On Finite NDC-Groups

Abstract

A finite p-group G is called a NDC-group if $G'/{G' \cap N}$ is cyclic or $N/{N \cap Z(G)}$ is cyclic for every normal subgroup N in G. In this paper, first we give some properties of NDC-groups and in particular, we give the order of $G'Z(G)/Z(G)$ for a NDC-group G so that we could investigate the structure of NDC-groups. Next, some necessary and sufficient conditions for a p-group G whose quotient group G / Z(G) is generated by two elements to be a NDC-group are given. Finally, we also give some further properties of p-groups with small derived subgroups by using the properties of capable p-groups with small generating systems, which maybe could apply some other research of p-groups.

1 Introduction

All groups considered in this paper are finite.

It is well known that the normality of subgroups in a group could give lots of information about the group. A typical result is the classification of Dedekindian groups in [8]. Recall that a non-abelian Dedekindian group is called Hamiltonian and a non-abelian group is called metahamiltonian if all non-abelian subgroups of it are normal. Many authors investigated metahamiltonian groups. Romalis and Sesekin investigated some properties of infinite metahamiltonian groups in [19,20,21], and Nagrebeckii in [14,15,16] studied non-nilpotent metahamiltonian groups. Recently, An and Zhang investigated metahamiltonian p-groups in [2], and Fang and An classified the metahamiltonian p-groups. Of course, there are still lots of open questions for p-groups in which many subgroups are assumed to be normal. On the other hand, in a p-group there are many natural normal subgroups, for example, every subgroup which is contained in the center or every subgroup containing the derived subgroup of the group must be normal in the group. It is because of this reason, Silberberg [24, 25] studied the structure of a p-group G in which every normal subgroup either contains the derived subgroup $G'$ or is contained in the center Z(G). So it is quite interesting to investigate the p-groups in which every normal subgroup of a p-group is not “far away” from the center or the derived subgroup. For convenience, we give the following definition.

Definition 1

A p-group G is said to be a NDC-group if $G'/{G' \cap N}$ is cyclic or $N/{N \cap Z(G)}$ is cyclic for every normal subgroup N of G.

NDC-groups contain many important classes of p-groups. For example (see Proposition 1), all metahamiltonian p-groups are NDC-groups and therefore, p-groups all of whose subgroups of index $p^2$ are abelian, which have been extensively investigated (for example, see [6, 12, 23, 29]), are NDC-groups. Another example are p-groups G in which $C_G(A) = N_G(A)$ for any non-normal abelian subgroup A in G (see [1]).

Some authors have investigated the structure of p-groups G by the property that the normal closures of non-normal subgroups are “close to” G or “ far away” G. For example, a p-group G is said to be a $C(p^m)$-group for a fixed positive integer m if $[G : \langle a\rangle ^G]\le p^m$ for every non-normal cyclic subgroup $\langle a\rangle$ in G. Zhao and Guo in [31] proved that the order of a non-Dedekindian $C(p^2)$-group is bounded above by $p^6$ for $p>2$, and therefore, every $C(p^2)$-group must be a NDC-group if $p>2$. We also noticed that every $\mathcal {N}_{p^2}$ group must be a NDC-group if $p>2$ (a p-group is said to be an $\mathcal {N}_{p^m}$ if every non-normal cyclic subgroup has index no more than $p^m$ in their normalizers for a positive integer m, see [30]).

In this paper, first we give some properties of NDC-groups and in particular, we give the order of $G'Z(G)/Z(G)$ for a NDC-group G (see Theorem 3) so that we could investigate the structure of NDC-groups (See Sects. 5, 6). Next, some necessary and sufficient conditions for a p-group G whose quotient group G / Z(G) is generated by two elements to be a NDC-group are given. Finally, we also give some further properties of p-groups with small derived subgroups by using the properties of capable p-groups with small generating systems, which maybe could apply some other research of p-groups.

2 Preliminaries

In this section, we recall some basic concepts and notations and then give some basic results which are useful in the sequel.

We use $D_{2^n}$, $Q_{2^n}$, $SD_{2^n}$, $C_{p^n}$ and $C_p^n$ to denote the dihedral group of order $2^n$, the generalized quaternion group of order $2^n$, the semi-dihedral group of order $2^n$, the cyclic group of order $p^n$ and the elementary abelian group of order $p^n$, respectively. We use $A \times B$, $A *B$ and $A \backslash B$ to denote the direct product, the central product and the set $\{x | x\in A$, but $x\not \in B\}$ for a group A and a group B. We also use d(G), c(G) and $G_i$ to denote the minimal number of generators of a group G, the nilpotent class of a group G and the ith term of the lower central series of a nilpotent group G, respectively. If G is a p-group, then $\Omega _{i}(G) = \langle g \in G \mid g^{p^i} = 1\rangle$ and $\mho _{i}(G)$= $\langle g^{p^i} \mid g \in G\rangle$. Furthermore, if H is a subgroup of a p-group G and a is an element of G, then $[a, H] = \langle [a, b] \mid b \in H \rangle$ and $H(n) = \mho _{n}(H') \mho _{n-1}(H_p) \mho _{n-2}(H_{p^2}) \cdots H_{p^n}$, where $H_{p^i}$ is the $p^i$th term of the lower central series of H, $1 \le i \le n$ and $n \ge 1$. In addition, following Rédei [18], we use $M_p(n,m,1)$ to denote the group

$$\begin{aligned} \langle a, b \mid a^{p^n}=b^{p^m}=c^p=1, [a,b]=c, [c,a]=[c,b]=1\rangle , \end{aligned}$$

where $n\ge {m}\ge 1$, and if $p = 2$, then $m+n\ge 3$. All other terminology and notation not mentioned here are standard.

Definition 2

Let n be a positive integer. A group G is called n-abelian if for any two elements $a, b \in G$, $(ab)^{n} = a^{n}b^{n}$.

Definition 3

[9] A group G is called capable if there exists a group H such that $G \cong H/Z(H)$.

Definition 4

[5, $\S$7, Definition 1] A p-group G is called to be regular if for any x, $y \in G$, there is $c \in \mho _1(\langle x, y \rangle ')$ such that $x^py^p = (xy)^pc$.

Definition 5

[4, Definition 1.2] A p-group G is called to be $\Phi$-regular if for any x, $y \in G$,

$$\begin{aligned} (xy)^p = x^py^pc_3^p \cdots c_m^p, \quad \mathrm{where} ~ c_i \in \Phi (\langle x, y\rangle ), \quad i = 3, \cdots , m. \end{aligned}$$

Lemma 1

[2, Theorem 3.5] Let G be a p-group. Then, G is metahamiltonian if and only if $G'$ is contained in every non-abelian subgroup of G.

Lemma 2

[5, §1, Lemma 1.4] Let G be a p-group and $N \unlhd G$. If N has no abelian G-invariant subgroups of type (p, p), then N is either cyclic or isomorphic to one of the groups $D_{2^n}$, $Q_{2^n}$ and $SD_{2^n}$. If, in addition, $N \le \Phi (G)$, then N is cyclic.

Lemma 3

[3, Lemma 2.13] Let G be a p-group. If $G_m \cong C_{p^t} \times C_p$ with $t \ge 2$, then $\exp (G_m) > \exp (G_{m + 1})$.

Lemma 4

[11, Chapter VIII, Lemma 1.1]

• (1) Let G be a p-group, let x and y be elements of G, and let $K = \langle x, y\rangle$. Then, for all $n \ge 1$,

  $$\begin{aligned} (xy)^{p^{n}} \equiv x^{p^n}y^{p^n} (\mathrm{mod}~ K(n)), \end{aligned}$$

  where $K(n) = \mho _{n}(K') \mho _{n-1}(K_p) \mho _{n-2}(K_{p^2}) \cdots K_{p^n}$.

• (2) Let G be a p-group, let x and y be elements of G, and let $L = \langle x, [x, y]\rangle$. Then, for all $n \ge 1$,

  $$\begin{aligned}{}[x^{p^n}, y] \equiv [x, y]^{p^n} (\mathrm{mod}~ L(n)), \end{aligned}$$

  where $L(n) = \mho _{n}(L') \mho _{n-1}(L_p) \mho _{n-2}(L_{p^2}) \cdots L_{p^n}$.

Lemma 5

[3, Lemma 2.3] Let s and t be positive integers with $s \le t$, and let G be a $p^s$-abelian p-group. Then, $\mho _s(G) \le Z(G)$ and G is also a $p^t$-abelian p-group.

Lemma 6

[9, Lemma 3.1] Let G be a capable group and let $\mathcal {G}$ be a generating system for G. Then,

$$\begin{aligned} \bigcap \{ \langle g \rangle \mid g \in \mathcal {G} \} = 1. \end{aligned}$$

Lemma 7

Let G be a p-group. If for any elements x and y in G, $\langle x, y\rangle _p \le \mho _1(\Phi (\langle x, y\rangle ))$, then G is $\Phi$-regular.

Proof

For any elements x and y in G, set $K = \langle x, y\rangle$. By Huppert [10, Satz.III, Theorem 9.4], we have

$$\begin{aligned} (xy)^p = x^py^pc_3^p \cdots c_{p - 1}^pd, \quad \mathrm{where} ~ c_i \in K' \le \Phi (K), \quad d \in K_p. \end{aligned}$$

Since $K_p \le \mho _1(\Phi (K))$, there exist $c_p, \ldots , c_n \in \Phi (K)$, such that $d = c_p^p \cdots c_n^p$. Hence,

$$\begin{aligned} (xy)^p = x^py^pc_3^p \cdots c_{p - 1}^pc_p^p \cdots c_n^p. \end{aligned}$$

Therefore, G is $\Phi$-regular. $\square$

Lemma 8

Let G be a two-generator p-group. If $G'$ is a non-cyclic subgroup and $G'Z(G) / Z(G)$ is cyclic, then $c(G / \mho _1(G')) \le 3$ and $C_p^2 \lesssim G' / \mho _1(G') \lesssim C_p^3$. In addition, if p is odd, then $\mho _{1}(G) / \mho _1(G') \le Z(G / \mho _1(G'))$.

Proof

Let $\widetilde{G} = G / \mho _1(G')$. Since $G'Z(G) / Z(G)$ is cyclic, we see $c(\widetilde{G}) \le 3$. Also $d(G) = 2$ and $G'$ is non-cyclic. It follows that $C_p^2 \lesssim \widetilde{G'} \lesssim C_p^3$ and therefore $\mho _{1}(\widetilde{G}) \le Z(\widetilde{G})$ if $p \ge 3$ by Bai and Guo [3, Lemma 2.15(a)]. $\square$

3 Properties of NDC-Groups

In this section we give some properties of NDC-groups.

Theorem 1

Let N be a normal subgroup of a NDC-group G. Then G / N is also a NDC-group.

Proof

Let $L / N = Z(G / N)$ and M / N be a normal subgroup of G / N. Since G is a NDC-group, we see $G'/{G' \cap M}$ is cyclic or $M/{M \cap Z(G)}$ is cyclic. Noticing that

$$\begin{aligned} (G'N / N) /(G'N / N \cap M / N) \cong G'N/ G'N\cap M \cong G'/ G' \cap M \end{aligned}$$

and

$$\begin{aligned} (M / N) / (M / N \cap Z(G / N)) \cong M / M \cap L \cong M / (M \cap Z(G)) \diagup (M \cap L) / (M \cap Z(G)), \end{aligned}$$

we see $(G'N / N) /(G'N / N \cap M / N)$ is cyclic or $(M / N) / (M / N \cap Z(G / N))$ is cyclic. Therefore G / N is a NDC-group. $\square$

Theorem 2

Let G be a p-group. Then G is not a NDC-group if and only if there exist normal subgroups M and N in G such that $C_p^2 \cong M / N \le G'N / N$ and NZ(G) / Z(G) is isomorphic to one of $C_p^2$ and dihedral groups.

Proof

By the definition of NDC-groups, the sufficiency holds and so we need only to prove the necessity. Since G is not a NDC-group, G has a normal subgroup L such that $d(G' / G' \cap L) \ge 2$ and $d(L / L \cap Z(G)) \ge 2$.

We first prove the existence of N. If LZ(G) / Z(G) has no abelian G / Z(G)-invariant subgroups of type (p, p), then by Lemma 2, LZ(G) / Z(G) is one of the groups $D_{2^n}$, $Q_{2^n}$ and $SD_{2^n}$ for a suitable integer n and therefore LZ(G) / Z(G) is a dihedral group by Shahriari [22, (4.2)Theorem and (4.3)Theorem]. If LZ(G) / Z(G) has an abelian G / Z(G)-invariant subgroup K / Z(G) of type (p, p), then $K \le LZ(G)$. Set $F = L \cap K$. Then $K = K \cap LZ(G) = FZ(G)$ and so $FZ(G) / Z(G) \cong C_p^2$. Thus L has a G-invariant subgroup N such that $NZ(G) / Z(G) \cong C_p^2$ or $D_{2^n}$.

Now we prove the existence of M. Noticing that $G' \cap N \le G' \cap L$, we see $G' / (G' \cap N)$ is not cyclic and so $G'N / N$ is not cyclic. Therefore there exists a normal subgroups M / N in G / N such that $C_p^2 \cong M / N \le G'N / N$ by Lemma 2. $\square$

Theorem 2 is very useful for deciding whether a p-group is a NDC-group.

Remark 1

A subgroup of a NDC-group G may be not a NDC-group. For example,

Let $G = \langle a, b \mid a^8 = b^{16} = c^8 = d^4 = 1, c^2 = a^4b^4d^2, [b, a] = c, [c, a] = d, [c, b] = [d, a] = [d, b] = 1 \rangle$. Then G is a NDC-group, but $\langle a^2, ba^{-1}, c \rangle$ is not a NDC-group.

Proposition 1

Let G be a p-group. Then each of the following conditions is sufficient to guarantee that G is a NDC-group:

• (1) G is metahamiltonian;

• (2) G / Z(G) is metacyclic, which forces $G'$ to be cyclic;

• (3) The order of G / Z(G) is at most $p^4$;

• (4) The order of G is at most $p^6$;

• (5) For any non-normal abelian subgroup A in G, $C_G(A) = N_G(A)$;

• (6) All non-normal abelian subgroups of G are cyclic.

Proof

Let N be a normal subgroup of G.

• (1) If $N \le Z(G)$, then $d(N/{N \cap Z(G)}) = 1$. Now assume $N \not \le Z(G)$ and so there exist $x \in N \backslash Z(G)$ and $g \in G$ such that $\langle x, g \rangle$ is non-abelian. Since G is metahamiltonian, $G' \le \langle x, g \rangle$ by Lemma 1. Noticing that

  $$\begin{aligned} G' / G' \cap \langle x \rangle ^G \cong G'\langle x \rangle ^G / \langle x \rangle ^G \le G' \langle x ,g \rangle / \langle x \rangle ^G = \langle x ,g \rangle / \langle x \rangle ^G, \end{aligned}$$

  we see $G' / G' \cap \langle x \rangle ^G$ is cyclic and so $G'/G' \cap N$ is cyclic. Therefore G is a NDC-group.

• (2) Since G is a p-group and G / Z(G) is metacyclic, we may assume $G = \langle a, b, Z(G)\rangle$, $G/Z(G) = \langle aZ(G), bZ(G) \rangle$ and $[aZ(G), bZ(G)] = a^{i}Z(G)$ with $p \mid i$. Thus there exists $z \in Z(G)$ such that $[a, b] = za^{i}$ and so $G' = \langle (za^{i})^{g_1} \mid g_1 \in G \rangle$ by Huppert [10, Satz.III, Lemma 1.11(a)]. Noticing that $(za^{i})^{g_2} \in \langle za^{i} \rangle$ for any $g_2 \in G$, we see $G' = \langle za^{i} \rangle$. Therefore G is a NDC-group.

