < \infty$ for all $n\in\mathbb{N}$ (see [12, Definition $1.1$]). The group $G$ is called uniformly bounded if $\Delta(G) < \infty$ (see [12, Definition $1.1$]).

1.2.   Motivation and statements of results

As we have seen in the previous section, word norms and their diameters emerge in the study of boundedness. Since any finite group is known to be uniformly bounded, the study of boundedness mainly focuses on infinite groups. However, word norms and their diameters are also of interest in finite group theory. For example, conjugacy diameters of finite groups can be used to study conjugacy class sizes (see [12, Proposition 7.1]). Recently, conjugacy diameters were studied for several classes of finite groups. For example, Kȩdra et al. showed that $\Delta(PSL(n, q))\leq 12(n-1)$ for any $n\geq 3$ and any prime power $q$ (see [12, Example 7.2]). Also, Libman and Tarry proved that $\Delta(S_n) = n-1$ for any $n\geq 2$ (see [14, Thereom 1.2]), and that, if $G$ is a non-abelian group of order $pq$, where $p$ and $q$ are prime numbers with $p < q$, then $\Delta(G) = \mathrm{max} \lbrace \frac{p-1}{2}, 2 \rbrace$ (see [14, Thereom 1.1]). The first author determined the conjugacy diameters of finite dihedral groups as described below.

Theorem 1.1. ([1, Theorem 6.0.2]) Let $n\geq 3$ be a natural number and $G: = D_{2n} = \langle a, b|a^{n} = 1 = b^{2}, bab = a^{-1}\rangle$ be the dihedral group of order $2n.$ Then,

For the case of the infinite dihedral group $D_\infty = \langle a, b|b^{2} = 1, b a b = a^{-1}\rangle$, we have $\Delta(D_\infty)\leq 4$ (see [12, Example 2.8]).

The goal of this paper is to take Theorem 1.1 further. For any natural number $n\geq 3,$ the dihedral group $D_{2n}$ is non-abelian and has a cyclic maximal subgroup. Thus, one way to extend Theorem 1.1 would be to prove further results about conjugacy diameters of non-abelian finite groups with cyclic maximal subgroups. Instead of looking at arbitrary non-abelian finite groups with cyclic maximal subgroups, let us restrict our attention to non-abelian finite $p$-groups with cyclic maximal subgroups.

Definition 1.2. Let $n$ be a natural number.

(i) For $n\geq 4,$ define

We call $SD_{n}$ the semidihedral group of order $2^{n}.$

(ii) For $n\geq 3,$ define

We call $Q_{n}$ the (generalized) quaternion group of order $2^{n}.$

(iii) Let $p$ be a prime number and assume that $n\geq 4$ if $p = 2$ and $n\geq 3$ if $p$ is odd. Define

We call $M_{n}(p)$ the modular $p$-group of order $p^{n}.$

The groups introduced in Definition 1.2 turn out to be important in several contexts. For example, they appear in ^[6,7,8,9,16], where various spectra and energies of commuting and non-commuting graphs of these groups have been computed.

The non-abelian finite $p$-groups with cyclic maximal subgroups are fully classified by applying the following theorem.

Theorem 1.3. (see [10, Chapter 5, Theorem 4.4]) Let $P$ be a non-abelian $p$-group of order $p^{n}$ which contains a cyclic subgroup $H$ of order $p^{n-1}.$ Then, the following holds:

(i) If $n = 3$ and $p = 2,$ then $P$ is isomorphic to $D_8$ or $Q_3.$

(ii) If $n > 3$ and $p = 2,$ then $P$ is isomorphic to $M_n(2), D_{2^{n}}, Q_{n},$ or $SD_{n}.$

(iii) If $p$ is odd, then $P$ is isomorphic to $M_n(p).$

No two groups among $D_{2^{n}}, Q_{n}, SD_{n},$ or $M_{n}(p)$ are isomorphic to each other (see [10, Chapter 5, Theorem 4.3(ⅲ)]). We will prove the following results:

Theorem 1.4. Let $n\geq 4$ be a natural number and $G: = SD_{n}$. Then,

Theorem 1.5. Let $n\geq 3$ be a natural number and $G: = Q_{n}$. Then,

Theorem 1.6. Let $p$ be a prime number and $n$ be a natural number, where $n\geq 4$ if $p = 2$ and $n\geq 3$ if $p$ is odd. Let $G: = M_n{(p)}$. Then,

With Theorems 1.1 and 1.4–1.6, we are able to calculate all of the conjugacy diameters of non-abelian finite $p$-groups with cyclic maximal subgroups.

2.   Preliminaries

In this section, we present some results and the notation needed for the proofs of Theorems 1.4–1.6.

Definition 2.1 ([12, Section 2]) Let $X$ be a subset of a group $G$. For any $n\geq 0,$ we define $B_X(n)$ to be the set of all elements of $G$ which can be expressed as a product of at most $n$ conjugates of elements of $X$ and their inverses.

By Definition 2.1, we have

The next result follows from the above definition.

Lemma 2.2. ([12, Lemma 2.3 (iii) ]) Let $G$ be a group, $X, Y\subseteq G,$ and $n, m\in \mathbb{N}$. Then, $B_{X}(n)B_{X}(m) = B_{X}(n + m).$

The following terminology is not standard, but it will be useful in this paper.

Definition 2.3 Let $n$ be a natural number.

(i) If $n\geq 4$ and $G = SD_n,$ then a pair $(a, b)$ of elements $a, b\in G$ is called a standard generator pair if $ord(a) = 2^{n-1}, ord(b) = 2, bab = a^{2^{n-2}-1}.$

(ii) If $n\geq 3$ and $G = Q_n,$ then a pair $(a, b)$ of elements $a, b\in G$ is called a standard generator pair if $ord(a) = 2^{n-1}, b^2 = a^{2^{n-2}}, b^{-1}ab = a^{-1}.$

(iii) If $G = M_n(p),$ where $n\geq 4$ and $p = 2$ or $n\geq 3$ and $p$ is an odd prime, then a pair $(a, b)$ of elements $a, b\in G$ is called a standard generator pair if $ord(a) = p^{n-1}, ord(b) = p, b^{-1}ab = a^{p^{n-2}+1}.$

Lemma 2.4. Let $G = SD_n$ for some $n\geq 4$ or $G = Q_n$ for some $n\geq 3,$ and let $(a, b)$ be a standard generator pair of $G.$ Then, the subgroups $\langle a\rangle, \langle a^{2}\rangle \langle ab\rangle\, , \langle a^{2}\rangle \langle b\rangle$ are proper normal subgroups of $G.$

Proof. We have

So, each of these subgroups is a maximal subgroup of $G.$ This proves the lemma since each maximal subgroup of a finite $2$-group is a proper normal subgroup. □

Remark 2.5. Let $n\geq 4$ be a natural number and $G = SD_n.$ Let $(a, b)$ be a standard generator pair of $G.$ If $m\in \mathbb{Z},$ then one can easily show that

Lemma 2.6. Let $n\geq 4$ be a natural number and $G = SD_n.$ Let $(a, b)$ be a standard generator pair of $G.$ Then, the following holds:

(i) The conjugacy class of $a^{2}b$ is $\big\{a^{r}b\big| \, r\, \mathit{\mbox{is even and}} \, \, 0\leq r < 2^{n-1}\big\}.$

(ii) The conjugacy class of $ab$ is $\big\{a^{r}b\big| \, r\, \mathit{\mbox{is odd and}} \, \, 0 < r < 2^{n-1}\big\}.$

(iii) If $0\leq m < 2^{n-1},$ then the conjugacy class of $a^{m}$ is $\big\{a^{m}, (a^{-1+2^{n-2}})^{m}\big\}.$

(iv) If $0\leq m < 2^{n-1}$ is even, then $(a^{m}b)^{-1} = a^{{m}^{\prime}}b$ for some even $0\leq m^{\prime} < 2^{n-1}.$

(v) If $0 < m < 2^{n-1}$ is odd, then $(a^{m}b)^{-1} = a^{{m}^{\prime}}b$ for some odd $0 < m^{\prime} < 2^{n-1}.$

Proof. Let $0\leq m < 2^{n-1}$ and $\ell\in \mathbb{Z}.$ For (ⅰ) and (ⅱ), we have

Thus, if $m$ is even, then the conjugacy class of $a^{m}b$ is $\big\{a^{r}b\big| \, r\, \mathit{\mbox{is even and}} \, \, 0\leq r < 2^{n-1}\big\}.$ In particular, (ⅰ) holds. If $m$ is odd, then the conjugacy class of $a^{m}b$ is $\big\{a^{r}b\big| \, r\, \mathit{\mbox{is odd and}} \, \, 0 < r < 2^{n-1}\big\}.$ In particular, (ⅱ) holds.

