Proof. Suppose that for a divisor m of e, the power subgroup $G^m$ is cyclic of order $p^k$ . Let $q\ne p$ be a prime divisor of $|G|$ and let y be an element of order $q^\ell $ , where $q^\ell $ is the exponent of a Sylow q-subgroup. If $q^h$ is the highest power of q that divides m, there exist integers s and t such that $q^h = sq^\ell + tm$ . Therefore, $y^{q^h} = y^{tm}\in G^m$ ; thus, $y^{q^h} = 1$ and hence, $h = \ell $ . This proves that $m = p^jd$ for some j, where d is the product of the exponents of the Sylow q-subgroups with $q\ne p$ . Therefore, for all $g\in G$ , $g^d$ is a p-element and it follows that $p^{k+j}$ is the exponent of the Sylow p-subgroups of G. Thus, $m = e/p^k$ .

Proof. Let P be a Sylow p-subgroup of G and suppose that P is not cyclic. By a theorem of Burnside [Reference Burnside5, Section 105], for all m where $1\le m\le f$ , P contains more than one subgroup of order $p^m$ . Miller [Reference Miller11] proved that for $m> 1$ , the number of cyclic subgroups of order $p^m$ is divisible by p and the number of subgroups of order p is congruent to $1+p$ modulo $p^2$ (see also Berkovich [Reference Berkovich3, Theorem 1.10]). Therefore, P has at least p subgroups of order $p^i$ for $2\le i\le f$ and at least $p+1$ subgroups of order p. Furthermore, from Theorem 3.1, P has exactly one cyclic power subgroup of order $p^i$ for $1\le i\le k$ . Therefore, G has at least $p + (k-1)(p-1) + p(f-k) = pf-k+1$ cyclic nonpower p-subgroups.

Proof. Let $G = G^{(r)}_{p,n;\,p,m}$ . Since $G'$ is cyclic, from [Reference Huppert9, III, Section 10], G is a regular p-group. Then, from [Reference Hall7, Theorem 4.21], $G^{p^s}$ is the set $\{\,g^{p^s}\mid g\in G\,\}$ . Thus, there are exactly $\max (n,m)+1$ power subgroups in G. This means that $\operatorname {\mathrm {ps}}(G)=\operatorname {\mathrm {ps}}(C_{p^n}\times C_{p^m})$ . It follows from [Reference Mann, Lewis, Navarro, Passman and Wolf10, Proposition 1] that G is lattice-isomorphic to $C_{p^n}\times C_{p^m}$ . Thus, $s(G) = s(C_{p^n}\times C_{p^m})$ and, therefore, $\operatorname {\mathrm {nps}}(G) = \operatorname {\mathrm {nps}}(C_{p^n}\times C_{p^m})$ .

Proof. For $0\le i < k$ and $0\le j < m$ , let $P_i = \langle a^{p^i} \rangle $ , $Q_j = \langle b^{q^{m-j}} \rangle $ and $H_{ij} = P_iQ_j$ . From the presentation of G, we have $Q_j\trianglelefteq G$ and $Z(G)=\langle a^{p^k} \rangle $ . Furthermore, $P_0\in \operatorname {\mathrm {Syl}}_p(G)$ and G is the semidirect product $Q\rtimes P_0$ , where $Q\in \operatorname {\mathrm {Syl}}_q(G)$ . It is clear that ${P_0\subseteq N_G(P_i)}$ and we claim that for $0\le i< k$ , $N_G(P_i) = P_0$ . Indeed, if $x\in Q\cap N_G(P_i)$ , then $[P_i,\langle x \rangle ]\subseteq P_i\cap Q = 1$ ; that is, $x\in C_G(P_i)$ . It follows from [Reference Gorenstein6, Theorem 5.2.4] that $x\in C_G(P_0)$ and hence, $x\in Z(G)\cap Q = 1$ . This establishes the claim.