• (3) Assume NZ(G) / Z(G) is non-cyclic. In the following, we will prove that $G' / G' \cap N$ is cyclic. Since $|G / Z(G)| \le p^4$, we see $2 \le d(G / Z(G)) \le 4$ and $|G / NZ(G)| \le p^2$. Hence $G' \le NZ(G)$, which indicates $G_3 = [G', G] \le [NZ(G), G] \le N$. Let $\overline{G} = G / G_3$. If $d(G / Z(G)) = 2$, then $\overline{G'}$ is cyclic. Also $G_3 \le N$. So $G' / G' \cap N$ is cyclic. If $d(G / Z(G)) = 3$, then, since $|G / NZ(G)| \le p^2$, we see that G has a generator u such that $u \in N \backslash Z(G)$. This forces $\overline{G'} / \overline{G'} \cap \langle \overline{u} \rangle ^{\overline{G}}$ to be cyclic, which implies that $\overline{G'} / \overline{G'} \cap \overline{N}$ is cyclic. It follows from $G_3 \le N$ that $G' / G' \cap N \cong \overline{G'} / \overline{G'} \cap \overline{N}$ is cyclic. If $d(G / Z(G)) = 4$, then $G / Z(G) \cong C_p^4$. Since $|G / NZ(G)| \le p^2$, we see that G has generators v and w such that $\langle v, w\rangle \le N$ and $\langle v, w\rangle Z(G) / Z(G) \cong C_p^2$. Noticing that $c(G) = 2$, we see $G' / \langle v, w\rangle ^G \cap G'$ is cyclic and so $G' / G' \cap N$ is cyclic. Therefore G is a NDC-group.

• (4) Follows from [3, Theorem 3.5], (5) follows from [1, Theorem 3.1], and (6) follows from (4) and [28, Theorems 3.3 and 3.6].

$\square$

Theorem 3

Let G be a NDC-group. Then either $G'Z(G)/Z(G)$ is cyclic, or $G'Z(G)/Z(G) \cong C_p^3$, or $G'Z(G)/Z(G) \cong C_{p^m} \times C_p$ for a positive integer m.

Proof

Let $\overline{G} = G / Z(G)$. If $\overline{G'}$ is cyclic, then the theorem is clear. Now assume $\overline{G'}$ is non-cyclic and then $Z(\overline{G'})$ is non-cyclic by Huppert [10, Satz.III, Theorem 7.8(b)]. Set $X / Z(G) = Z(\overline{G'})$. Since G is a NDC-group, $\overline{G'} / \overline{G'} \cap Z(\overline{G'}) \cong G' / G' \cap X$ is cyclic and so $\overline{G'}$ is abelian. Thus there exists an abelian $\overline{G}$-invariant subgroup $\overline{M}$ of type (p, p) in $\overline{G'}$, which implies $G' / G' \cap MZ(G)$ is cyclic. It follows that $\overline{G'} / \overline{M} \cong C_{p^r}$ for a positive integer r. Hence $\overline{G'} \lesssim C_{p^{r + 1}} \times C_p^2$. However, if $d(\overline{G'}) = 3$ and $\mho _1(\overline{G'}) \ne \overline{1}$, then $\overline{G'}$ has a $\overline{G}$-invariant subgroup $\overline{L}$ of type (p, p) such that $\overline{L} \cap \mho _1(\overline{G'}) \ne \overline{1}$ and so $G' / G' \cap LZ(G) \cong \overline{G'} / \overline{G'} \cap \overline{L}$ is not cyclic, which indicates that G is not a NDC-group, a contradiction. Therefore $\overline{G'} \cong C_p^3$ or $\overline{G'} \cong C_{p^{r + 1}} \times C_p$. $\square$

4 Capable p-Groups with Small Generating Systems

In this section we give some properties of capable p-groups with small generating systems, which are useful for describing the structure of NDC-groups.

Lemma 9

Let p be an odd prime and G be a two-generator capable p-group whose derived subgroup is cyclic. Write $G' \cong C_{p^t}$ and $\exp (G) = p^m$. Then $[\mho _{m - 1}(G), G] \le \Omega _1(G')$, G is $p^t$-abelian and for any elements a, b such that $G = \langle a, b\rangle$, we have $[a^{p^{t - 1}}, G] = \Omega _1(G') = [b^{p^{t - 1}}, G]$ and $o(a) = o(b) = \exp (G)$. In addition, if $t \ge 2$, $\mho _1(G') \le \langle a^p, b^p\rangle$ and $G' \not \le \langle a^p, b^p\rangle$, then $m > t$ and $\Omega _1(G) \cong C_p^3$.

Proof

Since $G'$ is cyclic, we see G is regular by Berkovich [5, §7, Exercise 14]. Let $G = \langle a, b\rangle$. Then by Berkovich [5, §7, Theorem 7.2(e)], $[a^{p^{t - 1}}, G] = \Omega _1(G') = [b^{p^{t - 1}}, G]$ and $[\mho _{m - 1}(G), G] \le \Omega _1(G')$. According to [10, Satz.III, Theorem 10.8(g)], G is $p^t$-abelian and hence by Lemma 6, we have $\langle a \rangle \cap \langle b \rangle = 1$ and $o(a) = o(b) = \exp (G)$.

Let $[a, b] = c$. Since G is regular, we see $\mho _1(G) = \langle a^p, b^p, c^p \rangle$. If $c^p \in \langle a^p, b^p \rangle$, then $\mho _1(G) = \langle a^p, b^p\rangle$. Noticing that $\langle a^p, b^p \rangle ' \le \langle c^{p^2} \rangle$, we see $\Phi (\mho _1(G)) = \langle a^{p^2}, b^{p^2}, c^{p^2}\rangle \le \mho _2(G)$ and so $\mho _1(G)$ is metacyclic by Huppert [10, Satz.III, Theorem 11.4], which implies $\mho _{m - 1}(G) = \langle a^{p^{m - 1}}, b^{p^{m - 1}}\rangle$. If $t \ge 2$, then, since $\langle a \rangle \cap \langle b \rangle = 1$, $o(a) = o(b) = \exp (G)$ and $[\mho _{m - 1}(G), G] \le \Omega _1(G')$, we see $\langle a^{p^{m - 1}}, b^{p^{m - 1}}\rangle \cong C_p^2$. Also, $[a^{p^{t - 1}}, b] \ne 1$ and $[b^{p^{t - 1}}, a] \ne 1$. So $a^{rp^{t - 1}}b^{sp^{t - 1}} \not \in Z(G)$ for any integers r, s such that $(r, p) = 1$ or $(s, p) = 1$, which indicates $m > t$. It follows from $[a^{p^{t - 1}}, G] = \Omega _1(G') = [b^{p^{t - 1}}, G]$ that $C_p^2 \cong \mho _{m - 1}(G) \le Z(G)$. Noticing that $|G/\mho _1(G)| = p^3$, we see $|\Omega _1(G)| = p^3$ by Berkovich [5, §7, Theorem 7.2(d)] and so $\Omega _1(G) \cong C_p^3$. $\square$

Lemma 10

Let p be an odd prime and G be a p-group whose derived subgroup is a non-cyclic group of order at most $p^3$, and let x and y be elements of G. Then $G'$ is abelian, G is $p^2$-abelian, $\mho _2(G) \le Z(G)$, and $[\mho _1(G), G] \le \mho _1(G')G_4 \le Z(G)$. Furthermore,

• (1) $[x^p, y] = [x, y]^p$ if $G_{p + 1} = 1$.

• (2) $(xy)^{p} \equiv x^{p}y^{p} (\mathrm{mod}~ \mho _1(G'))$ if $p \ge 5$.

Proof

Let $K = \langle x, y \rangle$ and $L = \langle x, [x, y]\rangle$. Then $L' \le G_3$. By Bai and Guo [3, Lemma 2.15], $G'$ is abelian and therefore $G' \cong C_{p^2} \times C_p$ or $C_p^t$ with $2 \le t \le 3$. It follows from Lemma 3 that $\exp (G_3) \le p$, which indicates that $K(2) = 1$, $L(2) = 1$, $L(1) = L_p \le G_4 \le Z(G)$. In addition, $K(1) \le \mho _1(G')$ if $p \ge 5$, and $L_p = 1 = L(1)$ if $G_{p + 1} = 1$. Noticing the arbitrariness of x, y, we have the following results by Lemma 4.

$(xy)^{p^2} = x^{p^2}y^{p^2}$ and so G is $p^2$-abelian, which indicates $\mho _2(G) \le Z(G)$ by Lemma 5;

$[x^p, y] \equiv [x, y]^p (\mathrm{mod}~ L_p)$ and so $[\mho _1(G), G] \le \mho _1(G')G_4 \le Z(G)$;

$(xy)^{p} \equiv x^{p}y^{p} (\mathrm{mod}~ \mho _1(G'))$ if $p \ge 5$;

$[x^p, y] = [x, y]^p$ if $G_{p + 1} = 1$. $\square$

The results of Lemma 10 will be used repeatedly.

Theorem 4

Let p be an odd prime and G be a two-generator capable p-group with $G' \cong C_p^2$. Then $G' / G' \cap N$ is cyclic for every non-cyclic normal subgroup N in G if and only if G is isomorphic to one of the following groups:

• (1) $\langle a, b \mid a^9 = b^3 = c^3 = 1, [a, b] = c, [c, a] = 1, [c, b] = a^{-3}\rangle$;

• (2) $\langle a, b \mid a^{p^{m + 1}} = b^{p^{m + 1}} = c^p = 1, [a, b] = c, [c, a] = b^{vp^m}, [c, b] = 1\rangle$, where $m \ge 1$. In addition, $v = -1$ if $p = 3$ and $m = 1$, v is equal to 1 or a fixed quadratic non-residue modulo p if $p \ge 5$ or $m \ge 2$;

• (3) $\langle a, b \mid a^p = b^p = c^p = d^p = 1, [a, b] = c, [c, a] = 1, [c, b] = d, [d, a] = [d, b] = 1\rangle$, where $p \ge 5$;

• (4) $\langle a, b \mid a^9 = b^3 = c^3 = d^3 = 1, [a, b] = c, [c, a] = d, [c, b] = 1, [d, a] = [d, b] = 1\rangle$.

Proof

For every group G of types from (1) to (4) listed in the theorem, we may find a suitable group H such that $H / Z(H) \cong G$, which indicates that G is a capable group. For example,

$(1')$:

If $H \cong \langle \alpha , \beta \mid \alpha ^9 = \beta ^3 = \gamma ^9 = 1,$$[\alpha , \beta ] = \gamma , [\gamma , \alpha ] = 1, [\gamma , \beta ] = \alpha ^{-3}, [\gamma ^3, \beta ] = 1\rangle$, then H / Z(H) is isomorphic to the type of (1);

$(2')$:

If $H \cong \langle \alpha , \beta \mid \alpha ^{p^{m + 1}} = \beta ^{p^{m + 1}} = \gamma ^{p^{m + 1}} = 1, [\alpha , \beta ] = \gamma ,$$[\gamma , \beta ] = 1, [\gamma , \alpha ] = \beta ^{vp^m}, [\gamma ^p, \alpha ]= 1 \rangle$, then H / Z(H) is isomorphic to the type of (2);

$(3')$:

If $H \cong \langle \alpha , \beta \mid \alpha ^p = \beta ^p = \gamma ^p = \delta ^p = \epsilon ^p = 1,$$[\alpha , \beta ] = \gamma , [\gamma , \beta ] = \delta , [\delta , \beta ] = \epsilon ,$$[\gamma , \alpha ] = [\delta , \alpha ] = [\delta , \gamma ] = [\epsilon , \alpha ] = [\epsilon , \beta ] = [\epsilon , \gamma ] = [\epsilon , \delta ] = 1\rangle$, then H / Z(H) is isomorphic to the type of (3);

$(4')$:

If $H \cong \langle \alpha , \beta \mid \alpha ^9 = \beta ^3 = \gamma ^3 = \delta ^3 = \epsilon ^3 = 1,$$[\alpha , \beta ] = \gamma , [\gamma , \alpha ] = \delta ,$$[\delta , \alpha ] = \epsilon , [\gamma , \beta ] = [\delta , \beta ] = [\epsilon , \alpha ] = [\epsilon , \beta ] = 1 \rangle$, then H / Z(H) is isomorphic to the type of (4).

It is also easy to see that G satisfies the remaining hypotheses of the theorem if G is one of types (1), (2), (3) and (4). So the sufficiency holds and we will prove the necessity in the following.

Since $G' \cong C_p^2$, we see that there exists a normal subgroup N of G such that $N\le G' \cap Z(G)$ and $|G' / N| = p$. It follows from $d(G) = 2$ that $d(G / N) = 2$ and so G / N is a minimal non-abelian p-group by Xu et al. [27, Lemma 2.2]. According to [13, Lemmas 8(2) and 9], $G / N \cong M_p(n, m, 1)$ and therefore all isomorphic types of G are contained in the following groups by Li et al. [13, Theorem 11]:

(a):

$\langle a, b \mid a^{p^{n + 1}} = b^{p^{m}} = c^p = 1, [a, b] = c,$$[c, a] = 1, [c, b] = a^{vp^n}\rangle$, $n \ge m \ge 1$. In addition, $m > 1 = v$ if $p = 2$, v is equal to 1 or a fixed quadratic non-residue modulo p if $p \ge 3$;

(b):

$\langle a, b\mid a^{p^{n + 1}} = b^{p^{m}} = c^p = 1,$$[a, b] = c, [c, a] = a^{p^n}, [c, b] = 1\rangle$, $n \ge m \ge 1$. In addition, $n \ge 2$ if $p = 2$;

(c):

$\langle a, b \mid a^8 = c^2 = 1, a^4 = b^2, [a, b] = c, [c, a] = a^4, [c, b] = 1\rangle$;

(d):

$\langle a, b \mid a^9 = c^3 = 1, a^3 = b^3, [a, b] = c, [c, a] = a^3, [c, b] = 1 \rangle$;

(e):

$\langle a, b \mid a^{p^n} = b^{p^{m + 1}} = c^p = 1,$$[a, b] = c, [c, a] = b^{vp^m}, [c, b] = 1\rangle$, $n > m \ge 1$. In addition, $v = 1$ if $p = 2$, v is equal to 1 or a fixed quadratic non-residue modulo p if $p \ge 3$;

(f):

$\langle a, b \mid a^{p^n} = b^{p^{m + 1}} = c^p = 1,$$[a, b] = c, [c, a] = 1, [c, b] = b^{p^m}\rangle$, $n > m \ge 1$. In addition, $m \ge 2$ if $p = 2$;

(g):

$\langle a, b \mid a^{p^n} = b^{p^m} = c^p = d^p = 1,$$[a, b] = c, [c, a] = 1, [c, b] = d, [d, a] = [d, b] = 1\rangle$, $n \ge m \ge 1$. In addition, $m \ge 2$ if $p = 2$, $n + m \ge 3$ if $p = 3$;

(h):

$\langle a, b \mid a^{p^n} = b^{p^m} = c^p = d^p = 1,$$[a, b] = c, [c, a] = d, [c, b] = 1, [d, a] = [d, b] = 1\rangle$, $n > m \ge 1$. In addition, $n + m \ge 4$ if $p = 2$.