For (ⅲ), we have

This completes the proof of (ⅲ).

For (ⅳ), we assume that $0\leq m < 2^{n-1}$ is even. We have $a^{m}b\in \langle a^{2} \rangle \langle b \rangle.$ Therefore, we have $(a^{m}b)^{-1}\in \langle a^{2} \rangle \langle b \rangle,$ and $(a^{m}b)^{-1}\notin \langle a^{2} \rangle$ as $a^{m}b\notin \langle a^{2} \rangle.$ It follows that $(a^{m} b)^{-1} = a^{{m}^{\prime}}b$ for some even $0\leq m^{\prime} < 2^{n-1}.$

For (ⅴ), we assume that $0 < m < 2^{n-1}$ is odd. It is clear that $(a^{m}b)^{-1}\notin \langle a \rangle.$ So we have $(a^{m}b)^{-1} = a^{{m}^{\prime}}b$ for some $0\leq m^{\prime} < 2^{n-1}.$ Since $a^{m}b\notin \langle a^{2} \rangle \langle b \rangle,$ we have that $m^{\prime}$ is odd. □

Remark 2.7. Let $n\geq 3$ be a natural number and $G = Q_n.$ Let $(a, b)$ be a standard generator pair of $G.$ If $m\in \mathbb{Z},$ then one can easily show that

(i) $b^{-1} a^{m} b = (b^{-1} a b)^{m} = a^{-m}.$

(ii) $b^{4} = 1.$

Lemma 2.8. Let $n\geq 3$ be a natural number and $G = Q_n.$ Let $(a, b)$ be a standard generator pair of $G.$ Then, the following holds:

(i) The conjugacy class of $a^{2}b$ is $\big\{a^{r}b\big| \, r\, \mathit{\mbox{is even and}} \, \, 0\leq r < 2^{n-1}\big\}.$

(ii) The conjugacy class of $ab$ is $\big\{a^{r}b\big| \, r\, \mathit{\mbox{is odd and}} \, \, 0 < r < 2^{n-1}\big\}.$

(iii) If $0\leq m < 2^{n-1},$ then the conjugacy class of $a^{m}$ is $\big\{a^{m}, a^{-m}\big\}.$

(iv) If $0\leq m < 2^{n-1}$ is even, then $(a^{m}b)^{-1} = a^{{m}^{\prime}}b$ for some even $0\leq m^{\prime} < 2^{n-1}.$

(v) If $0 < m < 2^{n-1}$ is odd, then $(a^{m}b)^{-1} = a^{{m}^{\prime}}b$ for some odd $0 < m^{\prime} < 2^{n-1}.$

Proof. Let $0\leq m < 2^{n-1}$ and $\ell\in \mathbb{Z}.$ For (ⅰ) and (ⅱ), we have

Thus, if $m$ is even, then the conjugacy class of $a^{m}b$ is $\big\{a^{r}b\big| \, r\, \mathit{\mbox{is even and}} \, \, 0\leq r < 2^{n-1}\big\}.$ In particular, (ⅰ) holds. If $m$ is odd, then the conjugacy class of $a^{m}b$ is $\big\{a^{r}b\big| \, r\, \mathit{\mbox{is odd and}} \, \, 0 < r < 2^{n-1}\big\}$. In particular, (ⅱ) holds.

For (ⅲ), we have

This completes the proof of (ⅲ).

For (ⅳ) and (ⅴ), we have similar arguments as in the proofs of Lemma 2.6(ⅳ) and (ⅴ). □

Lemma 2.9. Let $G = SD_n$ for some $n\geq 4$ or $G = Q_n$ for some $n\geq 3,$ and let $(a, b)$ be a standard generator pair of $G.$ Let $S$ be a normally generating subset of $G.$ Then, there are elements $x, y\in S$ such that $x = a^{\ell}b$ for some $0\leq\ell < 2^{n-1},$ and such that $y = a^{m}$ for some odd $0 < m < 2^{n-1}$ or $y = a^{m}b$ for some $0\leq m < 2^{n-1}$ with $\ell \not\equiv m \, \mathit{\mbox{mod}}\, \, 2.$ For each such $x$ and $y,$ we have $\{x, y\}\in \Gamma(G).$

Proof. If $S\subseteq \langle a\rangle,$ then $\langle\langle S\rangle \rangle \leq \langle a\rangle$ since $\langle a\rangle \trianglelefteq G$ (see Lemma 2.4), which is a contradiction to $S\in \Gamma(G).$ Thus, $S\nsubseteq \langle a\rangle.$

Let $x\in S\setminus \langle a \rangle.$ Then, $x = a^{\ell} b$ for some $0\leq \ell < 2^{n-1}.$ Assume that any element of $S$ has the form $a^{m},$ where $0\leq m < 2^{n-1}$ is even, or the form $a^{m}b$ with $\ell \equiv m \; {\mbox{mod}}\, \, 2.$ Then, $S\subseteq X_1: = \langle a^{2} \rangle \langle b\rangle$ or $S\subseteq X_2: = \langle a^{2} \rangle \langle ab\rangle.$ Since $X_1$ and $X_2$ are proper normal subgroups of $G$ by Lemma 2.4, it follows that $\langle\langle S\rangle \rangle \neq G,$ which is a contradiction. Consequently, there is some $y\in S$ such that $y = a^{m},$ where $0 < m < 2^{n-1}$ is odd, or $y = a^{m}b,$ where $0\leq m < 2^{n-1}$ and $\ell \not\equiv m\, \, \mathit{\mbox{mod}}\, 2.$

Let $x, y\in S$ be as above. We show that $X: = \{x, y\}\in \Gamma(G).$ Assume that $y = a^{m}$ for some odd $0 < m < 2^{n-1}.$ Then, $\langle a\rangle = \langle y\rangle\subseteq \langle\langle X\rangle \rangle.$ It follows that $G = \langle a\rangle \langle x\rangle\subseteq \langle\langle X\rangle\rangle.$ Thus, $G = \langle\langle X\rangle\rangle$ and we have $X\in \Gamma(G).$ Assume now that $y = a^{m}b,$ where $0\leq m < 2^{n-1}$ and $\ell \not\equiv m \, \mathit{\mbox{mod}}\, \, 2.$ Then, every element of $S\setminus \langle a\rangle$ is conjugate to $x$ or $y$ by Lemmas 2.6 and 2.8, so $S\setminus \langle a\rangle \subseteq \langle\langle X\rangle\rangle.$ Also, $a = ab\cdot b\in \langle\langle X\rangle\rangle$ if $G = SD_n$ and $a = a^{2^{n-2}+1}b\cdot b\in \langle\langle X\rangle\rangle$ if $G = Q_n.$ It follows that $G = \langle\langle X\rangle\rangle;$ hence, $X\in \Gamma(G).$ The proof is now complete. □