We also have $P_0\subseteq N_G(H_{ij})$ and it follows from the Frattini argument that ${N_G(H_{ij}) = Q_jN_G(P_i) = Q_j\rtimes P_0}$ . Therefore, $H_{ij}$ has $q^{m-j}$ conjugates in G and since for $0\le i < k$ and $0\le j < m$ every subgroup of order $p^{n-i}q^j$ is conjugate to $H_{ij}$ , we have found $k\sum _{j=0}^{m-1} q^{m-j} = kq(q^m-1)/(q-1)$ subgroups, all of which are nonpower subgroups. For $1\le i\le n$ , the power subgroup $G^{p^{n-i}}$ is the unique subgroup of order $p^iq^m$ . For $k\le s\le n$ and $0\le j < m$ , the power subgroup $G^{p^sq^{m-j}}$ is the unique subgroup of order $p^{n-s}q^j$ . This accounts for all subgroups of G. Therefore, $\operatorname {\mathrm {nps}}(G) = kq(q^m-1)/(q-1)$ .

Proof. There are $n+1$ power subgroups of $B_{n,p}^{1}$ and $B_{n,p}^{2}$ . We count all the subgroups of them by considering their exponents.

Notice that $\Omega _{n-1}(B^1_{n,p}) = \langle a \rangle \times \langle c \rangle \times \langle b^p \rangle \simeq C_p\times C_p\times C_{p^{n-1}}$ . By [Reference Tărnăuceanu and Tóth14, Corollary 2.2], the number of subgroups of exponent at most $p^{n-1}$ is $s(C_p{\kern-1pt}\times{\kern-1pt} C_p{\kern-1pt}\times{\kern-1pt} C_{p^{n-1}}) {\kern-1pt}={\kern-1pt} 2(n-1)p^2+ np+n+2$ . Furthermore, there are $p^2+p+1$ subgroups of exponent $p^n$ . Thus, $\operatorname {\mathrm {nps}}(B^1_{n,p}) = p^2(2n-1) + p(n+1) + 2$ .

Similar to $B^1_{n,p}$ , $\Omega _{n-1}(B^2_{n,p}) = \langle a \rangle \times \langle b^p \rangle \simeq C_{p^{2}}\times C_{p^{n-1}}$ . From [Reference Tărnăuceanu and Tóth14, Corollary 2.2], the number of subgroups of exponent at most $p^{n-1}$ is ${s(C_{p^2}{\kern-1pt}\times{\kern-1pt} C_{p^{n-1}}) {\kern-1pt}={\kern-1pt} (n{\kern-1pt}-{\kern-1pt}2)p^2{\kern-1pt}+{\kern-1pt}pn{\kern-1pt}+{\kern-1pt}n{\kern-1pt}+{\kern-1pt}2}$ . There are also $p^2{\kern-1pt}+{\kern-1pt}p{\kern-1pt}+{\kern-1pt}1$ subgroups of exponent $p^n$ . Thus, $\operatorname {\mathrm {nps}}(B^2_{n,p}) {\kern-1pt}={\kern-1pt} p^2(n-1) + p(n+1) + 2$ .

Proof. Similar to the calculation of $\operatorname {\mathrm {nps}}(G_{n,3})$ , we have $\operatorname {\mathrm {nps}}(G_{m,3}\times C_3) = 4m+6$ . As n increases, the number of subgroups will increase, but the number of power subgroups does not increase. Thus, $\operatorname {\mathrm {nps}}(G)\ge 4m+6$ . This completes the proof.

Proof. For all $n\ge 1$ , the exponent of G is $4$ and the only proper nontrivial power subgroup is $G^2$ of order $2$ . The group $Q_8\times C_2$ has $19$ subgroups, consequently, $\operatorname {\mathrm {nps}}(Q_8\times C_2) = 16$ and $\operatorname {\mathrm {nps}}(G)\ge 16$ for $n\ge 1$ .

Proof. Since $P\ne N_G(P)\ne G$ , both P and $N_G(P)$ have at least $p+1$ conjugates, and so $2(p+1)\le \operatorname {\mathrm {nps}}(G)\le 13$ ; hence, p is 2, 3 or 5.

First, we show that $p\ne 5$ . Suppose to the contrary that $p = 5$ . Then, the six conjugates of P and the six conjugates of $N_G(P)$ are nonpower subgroups; hence, every other subgroup is normal. In particular, if $Q\in \operatorname {\mathrm {Syl}}_3(G)$ , then $Q \trianglelefteq G$ . The permutation action of G on $\operatorname {\mathrm {Syl}}_5(G)$ defines a homomorphism $G \to \operatorname {\mathrm {Sym}}(6)$ with kernel K. Then, $|KP/K| = 5$ and $KQ/K$ is a normal 3-subgroup of $G/K$ . However, $\operatorname {\mathrm {Sym}}(6)$ does not contain a 3-subgroup normalised by a group of order 5; therefore, $p \ne 5$ .