Since $p \ge 3$, it is clear that G can not be the type of (c). If G is one of the types (a) except the case (1), (b), (d), (e) except the case (2), and (h) except the case (4), then, by Xu [26, Lemma 3] whenever $\exp (G) \le 3^2$ and Lemma 10 whenever $\exp (G) \ge 3^3$ or $p \ge 5$, we see $\langle a \rangle \cap \langle ab^{-1} \rangle \ne 1$, which contradicts to Lemma 6. Now assume that G is the type of (g). If $n > m = 1$, then by the same arguments as in the above, $\langle ab \rangle \cap \langle ab^{-1} \rangle \ne 1$, in contradiction to Lemma 6. If $m \ge 2$, then it follows from $G_{p + 1} = 1$ that $\langle a^p, b^p \rangle \le Z(G)$ by Lemma 10, and thus $d(\langle a^p, b^p \rangle ) = d(G' / G' \cap \langle a^p, b^p \rangle ) = 2$, which contradicts to the hypotheses. So the remain case is that G is the type of (3). Finally, we assume that G is the type of (f) and there is a group H such that $H / Z(H) \cong G$. Then H is generated by the set

$$\begin{aligned} \{\alpha , \beta , \gamma \} \cup Z(H), \end{aligned}$$

where $\alpha$, $\beta$ and $\gamma$ are preimages in H of the elements a, b and c from G.

Hence $[\alpha , \beta , \gamma ] = [\gamma , \alpha , \beta ] = 1$. By Hupper [10, Satz.III, Exercise 1], $[\beta , \gamma , \alpha ] = 1$ and so $\beta ^{p^m} \in Z(G)$, which contradicts to $b^{p^m} \ne 1$. This is the final contradiction and the theorem is now proved. $\square$

Theorem 5

Let p be an odd prime and G be a capable p-group with $G' \cong C_{p^m} \times C_p$ for $m \ge 2$. Then G is $p^m$-abelian. In addition, further suppose $d(G) = 2$ and $G' / G' \cap N$ is cyclic for every non-cyclic normal subgroup N in G. Then the following results hold.

• (1) If $m \ge 3$ or $G_{p + 1} = 1$, then $\Omega _1(G') \le \mho _{m}(G) \le Z(G)$, and for every generator x of G, $\langle x \rangle \cap G' \ne 1$ and $o(x) = \exp (G)$;

• (2) If $m = 2$ and $G_{p + 1} \ne 1$, then G is a 3-group of maximal class and of order $3^5$.

Proof

We first claim that $[u^{p^{m - 1}}, v] = [u, v]^{p^{m - 1}}$ for any elements u, $v \in G$ if $m \ge 3$. In fact, by the hypotheses, we may write $G' = \langle c \rangle \times \langle d \rangle$ with $o(c) = p^m$. Then $G_3 \le \langle c^p \rangle \times \langle d \rangle$ by Lemma 3 and $[d, g] \le \langle c^{p^{m - 1}}\rangle$ for any $g \in G$, which indicates that $G_4 \le \langle c^{p^2} \rangle$. It follows that $\mho _{m - i}(G_{2 + i}) = 1$ with $i \ge 0$. Let $L = \langle u, [u, v]\rangle$. Then $L_{2 + i} \le G_{3 + i}$, which implies for any $j \ge 0$,

$$\begin{aligned} \mho _{(m - 1) - j}(L_{2 + j}) \le \mho _{(m - 1) - j}(G_{3 + j}) = \mho _{m - (1 + j)}(G_{2 + (1 + j)}) = 1. \end{aligned}$$

If $j \ge 1$, then $p^j \ge 2 + j$ and so $\mho _{(m - 1) - j}(L_{p^j}) \le \mho _{(m - 1) - j}(L_{2 + j}) = 1$. Hence $L(m - 1) = 1$ and therefore $[u^{p^{m - 1}}, v] = [u, v]^{p^{m - 1}}$ by Lemma 4(2), the claim is proved.

Since $G' / \Omega _1(G')$ is cyclic, we see $G / \Omega _1(G')$ is regular by Berkovich [5, §7, Exercise 14], and so $(uv)^{p^{m - 1}} = u^{p^{m - 1}}v^{p^{m - 1}}(\mathrm{mod}~ \Omega _1(G'))$ for any elements u, $v \in G$ by Huppert [10, Satz.III, Theorem 10.8(g)]. It follows from the above claim that $(uv)^{p^m} = u^{p^m}v^{p^m}$ if $m \ge 3$. Thus G is $p^m$-abelian if $m \ge 3$. If $m = 2$, then G is $p^2$-abelian by Lemma 10.

Now we further assume $d(G) = 2$ and $G' / G' \cap N$ is cyclic for every non-cyclic normal subgroup N in G.

• (1) If $m \ge 3$, then $[u^{p^{m - 1}}, v] = [u, v]^{p^{m - 1}}$ for any elements u, $v \in G$ by the above claim. If $G_{p + 1} = 1$, then we also get $[u^{p^{m - 1}}, v] = [u, v]^{p^{m - 1}}$ for $m = 2$ by Lemma 10. Therefore

  $$\begin{aligned}{}[u^{p^{m - 1}}, v] = [u, v]^{p^{m - 1}} ~ \hbox {for any elements} ~ u, v \in G. \end{aligned}$$

  (4.1)

For any generator $\alpha$ of G, since $d(G) = 2$, there exists another generator $\beta$ such that $G = \langle \alpha , \beta \rangle$ and so $o([\alpha , \beta ]) = p^m$ by Huppert [10, Satz.III, Lemma 1.11(a)], which implies $\langle {\alpha }^{p^{m - 1}} \rangle ^{G} = \langle {\alpha }^{p^{m - 1}}, c^{p^{m - 1}}\rangle$ and $\langle {\beta }^{p^{m - 1}} \rangle ^{G} = \langle {\beta }^{p^{m - 1}}, c^{p^{m - 1}}\rangle$ by (4.1). It follows that

$$\begin{aligned} o(\alpha ) \ge p^m ~ \mathrm{and} ~ o(\beta ) \ge p^m. \end{aligned}$$

(4.2)

If $\langle \alpha \rangle \cap G' = 1$ or $\langle \beta \rangle \cap G' = 1$, then $d(\langle {\alpha }^{p^{m - 1}} \rangle ^{G})$ = $2 = d(G' / G' \cap \langle {\alpha }^{p^{m - 1}} \rangle ^G)$ or $d(\langle {\beta }^{p^{m - 1}} \rangle ^{G})$ = $2 = d(G' / G' \cap \langle {\beta }^{p^{m - 1}} \rangle ^G)$, which contradicts to the hypotheses, so

$$\begin{aligned} \langle \alpha \rangle \cap G' \ne 1 ~ \mathrm{and} ~ \langle \beta \rangle \cap G' \ne 1. \end{aligned}$$

(4.3)

If $o(\alpha ) = o(\beta ) = p^m$, then since $o([\alpha , \beta ]) = p^m$, we see $\langle \alpha \rangle \cap Z(G) = 1 = \langle \beta \rangle \cap Z(G)$ by (4.1), and so $\langle dc^{sp^{m - 1}} \rangle \le \langle \alpha \rangle \cap G'$ and $\langle dc^{tp^{m - 1}} \rangle \le \langle \beta \rangle \cap G'$ for suitable integers s, t by (4.3), which implies that $[d, \alpha ] = 1 = [d, \beta ]$. This indicates $d \in Z(G)$, in contradiction to $\langle \alpha \rangle \cap Z(G) = 1 = \langle \beta \rangle \cap Z(G)$. Therefore $o(\alpha ) > p^m$ or $o(\beta ) > p^m$ by (4.2). Since G is $p^m$-abelian, we see $\mho _m(G) = \langle {\alpha }^{p^m}, {\beta }^{p^m} \rangle$ and it follows from Lemma 6 that $\langle \alpha \rangle \cap \langle \beta \rangle = 1$ and $o(\alpha ) = o(\beta ) > p^m$. By (4.3), $\Omega _1(G') \le \mho _m(G)$ and by (4.1), $\mho _m(G) = \langle {\alpha }^{p^m} \rangle \times \langle {\beta }^{p^m} \rangle \le Z(G)$, which implies $o(\alpha ) = \exp (G)$. For every generator x of G, there exists an element y of $G'$ and integers k, l such that $x = {\alpha }^{k}{\beta }^{l}y$ with $(k, p) = 1$ or $(l, p) = 1$, and hence $x^{p^m} = {\alpha }^{kp^m}{\beta }^{lp^m}$, which indicates $o(x) = o(\alpha ) = o(\beta ) = \exp (G)$. Thus $\langle x^{p^{m - 1}}, c^{p^{m - 1}}\rangle \trianglelefteq G$ by (4.1). If $\langle x \rangle \cap G' = 1$, then $d(\langle x^{p^{m - 1}}, c^{p^{m - 1}}\rangle )$ = $2 = d(G' / G' \cap \langle x^{p^{m - 1}}, c^{p^{m - 1}}\rangle )$, which contradicts to the hypotheses, so $\langle x \rangle \cap G' \ne 1$.

• (2) Since $m = 2$ and $G_{p + 1} \ne 1$, it is easy to see that $p = 3$ and $c(G) = 4$, which implies $G' \cap Z(G) = \langle c^3 \rangle \cong C_3$. By Lemma 10, we also know $[\mho _1(G), G] \le G_4 = \mho _1(G')$. If there exists a generator a of G such that $o(a) \ge 3^2$ and $\langle a \rangle \cap G' = 1$, then $\langle a^3, c^3\rangle \trianglelefteq G$ and $d(\langle a^3, c^3\rangle ) = 2 = d(G' / G' \cap \langle a^3, c^3\rangle )$, which contradicts to the hypotheses. So, for every generator w of G, $o(w) = 3$ or $\langle w \rangle \cap G' \ne 1$. By Lemma 10 once more, G is $3^2$-abelian and $\mho _2(G) \le Z(G)$, it follows from Lemma 6 that $o(w) \le 3^2$ and so $\mho _1(G) \le G'$. This indicates $|G| = 3^5$ and therefore G is of maximal class.

$\square$

Theorem 6

Let p be an odd prime and G be a two-generator capable p-group with $G' \cong C_p^3$. Then $G' / G' \cap N$ is cyclic for every non-cyclic normal subgroup N in G if and only if G either has order $p^5$ or is isomorphic to a member of the following two group families:

$\langle a, b \mid a^{p^{n + 1}} = b^{p^{n + 1}} = c^p = 1, [a, b] = c, [c, a] = b^{\mu p^n}, [c, b] = a^{p^n}, [a^{p^n}, b] = [b^{p^n}, a] = 1\rangle$, where $n \ge 2$ and $\mu$ is equal to 1 or a fixed quadratic non-residue modulo p.

Proof

Let $T = \langle a, b \mid a^{p^{n + 1}} = b^{p^{n + 1}} = c^p = 1, [a, b] = c, [c, a] = b^{\mu p^n}, [c, b] = a^{p^n}, [a^{p^n}, b] = [b^{p^n}, a] = 1\rangle$ with $n \ge 2$, which is listed in theorem. Then there exists a group

$$\begin{aligned} H\cong & {} \langle \alpha , \beta \mid \alpha ^{p^{n + 1}} = \beta ^{p^{n + 1}} = \gamma ^{p^{n + 1}} = 1, [\alpha , \beta ] = \gamma ,\\ {[}\gamma , \alpha ]= & {} \beta ^{\mu p^n}, {[}\gamma , \beta ] = \alpha ^{p^n}, [\alpha ^{p^n}, \beta ] = [\alpha , \beta ^{p^n}] = \gamma ^{p^n}, [\gamma ^p, \alpha ] = [\gamma ^p, \beta ] = 1 \rangle \end{aligned}$$

such that $T \cong H / Z(H)$, which implies that T is a capable p-group.

If $G \cong T$, then $\Omega _1(G) = G'$ and so, for any non-cyclic normal subgroup N of G, $|N \cap G'| \ge p^2$ by Lemma 2. This forces $G' / G' \cap N$ to be cyclic. Now assume $|G| = p^5$. It follows from $|G'| = p^3$ that $\Phi (G) = G'$. In this case, for any non-cyclic normal subgroup N, if $N \le G'$, then $|G' / N| \le p$. If $N \not \le G'$, then there exists a generator u of G in N and so it follows from $d(G) = 2$ that $G' \le \langle u \rangle ^G \le N$. Hence the sufficiency holds and so we need only to prove the necessity. Since $|G'| = p^3$, we see $|G| \ge p^5$. Now assume $|G| \ge p^6$ and G satisfies the hypotheses of the theorem. In the following, we will prove $G \cong T$.

We first claim that $\langle w \rangle \cap G' \ne 1 \hbox {for any} ~ w \in \Phi (G) \backslash G'$. Assume the opposite. Then, since $d(G) = 2$ and $|G| \ge p^6$, we see $G' < \Phi (G)$ and so there exists $v \in \Phi (G) \backslash G'$ such that $\langle v \rangle \cap G' = 1$. Since $\Phi (G) = G'\mho _1(G)$, there exist $1 \ne x \in \mho _1(G)$ and $y \in G'$ such that $v = xy$. By Lemma 10, $[\mho _1(G), G] \le Z(G)$ and so $[\mho _1(G), G'] = 1$ by Huppert [10, Satz.III, Lemma 1.10(b)], which indicates $\langle x \rangle \cap G' = 1$. By Lemma 10 once more, $[x, G] \le G_4$. Also $G_4 \lesssim G_p$. So there exists $C_p \cong M \le Z(G) \cap G'$ such that $\langle x \rangle M \trianglelefteq G$. It follows from $\langle x \rangle \cap G' = 1$ that $d(\langle x \rangle M) = 2$ and $M = G' \cap \langle x \rangle M$ which implies $G' / {G' \cap \langle x \rangle M} \cong C_p^2$, which contradicts to the hypotheses. Thus the claim is true.

Now we prove that $\Omega _1(G) = G'$ and $G' \cap Z(G) \cong C_p^2$. In fact, it follows from $|G| \ge p^6$ that $G / G' > rsim C_{p^2} \times C_p$. Let $G = \langle s, t \rangle$ with $|\langle s \rangle G' / G'| \ge p^2$. Then $s^p \in \Phi (G) \backslash G'$ and so $\langle s^{p^2} \rangle \cap G' \ne 1$ by the above claim, which indicates $o(s) \ge p^3$. By Lemma 10, G is $p^2$-abelian. Also $\langle s \rangle \cap \langle t \rangle = 1$ by Lemma 6. So $o(s) = o(t) \ge p^3$. The arbitrariness of s and t implies that for every element $r \in G \backslash \Phi (G)$, $o(r) = o(t) \ge p^3$ and thus $\Omega _1(G) \le \Phi (G)$. Noticing that $G' \le \Omega _1(G)$, we see $G' = \Omega _1(G)$ by the above claim once more. By Lemma 10, we know $\mho _2(G) \le Z(G)$ and so $G' \cap Z(G) = \Omega _1(\langle s^{p^2}\rangle \times \langle t^{p^2}\rangle ) \cong C_p^2$.