Lemma 2.10. Let $n\geq 4$ be a natural number and $G = SD_n.$ Let $(a, b)$ be a standard generator pair of $G.$ If $m_1, m_2\in \mathbb{Z},$ then

(i) $a^{m_1} b \cdot a^{m_2} b = a^{m_1-m_2+2^{n-2} m_2}$;

(ii) $a^{m_1} b\cdot a^{m_2} = a^{m_1-m_2+2^{n-2} m_2}\, b$;

(iii) $a^{m_1}\cdot a^{m_2} b = a^{m_1+m_2} b.$

Proof. For (ⅰ), we have

For (ⅱ), we have

Statement (ⅲ) is clear. □

Lemma 2.11. Let $n\geq 3$ be a natural number and $G = Q_n.$ Let $(a, b)$ be a standard generator pair of $G.$ If $m_1, m_2\in \mathbb{Z},$ then

(i) $a^{m_1} b \cdot a^{m_2} b = a^{m_1-m_2+2^{n-2}}$;

(ii) $a^{m_1} b\cdot a^{m_2} = a^{m_1-m_2}b$;

(iii) $a^{m_1}\cdot a^{m_2} b = a^{m_1+m_2} b.$

Proof. For (ⅰ), we have

For (ⅱ), we have

Statement (ⅲ) is clear. □

We now prove some properties of modular $p$-groups. From now until Lemma 2.18, we work under the following hypothesis.

Hypothesis 2.12. Let $p$ be a prime number and $n$ be a natural number, where $n\geq 4$ if $p = 2$ and $n\geq 3$ if $p$ is odd. Set $G: = M_n(p);$ let $(a, b)$ be a standard generator pair of $G$ and $z: = a^{p^{n-2}}$.

Lemma 2.13. $\Phi(G) = Z(G)$ is the unique maximal subgroup of $\langle a\rangle.$ In particular, we have $z\in Z(G).$

Proof. This follows from [10, Chapter 5, Theorem 4.3(ⅰ)]. □

Lemma 2.14. For each $k \in \mathbb{Z}$, we have $b^{k} a b^{-k} = z^{-k} a$.

Proof. We show by induction that, for any non-negative integer $k$, we have $b^{k} a b^{-k} = z^{-k} a$ and $b^{-k} a b^{k} = z^{k} a$. This is clear for $k = 0$.

We now consider the case $k = 1$. We have $b^{-1}ab = a^{p^{n-2}+1} = za$. Since $z \in Z(G)$ by Lemma 2.13, we also have

and, thus, $bab^{-1} = z^{-1}a.$

Assume now that $k \ge 2$ is an integer such that $b^{k-1} a b^{-(k-1)} = z^{-(k-1)} a$ and $b^{-(k-1)} a b^{k-1} = z^{k-1} a$. Bearing in mind that $z \in Z(G)$ by Lemma 2.13 and that $bab^{-1} = z^{-1}a$, we see that

Similarly, we have

□

Lemma 2.15. Let $\ell, j \in \mathbb{Z}$. For each positive integer $k,$ we have

Proof. We prove the lemma by induction on $k$.

For $k = 1,$ we have

Suppose now that $k$ is a positive integer for which (2.1) holds. Then,

□

Lemma 2.16. Let $S\in \Gamma(G).$ Then, $S$ contains an element of order $p^{n-1}.$

Proof. Since $S\not\subseteq \langle a^p, b \rangle = \langle a^p \rangle \langle b \rangle \trianglelefteq G$, the set $S$ contains an element of the form $a^{\ell}b^j$, where $0 < \ell < p^{n-1}$ is not divisible by $p$ and $0 \le j < p$. By Lemma 2.15, we have

If $p = 2,$ then we have $(a^{\ell} b^{j})^{2} = (a^{\ell})^{2}$ or $(a^{\ell})^{2} z.$ In either case, $(a^{\ell} b^{j})^{2}$ is of order $2^{n-2},$ which implies that $a^{\ell} b^{j}$ is of order $2^{n-1}.$ If $p$ is odd, then $(a^{\ell} b^{j})^{p} = (a^{\ell})^{p}$ is of order $p^{n-2}$ and $a^{\ell} b^{j}$ is of order $p^{n-1}.$ □

Lemma 2.17. Let $x$ be an element of $G$ with order $p^{n-1}.$ Then, there is some $y\in G$ such that $(x, y)$ is a standard generator pair of $G.$

Proof. Since $x$ is of order $p^{n-1}$ and $G$ is of order $p^{n},$ the subgroup $\langle x\rangle$ is maximal in $G.$ By Lemma 2.13, we have $\langle z\rangle\subseteq Z(G) = \Phi(G)\subseteq \langle x\rangle.$ Because $\langle x\rangle$ is cyclic, $\langle x\rangle$ has only one subgroup of order $p.$ Hence, $\langle z\rangle$ is the only subgroup of $\langle x\rangle$ with order $p.$ Therefore, $\langle b\rangle$ is not contained in $\langle x\rangle.$ Hence, $\langle b\rangle$ is a complement of $\langle x\rangle$ in $G.$ Now, [13, Theorem 5.3.2] implies that there is some $y\in \langle b\rangle$ such that $(x, y)$ is a standard generator pair of $G.$ □

Lemma 2.18. Let $S\in \Gamma(G).$ Then, there is a standard generator pair $(x, y$) of $G$ such that $x\in S$ and $x^{\ell} y^{j}\in S$ for $0\leq \ell < p^{n-1}$ and $0 < j < p.$

Proof. By Lemma 2.16, $S$ contains an element $x$ of order $p^{n-1}.$ By Lemma 2.17, there is an element $y$ of $G$ such that $(x, y)$ is a standard generator pair of $G.$ Since $S\not\subseteq \langle x\rangle,$ there exist $0\leq \ell < p^{n-1}$ and $0 < j < p$ such that $x^{\ell} y^{j}\in S,$ as required. □

Given a group $X$ and an element $h$ of $X$, we write $\mathrm{Conj}_X(h)$ for the conjugacy class of $h$ in $X$.

Lemma 2.19. Let $0\leq \ell < p^{n-1}$ and $0 < j < p.$ Then, the following holds:

(i) $\mathit{\mbox{Conj}}_{G}(a) = \{ z^{r} a \big| 0\leq r < p\}.$

(ii) $\mathit{\mbox{Conj}}_{G}(a^{\ell} b^{j}) = \{ z^{r} a^{\ell} b^{j} \big| 0\leq r < p\}.$

(iii) $\mathit{\mbox{Conj}}_{G}(a^{-1}) = \{ z^{r} a^{-1} \big| 0\leq r < p\}.$

(iv) $\mathit{\mbox{Conj}}_{G}((a^{\ell} b^{j})^{-1}) = \{ z^{r} a^{-\ell} b^{-j} \big| 0\leq r < p\}.$

Proof. (ⅰ) We have $a^{b} = za.$ By induction, we conclude that $a^{{b}^{r}} = z^{r} a$ for all $r\geq 0.$ It follows that

(ⅱ) We have $b^{a} = a^{-1} b a = a^{-1} b a b^{-1} b = a^{-1} a^{b^{p-1}} b = a^{-1}z^{p-1}ab = a^{-1} z^{-1} ab = z^{-1} b.$ By induction, we conclude that $b^{{a}^{s}} = z^{-s} b$ for all $s \ge 0$. It follows that

(ⅲ) We have

(ⅳ) We have

and, hence,

From (ⅱ), we now see that

□

Lemma 2.20. Let $G = SD_n$ for some $n\geq 4$ , $G = Q_n$ for some $n\geq 3$ , $G = M_n(2)$ for some $n\geq 4,$ or $G = M_n(p)$ for some $n\geq 3$ and some odd prime $p$. Then, $\Delta_2(G) = \Delta(G).$