Claim. If $Q\in \operatorname {\mathrm {Syl}}_q(G)$ and $q\ne p$ , then $Q \trianglelefteq G$ .

If $N_G(Q){\kern-1pt}\ne{\kern-1pt} G$ , then $Q {\kern-1pt}={\kern-1pt} N_G(Q)$ , otherwise, G would have least $2(p{\kern-1pt}+{\kern-1pt}1){\kern-1pt}+{\kern-1pt}2(q{\kern-1pt}+{\kern-1pt}1) \ge 14$ nonpower subgroups.

There are at least $p+1$ conjugates of P and $N_G(P)$ , and at least $q+1$ conjugates of Q; therefore, $2(p+1) + (q+1)\le 13$ and so $q\le 5$ . If $q = 5$ , then $p = 2$ and ${|G:Q| = 6}$ . Then, $|P| = 2$ and G would have a normal subgroup H of index 2, which must contain the six conjugates of Q. However, $|H:Q| = 3$ , which is a contradiction and, therefore, $q\ne 5$ . It follows that $q\le 3$ , $G = PQ$ and $|P| = |G:Q|$ .

Suppose that $p = 2$ . Then, $q = 3$ , $|G : Q| = 4 = |P|$ and the permutation action of G on $\operatorname {\mathrm {Syl}}_3(G)$ defines a homomorphism $G\to \operatorname {\mathrm {Sym}}(4)$ with kernel $K\subseteq Q$ . It follows that $G/K\simeq \operatorname {\mathrm {Alt}}(4)$ and, therefore, $P\simeq C_2\times C_2$ . However, $N_G(P)\ne G$ ; therefore, $K\ne 1$ and so for each element $x\in P$ of order 2, it follows from Lemma 3.4 that $\langle x \rangle $ and $\langle x \rangle K$ are nonpower subgroups. Taking into account the three conjugates of P, the three conjugates of $N_G(P)$ and the four conjugates of Q, G would have at least 16 nonpower subgroups, contrary to our assumption.

This leaves the possibility that $p = 3$ , $q = 4$ and hence, $|G:Q| = 3 = |P|$ . Again, the permutation action of G on $\operatorname {\mathrm {Syl}}_3(G)$ defines a homomorphism $G\to \operatorname {\mathrm {Sym}}(4)$ , this time with kernel $K\subseteq N_G(P)$ . However, $P\nsubseteq K$ ; therefore, $K\subseteq Q$ and Lemma 4.7 implies $G/K\not \simeq \operatorname {\mathrm {Sym}}(4)$ . Consequently, $G/K\simeq \operatorname {\mathrm {Alt}}(4)$ and hence, $Q \trianglelefteq G$ .

If $p = 2$ , then $| G : N_G(P)| = 2k + 1$ for some k and so $2(2k+1)\le 13$ . Therefore, $q = |G : N_G(P)|$ is either 3 or 5 and we choose $Q\in \operatorname {\mathrm {Syl}}_q(G)$ . Similarly, if $p = 3$ , then $|G : N_G(P)| = 4$ . In this case, we let $q = 2$ and choose $Q\in \operatorname {\mathrm {Syl}}_2(G)$ . In all cases ${G = N_G(P)Q}$ .

Claim. $G = PQ\times A$ , where A is a cyclic group of order 1, r, $r^2$ or $rs$ , where r and s are primes different from p and q.