Finally, since $G' \cap Z(G) = C_p^2$, all isomorphic types of G are contained in the following groups by Qu and Hu[17, Lemma 2.5]:

(a):

$\langle a, b \mid a^{p^{n}} = b^{p^{m}} = c^p = d^p = e^p =1, [a, b] = c, [c, a] = d,$$[c, b] = e, [d, a] = [d, b] = [e, a] = [e, b] = 1\rangle$, where $n \ge m \ge 1$. In addition, $n \ge m \ge 2$ if $p = 2$;

(b):

$\langle a, b \mid a^{p^{n + 1}} = b^{p^{m}} = c^p = d^p = 1, [a, b] = c, [c, a] = a^{p^{n}},$$[c, b] = d, [d, a] = [d, b] = 1\rangle$, where $n \ge m \ge 1$. In addition, $n \ge m \ge 2$ if $p = 2$;

(c):

$\langle a, b \mid a^{p^{n + 1}} = b^{p^m} = c^p = d^p = 1, [a, b] = c, [c, a] = d,$$[c, b] = a^{\nu p^n}, [d, a] = [d, b] = 1\rangle$, where $n \ge m \ge 1$. In addition, $n \ge m \ge 2$ and $\nu = 1$ if $p = 2$, $\nu = -1$ if $p = 3$ and $n = m = 1$, $\nu$ is equal to 1 or a fixed quadratic non-residue modulo p if $p \ge 3$ except the case $p = 3$ and $n = m = 1$;

(d):

$\langle a, b \mid a^{p^{n}} = b^{p^{m + 1}} = c^p = d^p = 1, [a, b] = c, [c, a] = b^{\nu p^m},$$[c, b] = d, [d, a] = [d, b] = 1\rangle$, where $n > m \ge 1$. In addition, $n > m \ge 2$ and $\nu = 1$ if $p = 2$, $\nu$ is equal to 1 or a fixed quadratic non-residue modulo p if $p \ge 3$;

(e):

$\langle a, b \mid a^{p^n} = b^{p^{m + 1}} = c^p = d^p = 1, [a, b] = c, [c, a] = d,$$[c, b] = b^{p^m}, [d, a] = [d, b] = 1\rangle$, where $n > m \ge 1$. In addition, $n > m \ge 2$ if $p = 2$;

(f):

$\langle a, b \mid a^{9} = b^{9} = c^3 = d^3 = 1, a^3 = b^3, [a, b] = c, [c, a] = d,$$[c, b] = b^3, [d, a] = [d, b] = 1\rangle$;

(g):

$\langle a, b \mid a^{9} = b^{9} = c^3 = d^3 = 1, a^3 = b^3, [a, b] = c, [c, a] = d,$$[c, b] = b^6, [d, a] = [d, b] = 1\rangle$;

(h):

$\langle a, b \mid a^{p^{n + 1}} = b^{p^{m + 1}} = c^p = 1,$$[a, b] = c, [c, a] = b^{\mu p^m}, [c, b] = a^{\nu p^n}\rangle$, where $n > m \ge 1$. In addition, $n > m \ge 2$ and $\mu = \nu = 1$ if $p = 2$, $\mu$ and $\nu$ are equal to 1 or fixed quadratic non-residue modulo p if $p \ge 3$;

(i):

$\langle a, b \mid a^{p^{n + 1}} = b^{p^{n + 1}} = c^p = 1, [a, b] = c,$$[c, a] = b^{\mu p^n}, [c, b] = a^{p^n}\rangle$, where $n \ge 1$. In addition, $n \ge 2$ and $\mu = 1$ if $p = 2$, $\mu$ is equal to 1 or a fixed quadratic non-residue modulo p if $p \ge 3$;

(j):

$\langle a, b \mid a^{p^{n + 1}} = b^{p^{m + 1}} = c^p = 1, [a, b] = c,$$[c, a] = a^{r p^n}, [c, b] = b^{p^m}\rangle$, where $n \ge m \ge 1$. In addition, $n \ge m \ge 2$ if $p = 2$, $n + m \ge 3$ if $p = 3$, $1 \le r \le p - 1$ if $p \ge 3$ and $n > m$, $1 \le r \le p - 2$ if $p \ge 3$ and $n = m$;

(k):

$\langle a, b \mid a^{3^{n + 1}} = b^{3^{n + 1}} = c^3 = 1, [a, b] = c,$$[c, a] = a^{3^n}b^{3^n}, [c, b] = a^{3^n}\rangle$, where $n \ge 2$;

(l):

$\langle a, b \mid a^{3^{n + 1}} = b^{3^{n + 1}} = c^3 = 1, [a, b] = c,$$[c, a] = a^{2 \cdot 3^n}b^{2 \cdot 3^n}, [c, b] = a^{3^n}\rangle$, where $n \ge 2$;

(m):

$\langle a, b \mid a^{3^{n + 1}} = b^{3^{n + 1}} = c^3 = 1, [a, b] = c,$$[c, a] = a^{3^n}b^{3^n}, [c, b] = a^{2 \cdot 3^n}\rangle$, where $n \ge 2$;

(n):

$\langle a, b \mid a^{5^{n + 1}} = b^{5^{n + 1}} = c^5 = 1, [a, b] = c, [c, a] = a^{5^n}b^{5^n},$$[c, b] = a^{r \cdot 5^n}\rangle$, where $n \ge 1$, $1 \le r \le 4$ and $r \ne 2$;

(o):

$\langle a, b \mid a^{5^{n + 1}} = b^{5^{n + 1}} = c^5 = 1, [a, b] = c,$$[c, a] = a^{2 \cdot 5^n}b^{2 \cdot 5^n}, [c, b] = a^{2 \cdot 5^n}\rangle$, where $n \ge 1$;

(p):

$\langle a, b \mid a^{p^{n + 1}} = b^{p^{n + 1}} = c^p = 1, [a, b] = c,$$[c, a] = a^{p^n}b^{p^n}, [c, b] = a^{r \cdot p^n}\rangle$, where $n \ge 1$, $p > 5$ or $p = 2$. In addition, $r = 1$ if $p = 2$, $1 \le r \le p - 1$ and $r \ne 2$ if $p > 5$;

(q):

$\langle a, b \mid a^{p^{n + 1}} = b^{p^{n + 1}} = c^p = 1, [a, b] = c,$$[c, a] = a^{p^n}b^{-p^n}, [c, b] = b^{p^n}\rangle$, where $p > 5$ and $n \ge 1$.

Since $\Omega _1(G) = G'$, it is clear that G can not be one of the types (a) to (e). Since $|G| \ge p^6$, we also know that G can not be one of the types (f) and (g). If G is the type of (h), then $n > m$ and so $\langle a \rangle \cap \langle ab \rangle \ge \langle a^{p^n} \rangle \ne 1$ by Lemma 10, which contradicts to Lemma 6. It follows from type (i) that $G \cong T$, which is listed in theorem. In the following, we assume G is one of the types (j) to (q) and we work to obtain a contradiction. In fact, since G is a capable group, there exists a group H such that $H / Z(H) \cong G$, which indicates that H is generated by the set

$$\begin{aligned} \{\alpha , \beta , \gamma \} \cup Z(H), \end{aligned}$$

where $\alpha$, $\beta$ and $\gamma$ are preimages in H of the elements a, b and c from G.

If G is the type of (j) with $n > m$, then $\langle a \rangle \cap \langle ab \rangle \ge \langle a^{p^n} \rangle \ne 1$ by Lemma 10, which contradicts to Lemma 6. If G is the type of (j) with $n = m$, then it follows from $|G| \ge p^6$ that $n \ge 2$ and so $(ab)^{rp^n} = a^{rp^n}b^{rp^n}$ by Lemma 10, which implies $[\gamma , \alpha ][\gamma , \beta ]^r \in \langle \alpha \beta \rangle Z(H)$. Also $[\gamma , \alpha ] \in \langle \alpha \rangle Z(H)$ and $[\gamma , \beta ] \in \langle \beta \rangle Z(H)$. Hence $[\gamma , \alpha , \alpha ] = [\gamma , \beta , \beta ] = [[\gamma , \alpha ][\gamma , \beta ]^r, \alpha \beta ] = 1$ and so $[\gamma , \alpha , \beta ]^{r + 1} = [\gamma , \alpha , \beta ][[\gamma , \beta ]^r, \alpha ] = 1$. Noticing that $(r + 1, p) = 1$, we see $[\gamma , \alpha , \beta ] = 1$. By Huppert [10, Satz.III, Lemma 1.11(a)], we see $H_4 = 1$, which contradicts to $c(G) = c(H / Z(H)) = 3$.

If G is one of the types (k) to (p), then $[\alpha , \beta , \gamma ] = [\beta , \gamma , \alpha ] = 1$. By Huppert [10, Satz.III, Exercise 1], $[\gamma , \alpha , \beta ] = 1$ and so $\alpha ^{p^n} \in Z(G)$, which contradicts to $a^{p^n} \ne 1$.

If G is the type of (q), then, by Lemma 10, $(ab^{-1})^{p^n} = a^{p^n}b^{-p^n}$ and thus $[\gamma , \alpha ] \in \langle \alpha \beta ^{-1} \rangle Z(H)$. It follows that $[\gamma , \alpha , \alpha \beta ^{-1}] = 1$ and so $[\gamma , \alpha , \alpha ][\gamma , \alpha , \beta ]^{-1} = 1$, which implies $[\gamma , \alpha , \alpha ] = [\gamma , \alpha , \beta ]$. Also, it follows from $[\gamma , \alpha ][\gamma , \beta ] \in \langle \alpha \rangle Z(H)$ that $[[\gamma , \alpha ][\gamma , \beta ], \alpha ] = 1$ and thus $[\gamma , \alpha , \alpha ][\gamma , \beta , \alpha ] = 1$. So $[\gamma , \alpha , \alpha ] = [\gamma , \alpha , \beta ] = 1$. Noticing that $[\gamma , \beta , \beta ] = 1$, we see $H_4 = 1$ by Huppert [10, Satz.III, Lemma 1.11(a)], which contradicts to $c(G) = c(H / Z(H)) = 3$. This is the final contradiction and the theorem is now proved. $\square$

Theorem 7

Let p be an odd prime and G be a NDC-group with $d(G / Z(G)) = 2$ and $|G / Z(G)| \ge p^5$. Then the following results hold.

• (1) If $G'Z(G) / Z(G)$ is cyclic, then $G' Z(G) / G'$ is arbitrary;

• (2) If $G'Z(G) / Z(G) \cong C_p^2$, then $G' Z(G) / G'$ is cyclic. Further, $Z(G) \le \Phi (G)$;

• (3) If $G'Z(G) / Z(G) \cong C_p^3$ or $G'Z(G) / Z(G) \cong C_{p^m} \times C_p$ for a suitable positive integer $m \ge 2$, then $Z(G) \le G'$. In addition, if $G_4 \not \le Z(G)$, then $d(G' / G_4) = 2$.

Proof

We only need to prove (2) and (3). Let $\overline{G} = G / Z(G)$. Then $d(\overline{G}) = 2$. For convenience, we may write $G = \langle a, b, Z(G)\rangle$ and $[a, b] = c$. By Huppert [10, Satz.III, Lemma 1.11(a)], $o(\overline{c}) = \exp (\overline{G'})$ and so there exists an element d in $G'$ such that $C_p^2 \lesssim \langle \overline{c}\rangle \times \langle \overline{d} \rangle \le \overline{G'}$ by Berkovich [5, Introduction, Lemma 4(b)]. Since $\overline{G'} / \overline{G_3}$ is cyclic, without loss of generality, we may assume $d \in G_3 \backslash G_4$. We are ready to prove (2) and (3) in the following.

• (2) Since $|\overline{G}| \ge p^5$, there exists a generator a of G such that $\overline{\langle a^p, d \rangle ^G} = \langle \overline{a}^p \rangle \times \langle \overline{d}\rangle$ and $C_p^2 \lesssim \langle \overline{a}^p \rangle \times \langle \overline{d}\rangle \le Z(\overline{G})$ by Theorem 4. If $G' Z(G) / G'$ is non-cyclic or further, $Z(G) \not \le \Phi (G)$, then there exists an element $z_0 \in Z(G) \backslash G'$ such that $z_0 \not \in \langle a^p, d \rangle ^G \langle c^p \rangle$, which indicates that for any integers k and l with $(k, p) = 1$ or $(l, p) = 1$, $c^kd^l \not \in \langle a^p, dz_0 \rangle ^G \langle c^p \rangle$. Hence $G' / G' \cap \langle a^p, dz_0 \rangle ^G$ is non-cyclic. Also $\overline{\langle a^p, dz_0 \rangle ^G} = \langle \overline{a}^p \rangle \times \langle \overline{d}\rangle$ is non-cyclic. So G is not a NDC-group, a contradiction.

• (3) If $d(\overline{G'}) = 3$, then there exists $e_1 \in G_{c(G) - 1}$ such that $\overline{G_3} = \langle \overline{d}, \overline{e_1} \rangle \cong C_p^2$. If $d(\overline{G'}) = 2$, then $\Omega _1(\overline{G'}) = \langle \overline{d} \rangle \times \langle \overline{c}^{p^{m -1}} \rangle \trianglelefteq \overline{G}$ and hence, for notational convenience, we may also assume $e_2 = c^{p^{m - 1}}$. Since G is NDC-group, we see $\Omega _1(\overline{G'}) \le Z(\overline{G})$ or $\Omega _1(\overline{G'}) = \overline{G_3}$ by Theorem 5. So

  $$\begin{aligned}{}[\langle d, e_i \rangle Z(G), G] = [\langle d, e_i \rangle , G] = G_4 \hbox { or } \le Z(G) \mathrm{~ with ~} i = 1 \hbox { or } 2. \end{aligned}$$

For any $z \in Z(G)$, $\langle dz, e_i \rangle [\langle d, e_i\rangle , G] \trianglelefteq G$ and $\overline{\langle dz, e_i \rangle [\langle d, e_i \rangle , G]} = \langle \overline{d}, \overline{e_i}\rangle \cong C_p^2$. Also, G is a NDC-group. So $G' / {G' \cap \langle dz, e_i \rangle [\langle d, e_i \rangle , G]}$ is cyclic and so $G'\langle dz \rangle = \langle c, dz, e_i \rangle [\langle d, e_i \rangle , G]$, which indicates $G'\langle z \rangle = \langle c, dz, e_i \rangle [\langle d, e_i \rangle , G]$. Let $\widehat{G} = G / [\langle d, e_i \rangle , G]$. Then $\widehat{\langle z \rangle } \le \Phi (\widehat{\langle c, dz, e_i\rangle })$ and so $G' = \langle c, dz, e_i\rangle [\langle d, e_i\rangle , G]$. Take $z = 1$, then $G' = \langle c, d, e_i\rangle [\langle d, e_i\rangle , G]$ and thus $z \in G'$. The arbitrariness of z implies $Z(G) \le G' = \langle c, d, e_i\rangle [\langle d, e_i\rangle , G]$. In addition, assume $G_4 \not \le Z(G)$. In this case, if $\overline{G'} \cong C_p^3$, then $c(G) = 5$, which implies $e_1 \in G_4$ and $G_4 = [\langle d, e_1\rangle , G]$. It follows that $d(G' / G_4) = 2$. Now assume $\overline{G'} \cong C_{p^m} \times C_p$ with $m \ge 2$. Then $\overline{G_4} \le \mho _1(\overline{G'})$ and so $\mho _{m - 1}(\overline{G'}) \le \overline{G_4}$. Hence $\Omega _1(\overline{G'}) = \langle \overline{d} \rangle \times \langle \overline{e_2} \rangle \le \overline{G_3}$, which implies $[\langle d, e_2\rangle , G] \le G_4$. It follows that $d(G' / G_4) = 2$. $\square$

5 NDC-Groups Generated by Two Elements for $p \ge 3$

For convenience, in this section we always assume that p is a odd prime and G is a two-generator p-group with $|G / Z(G)| \ge p^5$ and $d(G/Z(G)) = 2$ by Proposition 1(3). So we may write $\overline{G} = G / Z(G)$, $G = \langle a, b \rangle$ and $[a, b] = c$.