Proof. Since $\Gamma_2(G)\subseteq \Gamma(G)$, we have $\Delta_2(G)\leq\Delta(G).$ Now, we want to show that $\Delta(G)\leq \Delta_2(G).$ Let $S\in \Gamma(G).$ We see from Lemmas 2.9 and 2.18 that there is a subset $T$ of $S$ such that $T\in \Gamma_2(G).$ Since $\| G\|_{S}\leq \| G\|_{T},$ we have

□

3.   Proof of Theorem 1.4

Proof of Theorem 1.4. Let $n\geq 4$ be a natural number $, G: = SD_n,$ and $(a, b)$ be a standard generator pair of $G.$ Let $0 < \hat{o_1}, \hat{o_2} < 2^{n-1}$ be fixed and odd and $0\leq v_1 < 2^{n-1}$ be fixed and even. Set

From Lemmas 2.9 and 2.20, we see that we only need to consider $\|G\|_{S_j},$ where $1\leq j\leq 3,$ in order to determine $\Delta(G).$ In the sense of (1.1), and in view of Lemma 2.6, we have

For the reader's convenience, and to make the proof easy to follow, we set the following:

Next, we study $\|G\|_{S_j},$ where $1\leq j\leq 3.$

(ⅰ) We show that

For every $g\in C_1,$ we have

Now, suppose that $g\in \hat{C}_1\setminus{\{1\}}.$ Then, $g$ is either

where $0 < \hat{o} < 2^{n-1}$ is odd and $v = \hat{o}-\hat{o}_1,$

or

If the former holds, then $||g||_{S_1} = 2.$ Assume now $g\in \langle a \rangle \setminus \big\{ a^{\pm \hat{o}_1}, (a^{-1+2^{n-2}})^{\pm \hat{o}_1} \big\}.$

Case 1: $n = 4.$

It is easy to see that $\big\{ a^{\pm \hat{o}_1}, (a^{-1+2^{n-2}})^{\pm \hat{o}_1} \big\} = \big\{a, a^{3}, a^{5}, a^{7}\big\}.$ Hence, $g$ can be written as

where $0 < v < 8$ is even. Thus, $||g||_{S_1} = 2.$

Case 2: $n\geq 5.$

Let $0 < \ell < 2^{n-1}$ such that $g = a^{\ell}.$ If $\ell$ is even, then

and, hence, $||g||_{S_1} = 2.$ Assume now that $\ell$ is odd. One can see from Lemma 2.10 that $||a^{\ell}||_{S_{1}}\neq 2.$ But, we have

by Lemma 2.2; hence, $||g||_{S_1} = 3.$

Now, let $g = 1;$ then, by convention, $\{1\} = B_{S_{1}}(0)$ and, thus, $||g||_{S_1} = 0.$ Hence, $||G||_{S_1} = 2$ when $n = 4$ and $||G||_{S_1} = 3$ when $n\geq 5.$

(ⅱ) We show that

For every $g\in C_2,$ we have

Now, suppose that $g\in \hat{C}_2\setminus{\{1\}}.$ Then, $g$ is either

where $0\leq v < 2^{n-1}$ is even and $\hat{o} = v-\hat{o}_1,$

or

If the former holds, then $||g||_{S_2} = 2.$ Assume now $g\in \langle a \rangle \setminus \big\{ a^{\pm \hat{o}_1}, (a^{-1+2^{n-2}})^{\pm \hat{o}_1} \big\}.$

Case 1: $n = 4.$

As in (ⅰ) above, we have $\big\{ a^{\pm \hat{o}_1}, (a^{-1+2^{n-2}})^{\pm \hat{o}_1} \big\} = \big\{a, a^{3}, a^{5}, a^{7}\big\}.$ Hence, $g$ can be written as

where $0 < v < 8$ is even and $\hat{o} = v+1-2^{n-2}.$ Thus, $||g||_{S_2} = 2.$

Case 2: $n\geq 5.$

Let $0 < \ell < 2^{n-1}$ such that $g = a^{\ell}.$ If $\ell$ is even, then

where $\hat{o} = \ell+1-2^{n-1}$ and, hence, $||g||_{S_2} = 2.$ Assume now that $\ell$ is odd. One can see from Lemma 2.10 that $||a^{\ell}||_{S_{2}}\neq 2.$ But, we have

by Lemma 2.2; hence, $||g||_{S_2} = 3.$

Now, let $g = 1;$ then, by convention, $\{1\} = B_{S_{2}}(0)$ and, thus, $||g||_{S_2} = 0.$ Hence, $||G||_{S_2} = 2$ when $n = 4$ and $||G||_{S_2} = 3$ when $n\geq 5.$

(ⅲ) We show $||G||_{S_3} = 2.$

For every $g\in C_3,$ we have

Now, suppose that $g\in \hat{C}_3\setminus{\{1\}}.$ Then, $g$ can be written as

where $0 < \ell < 2^{n-1}.$

Now, let $g = 1;$ then, by convention, $\{1\} = B_{S_{3}}(0)$ and, thus, $||g||_{S_3} = 0.$ So, $||G||_{S_{3}} = 2.$

In view of (ⅰ)–(ⅲ) and Lemma 2.9, we have $\Delta_2(G) = 2$ if $n = 4$ and $\Delta_2(G) = 3$ if $n > 4.$ So, the result follows from that fact that $\Delta_2(G) = \Delta(G)$ (see Lemma 2.20). □

4.   Proof of Theorem 1.5

Proof of Theorem 1.5. Let $n\geq 3$ be a natural number $, G: = Q_n,$ and $(a, b)$ be a standard generator pair of $G.$ Let $0 < \hat{o_1}, \hat{o_2} < 2^{n-1}$ be fixed and odd and $0\leq v_1 \, < \, 2^{n-1}$ be fixed and even. Set

From Lemmas 2.9 and 2.20, we see that we only need to consider $\|G\|_{S_j},$ where $1\leq j\leq 3,$ in order to determine $\Delta(G).$ In the sense of (1.1), and in view of Lemma 2.8, we have

For the reader's convenience, and to make the proof easy to follow, we set the following:

Next, we study $\|G\|_{S_j},$ where $1\leq j\leq 3.$

(ⅰ) We show that

For every $g\in C_1,$ we have

Now, suppose that $g\in \hat{C}_1\setminus{\{1\}}.$ Then, $g$ is either

where $0 < \hat{o} < 2^{n-1}$ is odd and $v = \hat{o} -\hat{o}_1,$

or

If the former holds, then $||g||_{S_1} = 2.$ Assume now $g\in \langle a \rangle \setminus \big\{ a^{\pm \hat{o}_1}\big\}.$

Case 1: $n = 3.$

It is easy to see that $\big\{ a^{\pm \hat{o}_1} \big\} = \big\{a, a^{3}\big\}.$ Hence, $g$ can be written as

Thus, $||g||_{S_1} = 2.$

Case 2: $n\geq 4.$

Let $0 < \ell < 2^{n-1}$ such that $g = a^{\ell}.$ If $\ell$ is even, then

and, hence, $||g||_{S_1} = 2.$ Assume now that $\ell$ is odd. One can see from Lemma 2.11 that $||a^{\ell}||_{S_{1}}\neq 2.$ But, we have

by Lemma 2.2; hence, $||g||_{S_1} = 3.$

Now, let $g = 1;$ then, by convention, $\{1\} = B_{S_{1}}(0)$ and, thus, $||g||_{S_1} = 0.$ Hence, $||G||_{S_1} = 2$ when $n = 3$ and $||G||_{S_1} = 3$ when $n\geq 4.$