By Theorem 3, we see $\overline{G'} \cong C_p^3$ or $\overline{G'} \lesssim C_{p^m} \times C_p$ for a suitable positive integer m. By Theorem 7, we also see that if $\overline{G'} \cong C_p^3$ or $\overline{G'} \cong C_{p^n} \times C_p$ for a suitable positive integer $n \ge 2$, then $Z(G) \le G'$. It is clear that $\overline{G}$ is a capable group. So in this section we may give some necessary and sufficient conditions for a p-group to be a NDC-group by using the results in the above section.

Theorem 8

Suppose $\overline{G'} \cong C_p^3$.

• (A) If $|\overline{G}| = p^5$, then G is a NDC-group if and only if $Z(G) \le G_4$;

• (B) If $|\overline{G}| > p^5$, then G is a NDC-group if and only if $Z(G) \le G'$ and $\overline{G}$ is of one of the types listed in Theorem 6, which indicates $\Omega _1(\overline{G}) = \overline{G'}$.

Proof

For notational convenience, let $\widehat{G} = G / G_4$.

(A) $``\Rightarrow$” Noticing that $\widehat{G}'$ is abelian, we see $[[\widehat{\alpha }, \widehat{\beta }]^p, \widehat{\gamma }] = [\widehat{\alpha }, \widehat{\beta }, \widehat{\gamma }]^p$ for any $\alpha$, $\beta$, $\gamma \in G$. Since $\overline{G'} \cong C_p^3$, we get $\mho _1(G') \le Z(G)$ and so $[\widehat{\alpha }, \widehat{\beta }, \widehat{\gamma }]^p = \widehat{1}$, which indicates $\exp (\widehat{G_3}) = p$ by Huppert [10, Satz.III, Lemma 1.11(a)]. Set $\widehat{L} = \langle \widehat{\alpha }, [\widehat{\alpha }, \widehat{\beta }]\rangle$ and then $\widehat{L}(1) = \mho _1(\widehat{L}') \le \mho _1(\widehat{G}_3) = \widehat{1}$, which indicates $[\widehat{\alpha }^p, \widehat{\beta }] = [\widehat{\alpha }, \widehat{\beta }]^p$ by Lemma 4(2). By $|\overline{G}| = p^5$ and Lemma 10, respectively, we see $\mho _1(\overline{G}) \le \overline{G'}$ and $[\mho _1(\overline{G}), \overline{G}] \le \overline{G}_4$. It follows that $\mho _1(\overline{G}) \le \overline{G_3}$ and so $\widehat{\alpha }^p \le \widehat{G_3Z(G)}$, which indicates $[\widehat{\alpha }, \widehat{\beta }]^p = [\widehat{\alpha }^p, \widehat{\beta }] = \widehat{1}$. The arbitrariness of $\alpha$ and $\beta$ implies $\mho _1(\widehat{G'}) = \widehat{1}$, which implies $\widehat{G'} \cong C_p^2$ or $C_p^3$. Also $Z(G) \le G'$ by Theorem 7(3). If $G_4 \le Z(G)$, then, noticing that $\overline{G'} \cong C_p^3$, we see $Z(G) = G_4$. If $G_4 \not \le Z(G)$, then $\widehat{G'} \cong C_p^2$ by Theorem 7(3) once more, which indicates $Z(G) \le \mho _1(G')G_4 = G_4$. Therefore $Z(G) \le G_4$.

$``\Leftarrow$” Let $A \trianglelefteq G$ with $\overline{A} \cong C_p^2$. Since $|\overline{G}| = p^5$, we see $\Phi (\overline{G}) = \overline{G'}$, which indicates that $\overline{G}_3$ is the unique abelian normal subgroup of type (p, p) in $\overline{G}$. It follows that $\overline{A} = \overline{G}_3$ and so $AZ(G) = G_3Z(G)$, which implies $G_4 = [G_3Z(G), G] = [AZ(G), G] = [A, G]$. The normality of A implies $G_4 \le A$. It follows from $Z(G) \le G_4$ that $G_3 = A$ and so $G' / A = \overline{G'} / \overline{G_3} \cong C_p$. Therefore G is a NDC-group by Theorem 2.

(B) $``\Rightarrow$” $Z(G) \le G'$ follows from Theorem 7(3), and it follows from $|\overline{G}| \ge p^6$ that $\overline{G}$ is the type listed in Theorem 6, which indicates $\Omega _1(\overline{G}) = \overline{G'}$.

$``\Leftarrow$” Noticing that $c(\overline{G}) = 3$, we see $G_4 \le Z(G)$. Since $d(\overline{G}) = 2$, we get $d(\widehat{G}') = 3$ and it follows from $Z(G) \le G'$ that $\widehat{Z(G)} = \mho _1(\widehat{G'}) = \Phi (\widehat{G'})$.

Let $B \trianglelefteq G$ with $\overline{B} \cong C_p^2$. It follows from $\Omega _1(\overline{G}) = \overline{G'}$ that $\overline{B} = \overline{G_3}$ is the unique abelian normal subgroup of type (p, p) in $\overline{G}$ and so $BZ(G) = G_3Z(G)$, which implies $B \le G'$ and $G_4 = [G_3Z(G), G] = [BZ(G), G] = [B, G]$. The normality of B implies $G_4 \le B$ and so $G' / B \cong \widehat{G'} / \widehat{B}$. Noticing that $\widehat{B}\Phi (\widehat{G'})/ \Phi (\widehat{G'}) = \widehat{BZ(G)} / \widehat{Z(G)} \cong \overline{B}$, we see $\widehat{B}\Phi (\widehat{G'})$ is a maximal subgroup of $\widehat{G'}$ and so $(\widehat{G'} / \widehat{B}) / \Phi (\widehat{G'} / \widehat{B})$ is cyclic, which implies $\widehat{G'} / \widehat{B}$ is cyclic. This forces $G' / B$ to be cyclic and therefore G is a NDC-group by Theorem 2. $\square$

Theorem 9

Suppose $\overline{G'} \cong C_{p^m} \times C_p$ with $m \ge 2$.

• (A) If $m \ge 3$ or $G_{p + 2} = 1$, then G is a NDC-group if and only if there exists an element x in G such that $\langle x \rangle ^G = \Omega _1(G) \cong C_p^3$ and $Z(G) \le \mho _1(G') \le G' \cong C_{p^n} \times C_p$ with $n > m$.

• (B) If $m = 2$ and $G_{p + 2} \ne 1$, then G is a NDC-group if and only if $Z(G) \le G'$, $d(G' / G_4) = 2$ and $\overline{G}$ is a 3-group of maximal class and of order $3^5$.

Proof

Since $\overline{G'} \cong C_{p^m} \times C_p$ with $m \ge 2$, $o(\overline{c}) = \exp (\overline{G'}) = p^m$ by Huppert [10, Satz.III, Lemma 1.11(a)], and so there exists an element d in $G'$ such that $\overline{G'} = \langle \overline{c}\rangle \times \langle \overline{d}\rangle$ by Berkovich [5, Introduction, Lemma 4(b)].

(A) $``\Rightarrow ''$ We first claim that $\langle c^{p^{m - 1}}, d\rangle \trianglelefteq G$. In fact, by hypotheses, $\overline{G}$ is a group which satisfies the properties listed in Theorem 5 and so $\Omega _1(\overline{G'}) \le \langle \overline{a}^{p^m}, \overline{b}^{p^m}\rangle \le Z(\overline{G})$ by Lemma 6, which implies that there exist elements $z_1, z_2 \in Z(G)$ and integers $l_1$, $l_2$, $f_1$, $f_2$ such that

$$\begin{aligned} c^{p^{m - 1}} = z_1a^{l_1p^m}b^{f_1p^m} ~\mathrm{and }~ d = z_2a^{l_2p^m}b^{f_2p^m}. \end{aligned}$$

It follows from $\langle \overline{d} \rangle \le \Omega _1(\overline{G'})$ that $[b^{-1}, d^{-1}, a] = [d, a^{-1}, b^{-1}] = 1$ and so $[a, b, d]^{b^{-1}} = 1$ by Huppert [10, Satz.III, Theorem 1.4], which indicates that

$$\begin{aligned} G' ~ \hbox { is an abelian group}. \end{aligned}$$

(5.1)

Hence $G_3 = [c, G][d, G]$ and for any elements $h \in G'$ and $g \in G$, and any integer i, $[h^i, g] = [h, g]^i$. It follows that $\exp ([d, G]) \le p$ and $\exp ([c, G]) \le p^m$, which implies

$$\begin{aligned} \exp (G_3) \le p^m. \end{aligned}$$

(5.2)

By Lemma 3, $\overline{G_3} \le \langle \overline{c}^p \rangle \times \langle \overline{d} \rangle$ and so $G_4 \le [c^p, G][d, G] = \mho _1([c, G])[d, G]$, which implies $\exp (G_4) \le p^{m - 1}$. Let $L = \langle u, [u, v]\rangle$ for any $u, v \in G$. If $m = 2$, then $G_{p + 2} = 1$ by hypotheses and so $G_{p^2} = 1$. It follows that $L(2) \le \mho _{2}(G_3)\mho _{1}(G_4)G_{p^2} = 1$ and therefore $[u^{p^{2}}, v] = [u, v]^{p^{2}}$ by Lemma 4(2). If $m \ge 3$, then, noticing that $[\overline{d}, \overline{G}] \le \langle \overline{c}^{p^{m - 1}}\rangle$, we see $\overline{G_4} = [\overline{G_3}, \overline{G}] \le \langle \overline{c}^{p^2} \rangle$ and so $G_{r +3} \le [c^{p^r}, G] = \mho _{r}([c, G])$ with $r \ge 2$, which implies $\exp (G_{r +3}) \le p^{m - r}$. For any $j \ge 0$, we have $L_{2 + j} \le G_{3 + j}$ and so

$$\begin{aligned} \mho _{m - j}(L_{2 + j}) \le \mho _{m - j}(G_{3 + j}) = 1. \end{aligned}$$

If $j \ge 1$, then $p^j \ge 2 + j$ and so $\mho _{m - j}(L_{p^j}) \le \mho _{m - j}(L_{2 + j}) = 1$. Hence $L(m) = 1$ by (5.2) and thus $[u^{p^{m}}, v] = [u, v]^{p^{m}}$ by Lemma 4(2). It follows from (5.2) once more that $[u^{p^{m}}, v] = [u, v]^{p^{m}} \le \langle c^{p^m}\rangle \le Z(G)$ and so $\langle c^{p^{m - 1}}, d\rangle ^G = \langle z_1a^{l_1p^m}b^{f_1p^m}, z_2a^{l_2p^m}b^{f_2p^m}\rangle ^G \le \langle z_1a^{l_1p^m}b^{f_1p^m}, z_2a^{l_2p^m}b^{f_2p^m}, c^{p^m} \rangle \le \langle c^{p^{m - 1}}, d\rangle$ by (5.1), that is, $\langle c^{p^{m - 1}}, d\rangle \trianglelefteq G$, the claim is proved.

Now we prove that $Z(G) \le \mho _1(G') \le G' \cong C_{p^n} \times C_p$ with $n > m$. In fact, by the above claim, we have $\langle c^{p^{m - 1}}, d\rangle \trianglelefteq G$. Also, $\langle \overline{c}^{p^{m - 1}}, \overline{d} \rangle \cong C_p^2$. Hence $G' / G' \cap \langle c^{p^{m - 1}}, d\rangle$ is cyclic and so $G' = \langle c, d\rangle$. By Theorem 7(3), $Z(G) \le G'$ and so $Z(G) \le \mho _1(G')$ by (5.1). According to Theorem 5, we have

$$\begin{aligned} |\langle \overline{a} \rangle | = |\langle \overline{b} \rangle | = \exp (\overline{G}) > p^m \end{aligned}$$

(5.3)

and so $|\overline{G} / \overline{G'}| \ge p^3$ by Lemma 6, which implies G is not of maximal class. Also, G is non-metacyclic. So there exists a normal subgroup N of G such that $N \cong C_p^3$ by Lemma 2 and [7, Theorem 4.1]. It follows from $Z(G) \le G'$ that $\overline{N\mho _1(G')}$ is non-cyclic and so $G' / G' \cap N\mho _1(G')$ is cyclic. Noticing that $G' \cap N\mho _1(G') = \mho _1(G')(G' \cap N)$, we get $G' / G' \cap N$ is cyclic and so $G' \cong C_{p^n} \times C_p$ with $n > m$ by (5.1).

Finally, if G has a normal subgroup M such that $C_p^2 \cong M \not \le G'$, then $\overline{M\mho _1(G')}$ is non-cyclic and $G'/ G' \cap M\mho _1(G') = G'/ \mho _1(G')$ is non-cyclic, which implies that G is not a NDC-group, a contradiction. Therefore if we may prove that $|\Omega _1(G)| = p^3$, then $N = \Omega _1(G) \cong C_p^3$ and so there exists an element x in G such that $\langle x \rangle ^G = \Omega _1(G) \cong C_p^3$. In fact, $\Omega _1(\overline{G'}) \le \mho _m(\overline{G})$ by Theorem 5 and $\overline{G'} \not \le \mho _1(\overline{G})$ by Huppert [10, Satz.III, Theorem 11.4], too. Hence $\overline{G'} \mho _2(\overline{G})\mho _1(\overline{G'})/ \mho _2(\overline{G})\mho _1(\overline{G'}) \cong C_p$ and so $|\overline{G} / \mho _2(\overline{G})\mho _1(\overline{G'})| = p^5$ by (5.3) and Lemma 6, which indicates $\overline{G} / \mho _2(\overline{G})\mho _1(\overline{G'}) \cong M_p(2, 2, 1)$ by Xu et al. [27, Lemma 2.2]. It follows that $\Phi (\overline{G}) / \mho _2(\overline{G})\mho _1(\overline{G'}) \cong C_p^3$, and thus $\Phi (\overline{G}) / \mho _1(\Phi (\overline{G})) \cong C_p^3$. Also $Z(G) \le \mho _1(G')$. Hence $\Phi (\overline{G}) / \mho _1(\Phi (\overline{G})) \cong \Phi (G) / \mho _1(\Phi (G))$ and so $|\Phi (G) / \mho _1(\Phi (G))| = p^3$. Since $G' \cong C_{p^n} \times C_p$, we may assume $G' = \langle c \rangle \times \langle d \rangle$. Hence $o(c) = p^n$ by Huppert [10, Satz.III, Lemma 1.11(a)] and so $o(d) = p$. By Lemma 3, $G_3 \le \langle c^p \rangle \times \langle d \rangle$. Also, $[d, G] \le \langle c^{p^{n - 1}}\rangle$ clearly. we see $G_4 = [G_3, G] \le \langle c^{p^2} \rangle$. Thus for any $u, v \in G$, $[u^p, v] \equiv [u, v]^p (\mathrm{mod}~ \mho _1(G_3)G_4)$ by Lemma 4(2) and thus $[u^p, v] \le \langle c^p \rangle$. It follows from $\langle c^p\rangle ~\mathrm{char}~ G' \trianglelefteq G$ that $(\Phi (G))'$ is cyclic. By Berkovich [5, §7, Theorem 7.1(c)], $\Phi (G)$ is regular and so $|\Omega _1(\Phi (G))| = |\Phi (G) / \mho _1(\Phi (G))| = p^3$ by Berkovich [5, §7, Theorem 7.2(d)]. By Theorem 5 once more, $o(\overline{\alpha }) = \exp (\overline{G}) \ge p^m$ for every generator $\overline{\alpha }$ of $\overline{G}$, and so $\Omega _1(\overline{G}) \le \Phi (\overline{G})$; furthermore, it follows from $\overline{\Omega _1(G)} \le \Omega _1(\overline{G})$ that $\Omega _1(G)Z(G) \le \Phi (G)$, which indicates $\Omega _1(G) = \Omega _1(\Phi (G))$. Therefore $|\Omega _1(G)| = p^3$.

$``\Leftarrow ''$ Without loss of generality, we may assume $G' = \langle c \rangle \times \langle d \rangle \cong C_{p^n} \times C_p$. Then $o(c) = p^n$ by Huppert [10, Satz.III, Lemma 1.11(a)]. It follows from $Z(G) \le \mho _1(G')$ that $Z(G) = \langle c^{p^m} \rangle$. Let $A \trianglelefteq G$ with $\overline{A} \cong C_p^2$. If $\overline{A} \le \overline{G'}$, then $A = \langle c^{p^{m - 1}}, d \rangle$ and so $d(G' / A) = 1$, which indicates G is a NDC-group by Theorem 2. If $\overline{A} \not \le \overline{G'}$, then there exists $y \in A$ such that $\overline{y} \in \overline{A} \backslash \overline{G'}$. Noticing that $C_p \cong \langle \overline{c}^{p^{m - 1}}\rangle \le Z(\overline{G})$, we see $[c^{p^{m - 1}}, y] \le \langle c^{p^{n - 1}}\rangle$ and so there exists an integer s such that $C_p \cong \langle yc^{sp^{m - 1}} \rangle \not \le G'$. The hypotheses imply $\langle yc^{sp^{m - 1}} \rangle ^G = \Omega _1(G)$ and so $\langle y \rangle ^G = \langle y, c^{p^{n -1}}, d\rangle$, which implies $G'/ G' \cap \langle y\rangle ^G$ is cyclic. It follows from $G' \cap \langle y\rangle ^G \le G' \cap A$ that $G'/ G' \cap A$ is cyclic and so G is a NDC-group by Theorem 2.

(B) $``\Rightarrow ''$ By Theorem 5, $\overline{G}$ is a 3-group of maximal class and of order $3^5$, which implies $G_4 \not \le Z(G)$. Hence $Z(G) \le G'$ and $d(G' / G_4) = 2$ by Theorem 7(3).

$``\Leftarrow ''$ Let $B \trianglelefteq G$ with $\overline{B} \cong C_p^2$. Since $\overline{G}$ is of maximal class and order $3^5$, we see $\overline{G_4} = \mho _1(\overline{G'})$ and $\overline{B} = \overline{G_3}$, which implies $G_4Z(G) = \mho _1(G')Z(G)$ and $BZ(G) = G_3Z(G)$. Hence $G_4 = [G_3Z(G), G] = [BZ(G), G] = [B, G]$. The normality of B implies $G_4 \le B$. Let $\widehat{G} = G / G_4$. Then $\widehat{G'}$ is abelian. It follows from $Z(G) \le G'$ that $B \le G'$ and so $G' / B \cong \widehat{G'} / \widehat{B}$. If $G' / B$ is non-cyclic, then, since $d(\widehat{G'}) = 2$, we see $\widehat{B} \le \Phi (\widehat{G'}) = \mho _1(\widehat{G'})$, which indicates $B \le \mho _1(G')G_4$. It follows that $G_3Z(G) = BZ(G) \le \mho _1(G')G_4Z(G) = G_4Z(G)$, which is a contradiction. So $G' / B$ is cyclic and therefore G is a NDC-group by Theorem 2. $\square$

Theorem 10

Suppose $\overline{G'} \cong C_p^2$. Then G is a NDC-group if and only if one of the following holds:

• (A) $\mho _1(G')\mho _{m+1}(G) = Z(G)$ and $\overline{G}$ is isomorphic to (2) in Theorem 4;

• (B) $d(\Phi (G) / G') = 1$, $Z(G) = \mho _2(G)G_4$ and $\overline{G}$ is isomorphic to (4) in Theorem 4.

Proof

For convenience, if $\overline{G}$ is the type of (2) in Theorem 4, then we may assume

$$\begin{aligned} \overline{G}= & {} \langle \overline{a_1}, \overline{b_1} \mid \overline{a_1}^{p^{m + 1}} = \overline{b_1}^{p^{m + 1}} = \overline{c_1}^p = \overline{1}, [\overline{a_1}, \overline{b_1}] = \overline{c_1},\\ {[}\overline{c_1}, \overline{a_1}]= & {} \overline{b_1}^{vp^m}, [\overline{c_1}, \overline{b_1}] = \overline{1}\rangle ; \end{aligned}$$

and if $\overline{G}$ is the type of (4) in Theorem 4, then we may assume

$$\begin{aligned} \overline{G}= & {} \langle \overline{a_2}, \overline{b_2} \mid \overline{a_2}^9 = \overline{b_2}^3 = \overline{c_2}^3 = \overline{d_2}^3 = \overline{1}, {[}\overline{a_2}, \overline{b_2}] = \overline{c_2}, {[}\overline{c_2}, \overline{a_2}] = \overline{d_2},\\ {[}\overline{c_2}, \overline{b_2}]= & {} \overline{1}, [\overline{d_2}, \overline{a_2}] = [\overline{d_2}, \overline{b_2}] = 1\rangle . \end{aligned}$$

Since $\overline{G'} \cong C_p^2$, we have $c(G) \le 4$ and so $G'$ is abelian, which indicates $1 = [c^p, g] = [c, g]^p$ for any $g \in G$. So by Huppert [10, Satz.III, Lemma 1.11(a)],

$$\begin{aligned} \exp (G_3) = p. \end{aligned}$$

(5.4)

Noticing that $[b_i, G'] \le Z(G)$ with $i \in \{1, 2\}$, we have the following results for any $g \in G$ by Xu [26, Lemma 2] and (5.4).

• (i) $[b_1^{p^m}, g] = [b_1, g]^{p^m} [b_1, g, b_1, b_1]^{\left( {\begin{array}{c}p^m\\ 3\end{array}}\right) } = [b_1, g]^{p^m}$ and so $[b_1^{p^m}, G] = \langle c_1^{p^m} \rangle \cong C_p$;

• (ii) $[a_1^{p^m}, g] = [a_1, g]^{p^m} [a_1, g, a_1, a_1]^{\left( {\begin{array}{c}p^m\\ 3\end{array}}\right) } = [a_1, g]^{p^m}[b_1^{wp^m}, a_1]^{\left( {\begin{array}{c}p^m\\ 3\end{array}}\right) }$ for a suitable integer w and it follows from $a_1^{p^m} \not \in Z(G)$ that $[a_1^{p^m}, G] = \langle c_1^{p^m} \rangle \cong C_p$;

• (iii) $1 = [a_2, b_2^3] = [a_2, b_2]^3[a_2, b_2, b_2, b_2] = [a_2, b_2]^3 = c_2^3$ and so $\exp (G') = 3$ if $\overline{G}$ is the type of (4) in Theorem 4 by Huppert [10, Satz.III, Lemma 1.11(a)];

• (iv) $[a_2^3, g] = [a_2, g]^3[a_2, g, a_2, a_2] = [d_2^{u}, a_2]$ for a suitable integer u and so $[a_2^3, G] \le G_4$.

$``\Rightarrow$” By hypotheses, $\overline{G}$ is of one of the types (1) to (4) listed in Theorem 4. Since $|\overline{G}| \ge p^5$, we see $\overline{G}$ is either the type (2) or the type (4). If $\overline{G}$ is the type of (2), then $\mho _1(G')\mho _{m+1}(G) \le Z(G)$ by Lemma 10. Write $\widetilde{G} = G / \mho _1(G')\mho _{m+1}(G)$ and $U = \langle a_1^{p^m}, zb_1^{p^m} \rangle$ for any $z \in Z(G)$, then $C_p^2 \cong \widetilde{U} \le Z(\widetilde{G})$ by (i) and (ii), which implies $C_p^2 \cong \overline{U} \le Z(\overline{G})$. It follows that $G'U^G / U^G$ is cyclic and so $\widetilde{G'}\widetilde{U} / \widetilde{U}$ is cyclic. Noticing that $\widetilde{G'} \cap \widetilde{U} \lesssim C_p$, we see $\widetilde{G'} \lesssim C_p^2$. It follows from $\mho _1(G')\mho _{m+1}(G) \le Z(G)$ that $\widetilde{G'} \cong C_p^2$ and so $\widetilde{G'} \cap \widetilde{U} = \langle \widetilde{zb_1^{p^m}} \rangle = \widetilde{G_3} \cong C_p$. The arbitrariness of z indicates $\mho _1(G')\mho _{m+1}(G) = Z(G)$ and G is just (A) in the theorem. Now assume $\overline{G}$ is the type of (4). If $Z(G) \ne \mho _2(G)G_4$ or $d(\Phi (G) / G') \ge 2$, then there exists $z_1 \in Z(G) \backslash \langle a_2^9\rangle G_4$ such that $z_1^3 \in \langle a_2^9\rangle G_4$. Noticing that $\langle a_2^9\rangle G_4 \le Z(G)$, we write $\widehat{G} = G / \langle a_2^9 \rangle G_4$. Set $V = \langle a_2^3, d_2z_1\rangle$. It follows from (iii) that $\exp (G') = 3$ and so $C_3^2 \cong \widehat{V} \le Z(\widehat{G})$ by (iv), which implies $C_3^2 \cong \overline{V} \le Z(\overline{G})$. This forces $G'V^G / V^G$ to be cyclic and so $\widehat{G'}\widehat{V} / \widehat{V}$ is cyclic. Noticing that $\widehat{G'} \cap \widehat{V} \lesssim C_3$, we see $\widehat{G'} \lesssim C_3^2$. Since $\langle a_2^9\rangle G_4 \le Z(G)$, we get $\widehat{G'} \cong C_3^2$ and so $\widehat{G'} \cap \widehat{V} = \langle \widehat{d_2z_1} \rangle = \widehat{G_3} \cong C_3$. On the other hand, without loss of generality, we may assume $d_2 \in G_3$ and then $\langle \widehat{d_2} \rangle = \widehat{G_3}$, which indicates $\langle \widehat{d_2} \rangle = \langle \widehat{d_2z_1} \rangle$, in contradiction to $z_1 \not \in \langle a_2^9\rangle G_4$. Therefore G is just (B) in the theorem.

“$\Leftarrow$” Assume G satisfies (A) in Theorem 10. Let $W = \langle a_1^{p^m}, b_1^{p^m}, c_1\rangle$. Then by (i) and (ii), $\langle \overline{a_1}^{p^m}, \overline{b_1}^{p^m} \rangle \le Z(\overline{G})$ and so $\overline{W} \cong C_p^3$. If $m = 1$, then $\Omega _1(\overline{G} / \overline{G'}) = \langle \overline{a_1}^p, \overline{b_1} \rangle \overline{G'} / \overline{G'} \cong C_p^2$. Noticing that $\Omega _1(\overline{G})\overline{G'} / \overline{G'} \le \Omega _1(\overline{G} / \overline{G'})$, we see $\Omega _1(\overline{G}) \le \langle \overline{a_1}^p, \overline{b_1} \rangle \overline{G'} = \langle \overline{a_1}^p, \overline{b_1}, \overline{c_1}\rangle$. Also, $\langle \overline{a_1}^p, \overline{b_1}, \overline{c_1}\rangle$ is abelian. So $\Omega _1(\overline{G}) = \overline{W}$. If $m \ge 2$, then $\overline{G}_p \le \mho _2(\overline{G}) \le \mho _1(\Phi (\overline{G}))$ and for every $\overline{H} < \overline{G}$, $\overline{H'}$ is cyclic, which implies $\overline{H}_p \le \mho _1(\overline{H'}) \le \mho _1(\Phi (\overline{H}))$. Hence $\overline{G}$ is $\Phi$-regular by Lemma 7. Noticing that $|\overline{G} / \mho _1(\overline{G})| = p^3$, we see $|\Omega _1(\overline{G})| = p^3$ by Bannuscher [4, Theorem 2.4] and so $\Omega _1(\overline{G}) = \overline{W}$, too. Therefore $\langle \overline{a_1}^{p^m}, \overline{b_1}^{p^m}\rangle$ and $\langle \overline{c_1a_1^{up^m}}, \overline{b_1}^{p^m}\rangle$ with $0 \le u \le p - 1$ are all the abelian normal subgroups of type (p, p) in $\overline{G}$. Let $A \trianglelefteq G$ with $\overline{A} \cong C_p^2$. Then there exist $z_2, z_3, z_4, z_5\in Z(G)$ such that $\langle z_2a_1^{p^m}, z_3b_1^{p^m}\rangle ^G \le A$ or $\langle z_4c_1a_1^{up^m}, z_5b_1^{p^m}\rangle ^G \le A$. It follows from $c(G) \le 4$ that $G_{p^2} = 1$ and so $G / \mho _1(G')$ is $p^2$-abelian by Lemma 4(1). Also $G = \langle a_1, b_1\rangle$. We see $\mho _1(G')\mho _{m+1}(G) = \mho _1(G')\langle a_1^{p^{m + 1}}, b_1^{p^{m + 1}}\rangle = \langle c_1^p, a_1^{p^{m + 1}}, b_1^{p^{m + 1}}\rangle = Z(G)$ by (5.4). Hence $Z(G) \le \mho _1(W) \le \Phi (W)$ and it follows from $\overline{G'} \le \overline{W}$ that $G'Z(G) \le W$. So

$$\begin{aligned} G' \langle z_2a_1^{p^m}, z_3b_1^{p^m}\rangle ^G / \langle z_2a_1^{p^m}, z_3b_1^{p^m}\rangle ^G = \langle c_1, z_2a_1^{p^m}, z_3b_1^{p^m}\rangle / \langle z_2a_1^{p^m}, z_3b_1^{p^m}\rangle ^G \end{aligned}$$

and

$$\begin{aligned}&G' \langle z_4c_1a_1^{up^m}, z_5b_1^{p^m}\rangle ^G / \langle z_4c_1a_1^{up^m}, z_5b_1^{p^m}\rangle ^G\\&\quad =\langle c_1, z_4c_1a_1^{up^m}, z_5b_1^{p^m}\rangle ^{G_{p^{2u}}} / \langle z_4c_1a_1^{up^m}, z_5b_1^{p^m}\rangle ^G, \end{aligned}$$

that is

$$\begin{aligned} G' \langle z_2a_1^{p^m}, z_3b_1^{p^m}\rangle ^G / \langle z_2a_1^{p^m}, z_3b_1^{p^m}\rangle ^G ~ \hbox {is cyclic} \end{aligned}$$

and

$$\begin{aligned} G' \langle z_4c_1a_1^{up^m}, z_5b_1^{p^m}\rangle ^G / \langle z_4c_1a_1^{up^m}, z_5b_1^{p^m}\rangle ^G ~ \hbox { is cyclic}. \end{aligned}$$

This forces $G' / G' \cap A$ to be cyclic and therefore G is a NDC-group by Theorem 2.