(ⅱ) We show that

For every $g\in C_2,$ we have

Now, suppose that $g\in \hat{C}_2\setminus{\{1\}}.$ Then, $g$ is either

where $0\leq v < 2^{n-1}$ is even and $\hat{o} = v-\hat{o}_1,$

or

If the former holds, then $||g||_{S_2} = 2.$ Assume now $g\in \langle a \rangle \setminus \big\{ a^{\pm \hat{o}_1} \big\}.$

Case 1: $n = 3.$

As in (ⅰ) above, we have $\big\{ a^{\pm \hat{o}_1} \big\} = \big\{a, a^{3}\big\}.$ Hence, $g$ can be written as

Thus, $||g||_{S_2} = 2.$

Case 2: $n\geq 4.$

Let $0 < \ell < 2^{n-1}$ such that $g = a^{\ell}.$ If $\ell$ is even, then

where $\hat{o} = \ell+1-2^{n-2}$ and, hence, $||g||_{S_2} = 2.$ Assume now that $\ell$ is odd. One can see from Lemma 2.11 that $||a^{\ell}||_{S_{2}}\neq 2.$ But, we have

by Lemma 2.2; hence, $||g||_{S_2} = 3.$

Now, let $g = 1;$ then, by convention, $\{1\} = B_{S_{2}}(0)$ and, thus, $||g||_{S_2} = 0.$ Hence, $||G||_{S_2} = 2$ when $n = 3$ and $||G||_{S_2} = 3$ when $n\geq 4.$

(ⅲ) We show $||G||_{S_3} = 2.$

For every $g\in C_3,$ we have

Now, suppose that $g\in \hat{C}_3\setminus{\{1\}}.$ Then, $g$ can be written as

where $0 < \ell < 2^{n-1}.$

Now, let $g = 1;$ then, by convention, $\{1\} = B_{S_{3}}(0)$ and, thus, $||g||_{S_3} = 0.$ So, $||G||_{S_{3}} = 2.$

In view of (ⅰ)–(ⅲ) and Lemma 2.9, we have $\Delta_2(G) = 2$ if $n = 3$ and $\Delta_2(G) = 3$ if $n\geq 4.$ So, the result follows from the fact that $\Delta_2(G) = \Delta(G)$ (see Lemma 2.20). □

5.   Proof of Theorem 1.6

To establish Theorem 1.6, we need to prove a series of lemmas first. In what follows, we work under the following hypothesis.

Hypothesis 5.1. Let $p$ be a prime number and $n$ be a natural number, where $n\geq 4$ if $p = 2$ and $n\geq 3$ if $p$ is odd. Set $G: = M_n(p);$ let $(a, b)$ be a standard generator pair of $G$ and $z: = a^{p^{n-2}}$. Moreover, let $S: = \{a, a^{\ell} b^{j}\}$ for some $0\leq \ell < p^{n-1}$ and $0 < j < p.$

Note that $z \in Z(G)$ by Lemma 2.13.

Lemma 5.2. Let $g\in \langle a\rangle.$ Then, we have the following:

(i) If $p = 2,$ then $\| g\|_{S}\leq 2^{n-3}.$

(ii) If $p$ is odd and $(p, n)\neq (3, 3),$ then $\| g\|_{S}\leq \frac{p^{n-2}-1}{2}.$

(iii) If $p = 3 = n,$ then $\| g\|_{S}\leq 2.$

Proof. Let $0\leq r < p^{n-1}$ with $g = a^{r}.$ Assume that $r$ is divisible by $p^{n-2}$. Then, we have

So, with $s: = \frac{r}{p^{n-2}},$ we have $g = z^s = a^{-1}\cdot z^{s} a\in B_S(2)$ by Lemma 2.19. Thus, $\| g\|_{S}\leq 2$ and, hence, (ⅰ)–(ⅲ) hold when $r$ is divisible by $p^{n-2}$.

We assume now that $r$ is not divisible by $p^{n-2}$, and we treat the cases $p = 2$ and $p$ is odd separately.

Case 1: $p = 2$.

We may write $r$ in the form $r = m\cdot 2^{n-2}+s,$ where $m\in \{0, 1, 2\}$ and $-2^{n-3} < s \leq 2^{n-3}.$ Note that $s\neq 0$ by the choice of $r.$ Then, $g = a^{m\cdot 2^{n-2}+s} = (a^{{2}^{n-2}})^{m} a^{s} = z^{m} a^{s}.$

If $s > 0,$ then $g = z^{m} a\cdot a^{s-1}\in B_S(s)$ by Lemma 2.19; hence, $\| g\|_{S}\leq s \leq 2^{n-3}.$

If $s < 0,$ then $g = z^{m} a^{s} = z^{m} a^{-1} a^{s+1} = z^{m} a^{-1} (a^{-1})^{(-s-1)} \in B_S(-s)$ by Lemma 2.19 and, hence, $\| g\|_{S}\leq -s < 2^{n-3}.$

Case 2: $p$ is odd.

We may write $r$ in the form $r = m\cdot p^{n-2}+s,$ where $m\in \{0, 1, \dots, p\}$ and $-\frac{p^{n-2}-1}{2} \leq s \leq \frac{p^{n-2}-1}{2}.$ Note that $s\neq 0$ by the choice of $r$. Then, $g = a^{m\cdot p^{n-2}+s} = z^{m} a^{s}.$

If $s > 0,$ then $g = z^{m} a\cdot a^{s-1}\in B_S(s)$ by Lemma 2.19 and, hence, $\| g\|_{S}\leq \frac{p^{n-2}-1}{2}.$ In particular, if $(p, n) = (3, 3),$ then $\| g\|_{S} = 1 < 2.$

If $s < 0,$ then $g = z^{m} a^{s} = z^{m} a^{-1} a^{s+1} = z^{m} a^{-1} (a^{-1})^{(-s-1)}\in B_S(-s)$ by Lemma 2.19 and, hence, $\| g\|_{S}\leq -s \leq \frac{p^{n-2}-1}{2}.$ In particular, if $(p, n) = (3, 3),$ then $\| g\|_{S} = 1 < 2.$ □

Lemma 5.3. Let $g\in G\setminus \langle a\rangle.$ Then, the following holds:

(i) If $p = 2,$ then $\| g\|_{S}\leq 2^{n-3}+1$.

(ii) If $p$ is odd and $(n, p)\neq (3, 3),$ then $\| g\|_{S}\leq \frac{p^{n-2}+p-2}{2}$.

Proof. Assume that $p = 2.$ Since $g\in G\setminus \langle a\rangle,$ we have $g = a^{r}b$ for some $0\leq r < 2^{n-1}.$ By Lemma 5.2(ⅰ), we have $a^{r-\ell}\in B_S(2^{n-3}).$ Also, we have $a^{\ell}b\in S\subseteq B_S(1).$ Now, Lemma 2.2 implies that $g = a^{r}b = a^{r-\ell} a^{\ell} b\in B_S(2^{n-3}+1).$ In other words, $\| g\|_{S}\leq 2^{n-3}+1,$ so (ⅰ) holds.