Now assume G satisfies (B) in Theorem 10. Let $B \trianglelefteq G$ with $\overline{B} \cong C_3^2$. If $\overline{B} \not \le \Phi (\overline{G})$, then there exists $n \in B$ such that $\overline{n} \not \in \Phi (\overline{G})$ and so $n \not \in \Phi (G)$. Since $d(G) = 2$, the normality of B implies $G' \le \langle n\rangle ^G \le B$ and so $G' / (G' \cap B) = 1$. Now we assume $\overline{B} \le \Phi (\overline{G})$. By Lemma 10, $\Phi (\overline{G}) = \mho _1(\overline{G})\overline{G'} = \langle \overline{a_2}^3, \overline{c_2}, \overline{d_2}\rangle \cong C_3^3$, which implies that $\overline{B} = \langle \overline{a_2}^3, \overline{d_2}\rangle$ or $\langle \overline{c_2a_2^{3v}}, \overline{d_2}\rangle$ with $0 \le v \le p - 1$ by (iv). Hence there exist $z_6, z_7, z_8, z_{9}\in Z(G)$ such that $\langle z_6a_2^{3}, d_2z_7\rangle ^G \le B$ or $\langle z_8c_2a_2^{3v}, d_2z_{9}\rangle ^G \le B$. On the other hand, by (iii), $G'$ is elementary abelian and so $G' / G_4 \lesssim C_3^3$ according to $d(G) = 2$. By Lemma 10, $G / G_4$ is $3^2$-abelian and thus $Z(G) = \mho _2(G)G_4 = \langle a_2^9, b_2^9, G_4 \rangle$. Noticing that $\Phi (G) / G'$ is cyclic, we see $b_2^3 \in \langle a_2^3 \rangle G'$ and so $b_2^9 \in \langle a_2^9 \rangle$ by (iii) and (iv). It follows that $Z(G) = \langle a_2^9, G_4 \rangle$ and so $\Phi (G) = \langle a_2^3, c_2, d_2, G_4\rangle$, which indicates $Z(G) / G_4 \le \mho _1(\Phi (G) / G_4) \le \Phi (\Phi (G) / G_4)$. Hence

$$\begin{aligned} G' \langle z_6a_2^{3}, d_2z_7\rangle ^G / \langle z_6a_2^3, d_2z_7\rangle ^G = \langle c_2, z_6a_2^3, d_2z_7, G_4\rangle / \langle z_6a_2^3, d_2z_7, G_4\rangle ^G \end{aligned}$$

and

$$\begin{aligned}&G' \langle z_8c_2a_2^{3v}, d_2z_{9}\rangle ^G / \langle z_8c_2a_2^{3v}, d_2z_{9}\rangle ^G\\&\quad =\langle c_2, z_8c_2a_2^{3v}, d_2z_{9}, G_4\rangle ^{G_{p^{2v}}} / \langle z_8c_2a_2^{3v}, d_2z_{9}, G_4\rangle ^G, \end{aligned}$$

which implies

$$\begin{aligned} G' \langle z_6a_2^{3}, d_2z_7\rangle ^G / \langle z_6a_2^3, d_2z_7\rangle ^G ~ \hbox { is cyclic} \end{aligned}$$

and

$$\begin{aligned} G' \langle z_8c_2a_2^{3v}, d_2z_{9}\rangle ^G / \langle z_8c_2a_2^{3v}, d_2z_{9}\rangle ^G ~ \hbox { is cyclic}. \end{aligned}$$

This forces $G' / G' \cap B$ to be cyclic and therefore G is a NDC-group by Theorem 2. $\square$

Theorem 11

Suppose $\overline{G'} \cong C_{p^t}$ with $t \ge 1$.

• (A) If $G'$ is cyclic, then G is a NDC-group.

• (B) If $G'$ is non-cyclic, then set $\exp (\overline{G}) = p^m$ and $\widetilde{G} = G / \mho _1(G')$, and hence G is a NDC-group if and only if $\widetilde{G_3} \le \Omega _1(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle )$ and $\mho _1(\overline{G'}) \le \langle \overline{a}^p, \overline{b}^p\rangle$.

Proof

Clearly, if $G'$ is cyclic, we see $G' / G' \cap X$ is cyclic for any normal subgroup X in G and thus (A) is true. Now set $\widetilde{G} = G / \mho _1(G')$ and then we are ready to prove (B) in the following.

Since $G'$ is non-cyclic, we see $\overline{G}$ is non-metacyclic by Proposition 1(2). Since $\overline{G'}$ is cyclic, we see $\overline{G}$ is regular by Berkovich [5, §7, Exercise 14]. Noticing that $|\overline{G}| \ge p^5$, we also know $m \ge 2$.

“$\Rightarrow$” Let $U = \langle a^{p^{m - 1}}, b^{p^{m - 1}}\rangle ^G$. Then $\overline{U} > rsim C_p^2$ by Lemmas 6 and 9. Since G is a NDC-group, we see $G' / G' \cap U$ is cyclic and so $\widetilde{G'} / \widetilde{G'} \cap \widetilde{U}$ is cyclic. Hence $\widetilde{G_3} \le \Omega _1(\widetilde{U})$ and $\widetilde{U} = \langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle$ by Lemma 8, which implies $\widetilde{G_3} \le \Omega _1(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle )$.

Now we prove that $\mho _1(\overline{G'}) \le \langle \overline{a}^p, \overline{b}^p\rangle$. In fact, if $t = 1$, then $\mho _1(\overline{G'}) = \overline{1}$ and so there is nothing to be proved. Now assume that $t \ge 2$. By Lemma 6, $\langle \overline{a}\rangle \cap \langle \overline{b}\rangle = \overline{1}$. Without loss of generality, we may assume $\overline{\langle a^{p^{m - 1}}\rangle } \cap \Omega _1(\overline{G'}) = \overline{1}$. Then $\overline{\langle a^{p^{m - 1}}, c^{p^{t -1}}\rangle ^G} = \overline{\langle a^{p^{m - 1}}, c^{p^{t -1}} \rangle } \cong C_p^2$ by Lemma 9. If $\langle \widetilde{a^{p^{m - 1}}} \rangle \cap \widetilde{G'} < \widetilde{G_3}$, then $\widetilde{G'} / \widetilde{G'} \cap \langle \widetilde{a^{p^{m - 1}}}, \widetilde{c^{p^{t -1}}}\rangle ^{\widetilde{G}} = \widetilde{G'} / \widetilde{G'} \cap \widetilde{\langle a^{p^{m - 1}} \rangle } > rsim C_p^2$ by Lemma 8, which implies $G' / G' \cap \langle a^{p^{m - 1}}, c^{p^{t -1}}\rangle ^G$ is non-cyclic. This forces G to be a non-NDC-group, a contradiction. So $\widetilde{\langle a^{p^{m - 1}}\rangle } \cap \widetilde{G'} = \widetilde{G_3} \cong C_p$. Assume $o(\widetilde{b}) > o(\widetilde{a})$ and $\Omega _1(\widetilde{\langle b^{p^{m - 1}}\rangle }) = \widetilde{G_3} \cong C_p$. Then, since $\Omega _1(\widetilde{\langle a^{p^{m - 1}}\rangle }) = \widetilde{G_3}$, we see that there exists a suitable integer $\varepsilon$ such that $\widetilde{\langle b^{\varepsilon p}a^p \rangle } \cap \widetilde{G'} = \widetilde{1}$ by Lemma 8 and so $\widetilde{G'} / \widetilde{G'} \cap \widetilde{\langle b^{\varepsilon p}a^p \rangle }$ is non-cyclic, which indicates $G' / G' \cap \langle b^{\varepsilon p}a^p \rangle ^G$ is non-cyclic by Lemma 8. Noticing that $[\overline{b}^{\varepsilon p}\overline{a}^p, \overline{b}] = [\overline{a}^p, \overline{b}]$, we see $\langle [\overline{b}^{\varepsilon p}\overline{a}^p, \overline{b}] \rangle = \langle \overline{c}^p \rangle$ by Xu [26, Lemma 2] and so $\overline{\langle b^{\varepsilon p}a^p \rangle ^G} = \langle \overline{b^{\varepsilon p}a^p}, \overline{c^p}\rangle \ne \langle \overline{c^p} \rangle$. If $\langle \overline{c}^p \rangle \not \le \langle \overline{a}^p, \overline{b}^p\rangle$, then $\overline{\langle b^{\varepsilon p}a^p \rangle ^G}$ is non-cyclic, which implies that G is not a NDC-group, a contradiction. So $\langle \overline{c}^p \rangle \le \langle \overline{a}^p, \overline{b}^p\rangle$. Assume $o(\widetilde{b}) \le o(\widetilde{a})$ or $\Omega _1(\widetilde{\langle b^{p^{m - 1}}\rangle }) \cap \widetilde{G_3} = \widetilde{1}$. Then, since $\Omega _1(\widetilde{\langle a^{p^{m - 1}}\rangle }) = \widetilde{G_3} \cong C_p$, we see that there exists a suitable integer $\sigma$ such that $\widetilde{\langle a^{\sigma p}b^p \rangle } \cap \widetilde{G'} = \widetilde{1}$ by Lemma 8 and so $\widetilde{G'} / \widetilde{G'} \cap \widetilde{\langle a^{\sigma p}b^p\rangle }$ is non-cyclic, which indicates $G' / G' \cap \langle a^{\sigma p}b^p \rangle ^G$ is non-cyclic by Lemma 8. Noticing that $[\overline{a}^{\sigma p}\overline{b}^p, \overline{a}] = [\overline{b}^p, \overline{a}]$, we see $\langle [\overline{a}^{\sigma p}\overline{b}^p, \overline{a}] \rangle = \langle \overline{c}^p \rangle$ by Xu [26, Lemma 2] and so $\overline{\langle a^{\sigma p}b^p \rangle ^G} = \langle \overline{a^{\sigma p}b^p}, \overline{c^p}\rangle \ne \langle \overline{c^p} \rangle$. If $\langle \overline{c}^p \rangle \not \le \langle \overline{a}^p, \overline{b}^p\rangle$, then $\overline{\langle a^{\sigma p}b^p \rangle ^G}$ is non-cyclic, which implies that G is not a NDC-group, a contradiction. So $\langle \overline{c}^p \rangle \le \langle \overline{a}^p, \overline{b}^p\rangle$, too.

“$\Leftarrow$” For any $A \trianglelefteq G$ with $\overline{A} \cong C_p^2$, if we may prove that A has a G-invariant subgroup W such that $\widetilde{G'} / {\widetilde{G'} \cap \widetilde{W}}$ is cyclic, then $G' / {G' \cap W}$ is cyclic and so $G' / {G' \cap A}$ is cyclic, which forces G to be a NDC-group by Theorem 2. In the following, we find the suitable G-invariant subgroups W of A such that $\widetilde{G'} / {\widetilde{G'} \cap \widetilde{W}}$ are cyclic.

Assume $t = 1$. Then $\overline{G}$ is minimal non-abelian by Xu et al. [27, Lemma 2.2]. Since $\overline{G}$ is non-metacyclic, we see $\overline{G} \cong M_p(m, m, 1)$ by Lemma 6, which implies $\Omega _1(\overline{G}) = \langle \overline{a}^{p^{m - 1}}, \overline{b}^{p^{m - 1}}, \overline{c}\rangle$. If $\overline{G'} \le \overline{A}$, then there exists $z_1 \in Z(G)$ such that $cz_1 \in A$ and so take $W = \langle cz_1 \rangle ^G$. If $\overline{G'} \not \le \overline{A}$, then $\overline{A} = \langle \overline{c}^i\overline{a}^{p^{m - 1}}, \overline{c}^j\overline{b}^{p^{m - 1}}\rangle$ with $0 \le i, j \le p - 1$, which implies that there exist $z_2$, $z_3 \in Z(G)$ such that $\langle z_2c^ia^{p^{m - 1}}, z_3c^jb^{p^{m - 1}} \rangle \le A$. In this case, if $i \ne 0$ or $j \ne 0$, then we may assume $i \ne 0$ and hence take $W = \langle z_2c^ia^{p^{m - 1}}\rangle ^G$ by Lemma 8. If $i = j = 0$, then $\overline{A} = \langle \overline{a}^{p^{m - 1}}, \overline{b}^{p^{m - 1}}\rangle$. Noticing that $\widetilde{G_3} \le \Omega _1(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle )$, we see $\widetilde{Z(G)} \le \mho _1(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle )$ and so $\langle \widetilde{z_2a^{p^{m - 1}}}, \widetilde{z_3b^{p^{m - 1}}}\rangle = \langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle$, which indicates that we take $W = \langle z_2a^{p^{m - 1}}, z_3b^{p^{m - 1}} \rangle ^G$ by Lemma 8.

Assume $t \ge 2$. Since $\overline{G}$ is regular, we see $\mho _1(\overline{G}) = \langle \overline{a}^p, \overline{b}^p, \overline{c}^p \rangle$. It follows from $\overline{c}^p \in \langle \overline{a}^p, \overline{b}^p \rangle$ that $\mho _1(\overline{G}) = \langle \overline{a}^p, \overline{b}^p\rangle$. Let $\widehat{\mho _1(\overline{G})} = \mho _1(\overline{G}) / \langle \overline{c}^{p^2} \rangle$. Then $\widehat{\mho _1(\overline{G})}$ is abelian. Without loss of generality, we may assume $\widehat{\overline{1}} \ne \widehat{\overline{c}^p} \in \langle \widehat{{\overline{b}}^p} \rangle$ and then $\langle \overline{c}^p \rangle \le \langle \overline{b}^{p}\rangle$. By Lemma 9, $m > t$ and $\Omega _1(\overline{G}) \cong C_p^3$, which indicates that $\overline{c}^p \in \langle {\overline{b}}^{p^2} \rangle$ and there exists a positive integer k such that $o(\overline{c}\overline{b}^{kp}) = p$. Since $\overline{G}$ is non-metacyclic, we see $\overline{c} \not \in \mho _1(\overline{G})$ by Huppert [10, Satz.III, Theorem 11.4]. Also $\langle \overline{a}\rangle \cap \langle \overline{b}\rangle = \overline{1}$ by Lemma 6 and $o(\overline{a}) = o(\overline{b}) = p^m$ by Lemma 9 once more. So $\Omega _1(\overline{G}) = \langle \overline{a}^{p^{m - 1}}, \overline{c}\overline{b}^{kp}, \overline{b}^{p^{m - 1}}\rangle$. If $\overline{A} = \langle \overline{a}^{p^{m - 1}}, \overline{b}^{p^{m - 1}}\rangle$, then there are $z_4$, $z_5 \in Z(G)$ such that $\langle z_4a^{p^{m - 1}}, z_5b^{p^{m - 1}} \rangle \le A$. Noticing that $\widetilde{G_3} \le \Omega _1(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle )$, we see $\widetilde{Z(G)} \le \mho _1(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle )$ and so $\langle \widetilde{z_4a^{p^{m - 1}}}, \widetilde{z_5b^{p^{m - 1}}}\rangle = \langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle$, which indicates that we take $W = \langle z_4a^{p^{m - 1}}, z_5b^{p^{m - 1}} \rangle ^G$. If $\overline{A} \ne \langle \overline{a}^{p^{m - 1}}, \overline{b}^{p^{m - 1}}\rangle$, then there are integers w, u and v such that $\overline{a}^{wp^{m - 1}}(\overline{c}\overline{b}^{kp})^u \overline{b}^{vp^{m - 1}} \in \overline{A}$ with $(u, p) = 1$, and so there exists $z_{6} \in Z(G)$ such that $z_6a^{wp^{m - 1}}(cb^{kp})^ub^{vp^{m - 1}} \in A$. By Lemma 8, we take $W = \langle z_6a^{wp^{m - 1}}(cb^{kp})^ub^{vp^{m - 1}} \rangle ^G$. $\square$

6 NDC-Groups with Small Generating Systems for $p \ge 3$

For convenience, in this section we always assume that p is a odd prime and G is a p-group with $d(G) \ge 3$, $d(G/Z(G)) = 2$ and $|G / Z(G)| \ge p^5$ by Proposition 1(3). So we may write $\overline{G} = G / Z(G)$ and $G = H *T$, where $H = \langle a, b \rangle$, $[a, b] = c$ and T is abelian.