Assume now that $p$ is odd and $(n, p) \neq (3, 3)$. Since $g\in G \setminus \langle a\rangle,$ we have $g = a^{r}b^{s}$ for some $0\leq r < p^{n-1}$ and $0 < s < p.$ Because $0 < j < p$ and $\mathit{\mbox{ord}}(b) = p,$ we have $\mathit{\mbox{ord}}(b^j) = p$, and this easily implies that

Since $1 \neq b^{s} \in \langle b \rangle$, we have $b^{s} = b^{jk}$ or $b^{s} = b^{-jk}$ for some $0 < k\leq \frac{p-1}{2}.$ By Lemma 2.15, we have

and

Suppose that $b^{s} = b^{jk}.$ Then,

Set $u: = a^{k\ell} z^{-\frac{k(k-1)}{2} j\ell};$ then, $(a^{\ell} b^{j})^{k} = ub^{s}.$ We have $g = a^{r}b^{s} = a^{r} u^{-1} u b^{s} = a^{r} u^{-1} (a^{\ell} b^{j})^{k}.$ By Lemma 5.2(ⅱ), we have $a^{r} u^{-1} \in B_S(\frac{p^{n-2}-1}{2}).$ Moreover, $(a^{\ell} b^{j})^{k}\in B_S(\frac{p-1}{2})$ since $k\leq\frac{p-1}{2}.$ Applying Lemma 2.2, we conclude that $g\in B_S(\frac{p^{n-2}-1}{2}+\frac{p-1}{2}) = B_S(\frac{p^{n-2}+p-2}{2});$ hence, $\| g\|_{S}\leq \frac{p^{n-2}+p-2}{2}.$ This completes the proof of (ⅱ) for the case that $b^{s} = b^{jk}.$

Assume now that $b^{s} = b^{-jk}.$ Then,

With $u: = a^{-k\ell} z^{-\frac{k(k-1)}{2}j\ell},$ we have that $(a^{-\ell} b^{-j})^{k} = ub^{s}$ and $g = a^{r} b^{s} = a^{r} u^{-1} u b^{s} = a^{r} u^{-1} (a^{-\ell} b^{-j})^{k}.$ As in the case in which $b^{s} = b^{jk},$ we see that $a^{r} u^{-1} \in B_S(\frac{p^{n-2}-1}{2}),$ and, from Lemma 2.19(ⅳ), we see that $(a^{-\ell} b^{-j})^k\in B_S(k)\subseteq B_S(\frac{p-1}{2}).$ As in the case in which $b^{s} = b^{jk}$, it follows that $\| g\|_{S}\leq \frac{p^{n-2}+p-2}{2}.$ The proof is now complete. □

Lemma 5.4. Assume that $n = p = 3.$ If $g\in G\setminus \langle a\rangle,$ then $\| g\|_{S}\leq 2.$

Proof. Let $0\leq r < 9$ and $0 < s < 3$ such that $g = a^{r} b^{s}.$ Note that $\langle b^{s}\rangle = \langle b \rangle = \langle b^{j}\rangle$.

Case 1: $\ell$ is not divisible by $3.$

The elements $a^{\ell}, za^{\ell}, z^{2}a^{\ell}, a^{-\ell}, z a^{-\ell}, z^{2} a^{-\ell}$ are easily seen to be mutually distinct, and they are not contained in $\langle z\rangle.$ So, it follows that $\langle a\rangle \backslash \langle z\rangle = \{ a^{\ell}, za^{\ell}, z^{2}a^{\ell}, a^{-\ell}, z a^{-\ell}, z^{2} a^{-\ell}\}$. Likewise, one can see that $\langle a\rangle \backslash \langle z\rangle = \{ a, za, z^{2}a, a^{-1}, z a^{-1}, z^{2} a^{-1}\}$. Together with Lemma 2.19, it follows that

Case 1.1: $r$ is not divisible by $3, s = j.$

In this case, $a^{r}\notin \langle z\rangle.$ Therefore, by the above observations, we have $a^{r} = z^{m} a^{\ell}$ or $a^{r} = z^{m} a^{-\ell}$ for some $0\leq m < 3.$ If the former holds, then $g = z^{m} a^{\ell} b^{j}\in {\mbox{Conj}}_{G}(a^{\ell} b^{j})$ by Lemma 2.19(ⅱ) and, hence, $\| g\|_{S} = 1 < 2.$ If $a^{r} = z^{m} a^{-\ell},$ then $g = z^{m} a^{-\ell} b^{j} = a^{-2\ell} z^{m} a^{\ell} b^{j}.$ Since $a^{-2\ell} \in \langle a\rangle\backslash \langle z\rangle = {\mbox{Conj}}_{G}(a)\cup {\mbox{Conj}}_{G}(a^{-1})$ and $z^{m} a^{\ell} b^{j}\in {\mbox{Conj}}_{G}(a^{\ell} b^{j})$ by Lemma 2.19(ⅱ), it follows that $\| g\|_{S}\leq 2.$

Case 1.2: $r$ is not divisible by $3, s\neq j.$

In this case, $b^{s} = b^{-j}.$ Also, $a^{r} = z^{m} a^{\ell}$ or $z^{m} a^{-\ell}$ for some $0\leq m < 3.$ If the latter holds, then $g = z^{m} a^{-\ell} b^{-j}\in B_S(1)$ by Lemma 2.19(ⅳ) and, hence, $\| g\|_{S} = 1 < 2.$ If $a^{r} = z^{m} a^{\ell},$ then $g = z^{m} a^{\ell} b^{-j} = a^{2\ell} z^{m} a^{-\ell} b^{-j},$ and, since $a^{2\ell} \in \langle a \rangle \setminus \langle z \rangle = {\mbox{Conj}}_{G}(a)\cup {\mbox{Conj}}_{G}(a^{-1})$ and $z^{m} a^{-\ell} b^{-j} \in \mathrm{Conj}_G((a^{\ell}b^j)^{-1})$ by Lemma 2.19(ⅳ), it follows that $\| g\|_{S}\leq 2.$

Case 1.3: $r$ is divisible by $3.$

In this case, $g = z^{m} b^{j}$ or $g = z^{m} b^{-j}$ for some $0\leq m < 3.$ In the former case, we have $g = a^{-\ell} z^{m} a^{\ell} b^{j},$ and, in the latter case, we have $g = a^{\ell} z^{m} a^{-\ell} b^{-j}.$ We have $a^{\ell}, a^{-\ell} \in \langle a\rangle\backslash \langle z\rangle = {\mbox{Conj}}_{G}(a)\cup {\mbox{Conj}}_{G}(a^{-1});$ we also have that $z^{m} a^{\ell} b^{j}\in {\mbox{Conj}}_{G}(a^{\ell} b^{j})$ by Lemma 2.19(ⅱ) and $z^{m} a^{-\ell} b^{-j}\in {\mbox{Conj}}_{G}((a^{\ell} b^{j})^{-1})$ by Lemma 2.19(ⅳ), so it follows that $\| g\|_{S}\leq 2.$

Case 2: $\ell$ is divisible by $3.$

In this case, $a^{\ell} = z^{t}$ for some $0\leq t < 3.$ Note that $g = a^{r}b^{j}$ or $g = a^{r} b^{-j}$. Assume that $r$ is divisible by $3$. Then $g = z^mb^j$ or $g = z^m b^{-j}$ for some $0 \leq m < 3$. If the former holds, then $g = z^{m}b^{j} = z^{m-t}z^{t}b^{j} = z^{m-t}a^{\ell}b^{j}\in {\mbox{Conj}}_{G}(a^{\ell} b^{j})$ by Lemma 2.19(ⅱ) and, hence, $\| g\|_{S} = 1 < 2$. Otherwise, if $g = z^m b^{-j}$, then $g = z^{m} b^{-j} = z^{m+t} z^{-t} b^{-j} = z^{m+t} a^{-\ell} b^{-j}\in {\mbox{Conj}}_{G}((a^{\ell} b^{j})^{-1})$ by Lemma 2.19(ⅳ); hence, $\| g\|_{S} = 1 < 2.$ Assume now that $r$ is not divisible by $3$. Then, $r-1$ or $r+1$ is divisible by $3;$ thus, $a^{r-1}$ or $a^{r+1}$ lies in $\langle z\rangle.$ Hence, $a^{r} = z^{m} a$ or $z^{m} a^{-1}$ for some $0\leq m < 3.$ If $a^r = z^ma$, then either $g = az^{m}b^{j} = az^{m-t}z^tb^j = az^{m-t}a^{\ell}b^j$ or $g = az^{m}b^{-j} = az^{m+t}z^{-t}b^{-j} = az^{m+t}a^{-\ell}b^{-j};$ so, it follows from Lemma 2.19(ⅱ), (ⅳ) that $\| g\|_{S}\leq 2.$ If $a^r = z^ma^{-1}$, then we either have $g = a^{-1} z^{m}b^{j} = a^{-1}z^{m-t}z^tb^j = a^{-1}z^{m-t}a^{\ell}b^j$ or $g = a^{-1} z^{m}b^{-j} = a^{-1}z^{m+t}z^{-t}b^{-j} = a^{-1}z^{m+t}a^{-\ell}b^{-j};$ then it once again follows from Lemma 2.19(ⅱ), (ⅳ) that $\| g\|_{S}\leq 2.$ □

Lemma 5.5. Let $X_1 : = \langle z \rangle \cdot \lbrace a, a^{-1} \rbrace$, $X_2 : = \langle z \rangle \cdot \lbrace b, b^{-1} \rbrace,$ and $X : = X_1 \cup X_2$. Let $g\in G$ and $s, t$ be non-negative integers such that $g$ can be written in the form $g = x_1 \cdots x_{s+t}$ such that $x_i \in X$ for all $1 \le i \le s+t$, $|\lbrace 1 \le i \le s+t \mid x_i \in X_1 \rbrace| = s,$ and $|\lbrace 1 \le i \le s+t \mid x_i \in X_2 \rbrace| = t$. Then, $g = z^{r} a^{s_0} b^{t_0}$ for some $0\leq r < p,$ $-s\leq s_0\leq s,$ and $-t\leq t_0\leq t.$

Proof. We proceed by induction over $k : = s+t.$ If $k = 0,$ then $g = 1 = z^{0} a^{0} b^{0};$ thus the lemma is true for $k = 0.$ Assume now that $k \ge 1$ and suppose that the following holds: If $g' \in G$ and $s^{\prime}, t^{\prime}$ are non-negative integers with $s^{\prime} + t^{\prime} < k$ such that $g^{\prime}$ can be written in the form $g' = x_1^{\prime} \cdots x_{s^{\prime} + t^{\prime}}^{\prime}$, where $x_i^{\prime} \in X$ for all $1 \le i \le s' + t'$, $|\lbrace 1 \le i \le s^{\prime}+t^{\prime} \mid x_i^{\prime} \in X_1 \rbrace| = s^{\prime},$ and $|\lbrace 1 \le i \le s^{\prime}+t^{\prime} \mid x_i^{\prime} \in X_2 \rbrace| = t^{\prime}$, then $g' = z^{r'} a^{s_0^{\prime}} b^{t_0^{\prime}}$ for some $0 \le r' < p$, $-s' < s_0^{\prime} < s'$, $-t' < t_0^{\prime} < t'$ (induction hypothesis).

By the hypotheses of the lemma, we have $g = x_1 \cdots x_k$ for some $x_1, \dots, x_k \in X$, where $|\lbrace 1 \le i \le k \mid x_i \in X_1 \rbrace| = s$ and $|\lbrace 1 \le i \le k \mid x_i \in X_2 \rbrace| = t$. Using the induction hypothesis, we are going to prove that $g = z^{r} a^{s_0} b^{t_0}$ for some $0\leq r < p,$ $-s\leq s_0\leq s,$ and $-t\leq t_0\leq t.$ Set $h: = x_1\cdots x_{k-1}.$ We split the proof into two cases.

Case 1: $x_k\in X_1$.

In this case, $|\lbrace 1 \le i \le k-1 \mid x_i \in X_1 \rbrace| = s-1$ and $|\lbrace 1 \le i \le k-1 \mid x_i \in X_2 \rbrace| = t$. Therefore, by the induction hypothesis, $h = z^{{r}^{\prime}} a^{{s}^{\prime}_0} b^{{t}^{\prime}_0}$ for some $0\leq r^{\prime} < p, -(s-1)\leq s_0^{\prime}\leq s-1,$ and $-t\leq t_0^{\prime}\leq t.$ Since $x_k\in X_1,$ we have $x_k = z^{r^{\prime\prime}} a^{\varepsilon}$ for some $0 \le r^{\prime\prime} < p$ and $\varepsilon\in \{\pm 1\}.$ Thus, $g = h\cdot x_k = z^{{r}^{\prime}} a^{{s}^{\prime}_0} b^{{t}^{\prime}_0} z^{r^{\prime\prime}} a^{\epsilon} = z^{{r}^{\prime}+{r}^{\prime\prime}} a^{{s}^{\prime}_0} b^{{t}^{\prime}_0} a^{\epsilon}.$ From Lemma 2.14, one can easily see that $b^{{t}^{\prime}_0} a^{\epsilon} = z^{{r}^{\prime\prime\prime}} a^{\epsilon} b^{{t}^{\prime}_0},$ where $r^{\prime\prime\prime}\in \{t^{\prime}_0, -t^{\prime}_0\}.$ Then, it follows that $g = z^{r^{\prime}+r^{\prime\prime}} a^{{s}^{\prime}_0} b^{{t}^{\prime}_0} a^{\epsilon} = z^{r^{\prime}+r^{\prime\prime} +r^{\prime\prime\prime}} a^{{s}^{\prime}_0} a^{\epsilon} b^{{t}^{\prime}_0} = z^{r^{\prime}+r^{\prime\prime} +r^{\prime\prime\prime}} a^{{s}^{\prime}_0+\epsilon} b^{{t}^{\prime}_0}.$ Now, let $0\leq r < p$ with $r\equiv r^{\prime}+r^{\prime\prime} +r^{\prime\prime\prime}\, \mathit{\mbox{mod}}\, p,$ and set $s_0: = s^{\prime}_0+\epsilon, t_0: = t^{\prime}_0.$ Then, $g = z^{r} a^{{s}_0} b^{{t}_0},$ and we have that $-s\leq -(s-1)+\epsilon\leq s_0\leq (s-1)+\epsilon\leq s, -t\leq t_0\leq t.$ This completes the proof of the lemma for the case of $x_k\in X_1.$

Case 2: $x_k\in X_2$.

In this case, $|\lbrace 1 \le i \le k-1 \mid x_i \in X_1 \rbrace| = s$ and $|\lbrace 1 \le i \le k-1 \mid x_i \in X_2 \rbrace| = t-1$. Therefore, by the induction hypothesis, $h = z^{{r}^{\prime}} a^{{s}^{\prime}_0} b^{{t}^{\prime}_0}$ for some $0\leq r^{\prime} < p, -s\leq s_0^{\prime}\leq s,$ and $-(t-1)\leq t_0^{\prime}\leq t-1.$ Since $x_k\in X_2,$ we have that $x_k = z^{r^{\prime\prime}} b^{\varepsilon}$ for some $0\leq r^{\prime\prime} < p$ and $\varepsilon\in \{\pm 1\}$. Then, $g = hx_k = z^{{r}^{\prime}} a^{{s}^{\prime}_0} b^{{t}^{\prime}_0} x_k = z^{{r}^{\prime}+r^{\prime\prime}} a^{{s}^{\prime}_0} b^{{t}^{\prime}_0+\epsilon}.$ Let $0\leq r < p$ with $r\equiv r^{\prime}+r^{\prime\prime} \, \mathit{\mbox{mod}}\, p, s_0: = s^{\prime}_0,$ and $t_0: = t^{\prime}_0+\epsilon.$ Then, $g = z^{r} a^{{s}_0} b^{{t}_0},$ and we have that $-s\leq s_0\leq s$ and $-t\leq t_0\leq t.$ This completes the proof of the lemma for the case of $x_k\in X_2.$ □

Lemma 5.6. Let $T : = \lbrace a, b \rbrace$. Then,

(i) $\| G\|_{T} \geq 2^{n-3}+1$ if $p = 2,$

(ii) $\| G\|_{T} \geq \frac{p^{n-2}+p-2}{2}$ if $p$ is odd.