For any $t_1$ and $t_2$ of T, write $K = \langle at_1, bt_2\rangle$ and hence $G = K *T$. Noticing that $K' = G'$ and $Z(K) = K \cap Z(G)$, we know that for every normal subgroup N of K, $K' / N \cap K' = G' / N \cap G'$ and $N / N \cap Z(K) = N / N \cap K \cap Z(G) = N / N \cap Z(G)$. By Theorems 3 and 7, we also see $\overline{G'}$ is cyclic and so $K' Z(K)/ Z(K)$ is cyclic. This forces K to be listed in Theorem 11. Therefore in this section we may give some necessary and sufficient conditions for a p-group to be a NDC-group by analyzing all possible types of T.

Theorem 12

Suppose $\overline{G'} \cong C_{p^t}$ with $t \ge 1$.

• (A) If $G'$ is cyclic, then G is a NDC-group.

• (B) If $G'$ is non-cyclic, then $\overline{G}$ is non-metacyclic. Furthermore, set $\exp (\overline{G}) = p^m$, $\widetilde{G} = G / \mho _1(G')$ and the following hold:

  • (B-1) Assume $t = 1$. Then $\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle \cong C_{p^{e_1}} \times C_{p^{e_2}}$ with $e_1 \ge e_2 \ge 1$, and furthermore G is a NDC-group if and only if one of the following holds:

    • $(\alpha 1)$$\exp (\widetilde{T}) \le p^{e_1 - 1}$ and $\mho _{e_1 - 1}(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle ) = \widetilde{G_3}$;

    • $(\alpha 2)$$\mho _{e_2 - 1}(\widetilde{T}) \le \mho _{e_2}(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle )$ and $\widetilde{G_3} \le \mho _{e_2 - 1}(\langle \widetilde{a^{p^{m - 1}}}$, $\widetilde{b^{p^{m - 1}}}\rangle )$.

  • (B-2) Assume $t \ge 2$. Then $\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle \cong C_{p^{f_1}} \times C_{p^{f_2}}$ with $f_1 \ge f_2 \ge 0$ and $f_1 \ge 1$. Furthermore, G is a NDC-group if and only if $\mho _1(\overline{G'}) \le \langle \overline{a}^p, \overline{b}^p\rangle$, $\exp (\widetilde{T}) \le p^{f_1 - 1}$ and $\mho _{f_1 -1}(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle ) = \widetilde{G_3} \cong C_p$.

Proof

Clearly, there is nothing to be proved if $G'$ is cyclic. Therefore we only need to prove (B). In fact, since $G'$ is non-cyclic, we see $\overline{G}$ is non-metacyclic by Proposition 1(2). Since $\overline{G'}$ is cyclic, we see $\overline{G}$ is regular by Berkovich [5, §7, Exercise 14]. Noticing that $|\overline{G}| \ge p^5$, we also know $m \ge 2$ .

(B-1) Noticing that $\overline{G'} \cong C_p$, we see $\langle \overline{a^{p^{m - 1}}}, \overline{b^{p^{m - 1}}}\rangle \cong C_p^2$ by Lemma 6. Since $\mho _1(G') = \mho _1(H') \le Z(H)$, we see $\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle \cong C_{p^{e_1}} \times C_{p^{e_2}}$ with $e_1 \ge e_2 \ge 1$ by Lemma 8.

“$\Rightarrow$” Let $U = \langle t_1a^{p^{m - 1}}, t_2b^{p^{m - 1}}\rangle ^G$ for any $t_1$ and $t_2$ of T. Then $\overline{U} \cong C_p^2$ by Lemmas 6 and 9. Since G is a NDC-group, we see $G' / G' \cap U$ is cyclic and so $\widetilde{G'} / \widetilde{G'} \cap \widetilde{U}$ is cyclic. Hence $\widetilde{G_3} \le \Omega _1(\widetilde{U})$ and $\widetilde{U} = \langle \widetilde{t_1a^{p^{m - 1}}}, \widetilde{t_2b^{p^{m - 1}}}\rangle$ is abelian by Lemma 8. Take $t_1 = t_2 = 1$ and then $\widetilde{G_3} \le \Omega _1(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle )$.

If $C_p \cong \widetilde{G_3} = \mho _{e_1 - 1}(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle )$, then the arbitrariness of $t_1$ and $t_2$ implies ($\alpha 1$) applies. If $\widetilde{G_3} \cong C_p^2$ or $\widetilde{G_3} \ne \mho _{e_1 - 1}(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle )$, then $\widetilde{G_3} \le \mho _{e_2 - 1}(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle )$ and so the arbitrariness of $t_1$ and $t_2$ implies $\mho _{e_2 - 1}(\widetilde{T}) \le \Phi (\mho _{e_2 - 1}(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle )) = \mho _{e_2}(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle )$, which implies ($\alpha 2$) applies.

“$\Leftarrow$” For any $A \trianglelefteq G$ with $\overline{A} \cong C_p^2$, if we may prove that A has a G-invariant subgroup W such that $\widetilde{G'} / {\widetilde{G'} \cap \widetilde{W}}$ is cyclic, then $G' / {G' \cap W}$ is cyclic and so $G' / {G' \cap A}$ is cyclic, which forces G to be a NDC-group by Theorem 2. In the following, we find the suitable G-invariant subgroups W of A such that $\widetilde{G'} / {\widetilde{G'} \cap \widetilde{W}}$ are cyclic.

It follows from $t = 1$ that $\overline{G}$ is minimal non-abelian by Xu et al. [27, Lemma 2.2]. Since $\overline{G}$ is non-metacyclic, we see $\overline{G} \cong M_p(m, m, 1)$ by Lemma 6, which implies $\Omega _1(\overline{G}) = \langle \overline{a}^{p^{m - 1}}, \overline{b}^{p^{m - 1}}, \overline{c}\rangle$. If $\overline{G'} \le \overline{A}$, then there exists $z_1 \in Z(G)$ such that $cz_1 \in A$ and so take $W = \langle cz_1 \rangle ^G$. If $\overline{G'} \not \le \overline{A}$, then $\overline{A} = \langle \overline{c}^i\overline{a}^{p^{m - 1}}, \overline{c}^j\overline{b}^{p^{m - 1}}\rangle$ with $0 \le i, j \le p - 1$, which implies that there exist $z_2$, $z_3 \in Z(G)$ such that $\langle z_2c^ia^{p^{m - 1}}, z_3c^jb^{p^{m - 1}} \rangle \le A$. In this case, if $i \ne 0$ or $j \ne 0$, then we may assume $i \ne 0$ and hence take $W = \langle z_2c^ia^{p^{m - 1}}\rangle ^G$ by Lemma 8. If $i = j = 0$, then $\overline{A} = \langle \overline{a}^{p^{m - 1}}, \overline{b}^{p^{m - 1}}\rangle$. Noticing that $\widetilde{G_3} \le \Omega _1(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle )$, we see $\widetilde{Z(H)} \le \mho _1(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle )$, which indicates that $\mho _{e_1 - 1}(\widetilde{Z(G)}) = \mho _{e_1 - 1}(\widetilde{Z(H)}\widetilde{T}) = \widetilde{1}$ for ($\alpha 1$) and $\mho _{e_2 - 1}(\widetilde{Z(G)}) = \mho _{e_2 - 1}(\widetilde{Z(H)}\widetilde{T})\le \Phi (\mho _{e_2 - 1}(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle ))$ for ($\alpha 2$). Thus $\widetilde{G_3} \le \langle \widetilde{z_2a^{p^{m - 1}}}, \widetilde{z_3b^{p^{m - 1}}}\rangle$ and so we take $W = \langle z_2a^{p^{m - 1}}, z_3b^{p^{m - 1}} \rangle ^G$ by Lemma 8.

(B-2) By Lemmas 6 and 9, we see $\langle a^{p^{m - 1}}, b^{p^{m - 1}}\rangle Z(G) \mho _1(G') / Z(G) \mho _1(G')$ is non-trivial and thus $\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle$ is non-trivial. So $\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle \cong C_{p^{f_1}} \times C_{p^{f_2}}$ with $f_1 \ge f_2 \ge 0$ and $f_1 \ge 1$ by Lemma 8.

“$\Rightarrow$” Since $Z(G) = Z(H)T$ and H is a NDC-group, we see $\mho _1(\overline{G'}) = \mho _1(\overline{H'}) \le \langle \overline{a}^p, \overline{b}^p\rangle$ by Theorem 11(B). So we only need to prove $\exp (\widetilde{T}) \le p^{f_1 - 1}$ and $\mho _{f_1 -1}(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle = \widetilde{G_3} \cong C_p$.

By Lemma 9, $[\mho _{m - 1}(\overline{G}), \overline{G}] \le \Omega _1(\overline{G'})$ and so $\exp (\mho _{m - 1}(\overline{G})) = p$. If there exists $x \in \mho _{m - 1}(G)Z(G) \backslash Z(G)$ such that $\langle \overline{x} \rangle \cap \langle \overline{c^{p^{t -1}}} \rangle = \overline{1}$ and $\langle \widetilde{x} \rangle \cap \langle \widetilde{G'} \rangle < \widetilde{G_3}$, then $\overline{\langle x, c^{p^{t -1}}\rangle ^G} = \overline{\langle x, c^{p^{t -1}} \rangle } \cong C_p^2$ and by Lemma 8, $\widetilde{G'} / \widetilde{G'} \cap \langle \widetilde{x}, \widetilde{c^{p^{t -1}}}\rangle ^{\widetilde{G}} = \widetilde{G'} / \widetilde{G'} \cap \widetilde{\langle x \rangle } > rsim C_p^2$ which implies $G' / G' \cap \langle x, c^{p^{t -1}}\rangle ^G$ is non-cyclic. This forces G to be a non-NDC-group, a contradiction. So

$$\begin{aligned} \hbox {for any }~ g \in \mho _{m - 1}(G)Z(G) \backslash Z(G), \langle \overline{g} \rangle = \Omega _1(\overline{G'}) ~\mathrm{or }~ \langle \widetilde{g} \rangle \cap \widetilde{G'} = \widetilde{G_3} \cong C_p. \end{aligned}$$

(6.1)

We may assume $\overline{\langle a^{p^{m - 1}}\rangle } \cap \Omega _1(\overline{G'}) = \overline{1}$ by Lemma 6 and hence $\widetilde{\langle za^{p^{m - 1}}\rangle } \cap \widetilde{G'} = \widetilde{G_3} \cong C_p$ for any $z \in Z(G)$ by (6.1). The arbitrariness of z implies $\widetilde{G_3} = \Omega _1(\widetilde{\langle a^{p^{m - 1}}\rangle })$ and $\exp (\widetilde{Z(G)}) \le \frac{o(\widetilde{a^{p^{m - 1}}})}{p}$, which implies

$$\begin{aligned} \exp (\widetilde{T}) \le p^{f_1 - 1}. \end{aligned}$$

In this case, if $\overline{\langle b^{p^{m - 1}}\rangle } \cap \Omega _1(\overline{G'}) = \langle \overline{c^{p^{t -1}}} \rangle$, then $\widetilde{b^{p^{m - 1}}} \in \widetilde{Z(G)}$ and so

$$\begin{aligned} \mho _{f_1 -1}(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle ) = \mho _{f_1 -1}(\widetilde{\langle a^{p^{m - 1}}\rangle }) = \widetilde{G_3}. \end{aligned}$$

If $\overline{\langle b^{p^{m - 1}}\rangle } \cap \Omega _1(\overline{G'}) \ne \langle \overline{c^{p^{t -1}}} \rangle$, then $\overline{\langle b^{p^{m - 1}}\rangle } \cap \Omega _1(\overline{G'}) = \overline{1}$ and so $\widetilde{\langle b^{p^{m - 1}}\rangle } \cap \widetilde{G'} = \widetilde{G_3}$ by (6.1). Hence

$$\begin{aligned} \mho _{f_1 -1}(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle ) = \widetilde{G_3}. \end{aligned}$$

“$\Leftarrow$” Like (B-1), for any $B \trianglelefteq G$ with $\overline{B} \cong C_p^2$, we only need to find the suitable G-invariant subgroups V of B such that $\widetilde{G'} / {\widetilde{G'} \cap \widetilde{V}}$ are cyclic.

Since $\overline{G}$ is regular, we see $\mho _1(\overline{G}) = \langle \overline{a}^p, \overline{b}^p, \overline{c}^p \rangle$. It follows from $\overline{c}^p \in \langle \overline{a}^p, \overline{b}^p \rangle$ that $\mho _1(\overline{G}) = \langle \overline{a}^p, \overline{b}^p\rangle$. Let $\widehat{\mho _1(\overline{G})} = \mho _1(\overline{G}) / \langle \overline{c}^{p^2} \rangle$. Then $\widehat{\mho _1(\overline{G})}$ is abelian. Without loss of generality, we may assume $\widehat{\overline{1}} \ne \widehat{\overline{c}^p} \in \langle \widehat{{\overline{b}}^p} \rangle$ and then $\langle \overline{c}^p \rangle \le \langle \overline{b}^{p}\rangle$. By Lemma 9, $m > t$ and $\Omega _1(\overline{G}) \cong C_p^3$, which indicates that $\overline{c}^p \in \langle {\overline{b}}^{p^2} \rangle$ and there exists a positive integer k such that $o(\overline{c}\overline{b}^{kp}) = p$. Since $\overline{G}$ is non-metacyclic, we see $\overline{c} \not \in \mho _1(\overline{G})$ by Huppert [10, Satz.III, Theorem 11.4]. Also $\langle \overline{a}\rangle \cap \langle \overline{b}\rangle = \overline{1}$ by Lemma 6 and $o(\overline{a}) = o(\overline{b}) = p^m$ by Lemma 9 once more. So $\Omega _1(\overline{G}) = \langle \overline{a}^{p^{m - 1}}, \overline{c}\overline{b}^{kp}, \overline{b}^{p^{m - 1}}\rangle$. If $\overline{B} = \langle \overline{a}^{p^{m - 1}}, \overline{b}^{p^{m - 1}}\rangle$, then there are $z_4$, $z_5 \in Z(G)$ such that $\langle z_4a^{p^{m - 1}}, z_5b^{p^{m - 1}} \rangle \le B$. By Lemma 8, we see $\widetilde{G_3} \le \Omega _1(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle )$ and so $\widetilde{Z(H)} \le \mho _1(\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle )$, which implies that $\langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle \widetilde{Z(G)} = \langle \widetilde{a^{p^{m - 1}}}, \widetilde{b^{p^{m - 1}}}\rangle \widetilde{T}$. It follows from $\exp (\widetilde{T}) \le p^{f_1 - 1}$ that $\mho _{f_1 -1}(\langle \widetilde{z_4a^{p^{m - 1}}}, \widetilde{z_5b^{p^{m - 1}}}\rangle ) = \widetilde{G_3}$ by hypotheses and so we take

$$\begin{aligned} V = \langle z_4a^{p^{m - 1}}, z_5b^{p^{m - 1}} \rangle ^G. \end{aligned}$$

If $\overline{B} \ne \langle \overline{a}^{p^{m - 1}}, \overline{b}^{p^{m - 1}}\rangle$, then there are integers w, u and v such that

$$\begin{aligned} \overline{a}^{wp^{m - 1}}(\overline{c}\overline{b}^{kp})^u \overline{b}^{vp^{m - 1}} \in \overline{B} \end{aligned}$$

with $(u, p) = 1$, and so there exists $z_{6} \in Z(G)$ such that

$$\begin{aligned} z_6a^{wp^{m - 1}}(cb^{kp})^ub^{vp^{m - 1}} \in B. \end{aligned}$$

By Lemma 8, we take $V = \langle z_6a^{wp^{m - 1}}(cb^{kp})^ub^{vp^{m - 1}} \rangle ^G$. The theorem is now proved. $\square$