Proof. Let $X_1 : = \langle z \rangle \cdot \lbrace a, a^{-1} \rbrace$, $X_2 : = \langle z \rangle \cdot \lbrace b, b^{-1} \rbrace$ and $X : = X_1 \cup X_2$. From Lemma 2.19, we see that $X = \mathrm{Conj}_G(T^{\pm 1})$.

Assume that $p = 2$, and set $g: = a^{2^{n-3}}b.$ We show that $\| g\|_{T}\geq 2^{n-3}+1$. Let $k \geq 1$ and $x_1, \dots, x_k\in X$ with $g = x_1\cdots x_k.$ Set $s : = |\lbrace 1 \le i \le k \mid x_i \in X_1 \rbrace|$ and $t : = |\lbrace 1 \le i \le k \mid x_i \in X_2 \rbrace|$. Then, by Lemma 5.5, we have $a^{2^{n-3}}b = g = z^{r} a^{s_0} b^{t_0}$ for some $r\in\{0, 1\}$ and $-s \leq s_0\leq s$, $-t \leq t_0\leq t.$ This implies that $a^{2^{n-3}} = z^{r} a^{s_0}$ and $b = b^{t_0}.$ If $s < 2^{n-3}$, then $z^{r} a^{s_0}$ has the form $a^{m}, a^{2^{n-1}-m}, a^{2^{n-2}+m},$ or $a^{2^{n-2}-m}$ for some $0\leq m < 2^{n-3};$ hence, $z^{r} a^{s_0} \neq a^{2^{n-3}}.$ Thus, $s\geq 2^{n-3}.$ Also, $t\geq 1$ because, otherwise, $b = b^{t_0} = b^0 = 1$. So, we have $k = s+t\geq 2^{n-3}+1.$ Since $x_1, \dots, x_k$ were arbitrarily chosen elements of $X = \mathrm{Conj}_G(T^{\pm 1})$ with $g = x_1 \cdots x_k$, we can now conclude that $\| g\|_{T}\geq 2^{n-3}+1$. In particular, we have $\| G\|_{T}\geq 2^{n-3}+1$, completing the proof of (ⅰ).

Assume now that $p$ is odd, and set $g: = a^{{\frac{p^{n-2}-1}{2}}} b^{{\frac{{p-1}}{2}}}.$ We show that $\| g\|_{T}\geq \frac{p^{n-2}+p-2}{2}.$ Let $k \geq 1$ and $x_1, \dots\, , x_k\in X$ with $g = x_1\cdots\, x_k.$ Set $s : = |\lbrace 1 \le i \le k \mid x_i \in X_1 \rbrace|$ and $t : = |\lbrace 1 \le i \le k \mid x_i \in X_2 \rbrace|$. Then, by Lemma 5.5, we have $a^{\frac{p^{n-2}-1}{2}} b^{\frac{p-1}{2}} = g = z^{r} a^{s_0} b^{t_0}$ for some $0\leq r < p$ as well as some $-s \leq s_0\leq s,$ $-t \leq t_0\leq t.$ This implies that $a^{\frac{p^{n-2}-1}{2}} = z^{r} a^{s_0}$ and $b^{t_0} = b^{\frac{p-1}{2}}$. If $s < \frac{p^{n-2}-1}{2},$ then one can see, similarly as in the case in which $p = 2,$ that $z^{r}a^{s_0}$ cannot be $a^{\frac{p^{n-2}-1}{2}}.$ So, it follows that $s\geq \frac{p^{n-2}-1}{2}.$ Also, it follows from $b^{t_0} = b^{\frac{p-1}{2}}$ that $t\geq \frac{p-1}{2}.$ Consequently, we have $k = s+t\geq \frac{p^{n-2}+p-2}{2}$. Since $x_1, \dots, x_k$ were arbitrarily chosen elements of $X = \mathrm{Conj}_G(T^{\pm 1})$ with $g = x_1 \cdots x_k$, we can now conclude that $\| g\|_{T}\geq \frac{p^{n-2}+p-2}{2}$. In particular, we have $\| G\|_{T}\geq \frac{p^{n-2}+p-2}{2},$ completing the proof of (ⅱ). □

With the above lemmas at hand, we can now prove Theorem 1.6.

Proof of Theorem 1.6. Let $p$ be a prime number and $n$ be a natural number, where $n \ge 4$ if $p = 2$ and $n \geq 3$ if $p$ is odd, and let $G : = M_n(p)$. Let $S\in \Gamma_2(G)$. By Lemma 2.18, there is a standard generator pair $(a, b)$ of $G$ such that $S = \{a, a^{\ell} b^{j}\}$ for some $0\leq \ell < p^{n-1}$ and $0 < j < p.$

Assume that $p = 2.$ By Lemmas 5.2(ⅰ) and 5.3(ⅰ), we have $\| G\|_{S}\leq 2^{n-3}+1$. Because of Lemma 5.6, we even have equality when $S = \lbrace a, b \rbrace$. So, it follows that $\Delta_2(G) = 2^{n-3}+1$, and Lemma 2.20 implies that $\Delta(G) = 2^{n-3}+1.$

Assume now that $p$ is odd and $(n, p)\neq (3, 3).$ By Lemmas 5.2(ⅱ) and 5.3(ⅱ), we have $\| G\|_{S}\leq \frac{p^{n-2}+p-2}{2}$, and, because of Lemma 5.6, we even have equality when $S = \lbrace a, b \rbrace$. Applying Lemma 2.20, we conclude that $\Delta(G) = \Delta_2(G) = \frac{p^{n-2}+p-2}{2}.$

Assume now that $n = p = 3$. From Lemmas 5.2(ⅲ) and 5.4, we see that $\| G\|_{S}\leq 2,$ and, because of Lemma 5.6, we even have $\| G\|_{S} = 2$ when $S = \lbrace a, b \rbrace$. Applying Lemma 2.20, we conclude that $\Delta(G) = \Delta_2(G) = 2 = \frac{3^{3-2}+3-2}{2}$. □

6.   Conclusions

After the derivation of the conjugacy diameters of dihedral groups in [1, Theorem 6.0.2] and [12, Example 2.8], we proved further results about conjugacy diameters of non-abelian finite groups with cyclic maximal subgroups. Namely, we have determined the conjugacy diameters of the semidihedral $2$-groups, the generalized quaternion groups and the modular $p$-groups. In this manner, the conjugacy diameters of non-abelian finite $p$-groups with cyclic maximal subgroups have been comprehensively calculated. We believe that the strategies applied in the proofs of our results could also be used to study the conjugacy diameters of other finite groups. For example, we think that one could proceed similarly as in the proofs of Theorems 1.4 and 1.5 to study the conjugacy diameters of the generalized dihedral groups.

Our results also lead to a question concerning the relation between the conjugacy class sizes of a finite group and its conjugacy diameter. We found that the semidihedral $2$-groups and the generalized quaternion groups have relatively small conjugacy diameters, while the elements not lying in a cyclic maximal subgroup have, in relation to the group order, relatively large conjugacy class sizes. On the other hand, we found that the conjugacy diameters of the modular $p$-groups $M_n(p)$ grow fast as $n$ grows, while the conjugacy classes of $M_n(p)$ are relatively small (they have at most $p$ elements). Considering this observation, it would be interesting to study how the conjugacy diameters of finite groups are influenced by conjugacy class sizes. Note that [12, Proposition 7.1] addresses this question.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

We dedicate this paper to our former supervisor Professor Benjamin Martin. You have our deepest thanks.

Conflict of interest

The authors declare no conflict of interest.