Let R be a Sylow r-subgroup of G, where $r\ne p,q$ is a prime. Then, R acts by conjugation on $\operatorname {\mathrm {Syl}}_p(G)$ and in all cases, R fixes a conjugate of P. Thus, $R\subseteq N_G(P)$ and since $R \trianglelefteq G$ , we have $[P,R]\subseteq P\cap R = 1$ . Also $Q \trianglelefteq G$ and so $[Q,R]\subseteq Q\cap R =1$ . Let A be the product of the Sylow r-subgroups for $r\ne p,q$ . Then, $G = PQ\times A$ and from Lemma 3.3,

$$ \begin{align*} \operatorname{\mathrm{nps}}(G) = \operatorname{\mathrm{nps}}(PQ)s(A) + \operatorname{\mathrm{ps}}(PQ)\operatorname{\mathrm{nps}}(A). \end{align*} $$

We have $\operatorname {\mathrm {nps}}(PQ)\ge 3$ and $\operatorname {\mathrm {ps}}(PQ)\ge 3$ . Thus, A is cyclic, otherwise, $s(A)\ge 3$ and from Theorem 1.1, $\operatorname {\mathrm {nps}}(A)\ge 3$ and so $\operatorname {\mathrm {nps}}(G)> 13$ . Since A is cyclic, we have ${\operatorname {\mathrm {nps}}(A) = 0}$ and, therefore, $\operatorname {\mathrm {nps}}(G) = \operatorname {\mathrm {nps}}(PQ)s(A)\le 13$ . Thus, $s(A)\le 4$ and $|A|$ is 1, r or $rs$ , where $r\le s$ are primes.

Claim. If $p = 2$ , then G is isomorphic to $\operatorname {\mathrm {Sym}}(3)\times C_3$ , $G_{n,5}\times C_r$ , where $r\ne 2,5$ is a prime, or $G_{n,3}\times C_{rs}$ , where r and s are primes such that $3 < r < s$ .

We have $G = PQ\times A$ , where A is cyclic and $\operatorname {\mathrm {nps}}(G) = \operatorname {\mathrm {nps}}(PQ)s(A)$ . If $A\ne 1$ , then $s(A)\ge 2$ and, therefore, $\operatorname {\mathrm {nps}}(G)\le 6$ . It follows from Theorems 1.1 and 1.2 that for some n, $G\simeq G_{n,3}$ or $G_{n,5}$ . If $s(A) = 2$ , then $A\simeq C_r$ ; if $s(A) = 3$ , then $A\simeq C_{r^2}$ ; if ${s(A) = 4}$ , then $A\simeq C_{rs}$ for some primes r and s.

This leaves the possibility that $A = 1$ and $G = PQ$ . If $|G:N_G(P)| = 5$ , then P and $N_G(P)$ each have five conjugates and if Q is not cyclic, it follows from Theorem 3.1 that Q contains at least five nonpower subgroups of order 5. This contradicts the assumption that $\operatorname {\mathrm {nps}}(G)\le 13$ and, therefore, $|G:N_G(P)| = 3$ .

We have $[N_Q(P),P] \subseteq P\cap Q = 1$ and, therefore, $N_Q(P) = C_Q(P)$ . Consequently, $N_G(P) = P\times C_Q(P)$ . Let K be the kernel of the permutation action on $\operatorname {\mathrm {Syl}}_2(G)$ . Then, $G/K\simeq \operatorname {\mathrm {Sym}}(3)$ and $K\subseteq N_G(P)$ ; hence, $K\cap Q = C_Q(P)$ and .

If P is not cyclic, then $\operatorname {\mathrm {nps}}(P)\ge 3$ and for each nonpower subgroup H of P, it follows from Lemma 3.4 that H, $HQ$ and $HC_Q(P)$ are nonpower subgroups of G; hence, $\operatorname {\mathrm {nps}}(G)\ge 15$ , which contradicts the assumption $\operatorname {\mathrm {nps}}(G)\le 13$ . Therefore, P is cyclic and so $N_G(P) = C_G(P)$ .

Let $P = \langle a \rangle $ . Then, $P\cap K = \langle a^2 \rangle $ and $K = \langle a^2 \rangle \times C_Q(P)$ . Thus, $\langle a^2 \rangle $ is a characteristic subgroup of K and, therefore, $\langle a^2 \rangle \trianglelefteq G$ . It follows that $z = a^2\in Z(G)$ . Let b be a conjugate of a such that $b\ne a$ and let $x = ab^{-1}$ . Then, $b^2 = z$ , $b^{-1} = bz^{-2}$ and ${a^{-1}xa = x^{-1}}$ . Since $x\in \langle z \rangle Q$ and P is cyclic, $y = x^{2^k}\ne 1$ belongs to Q for some k. Furthermore, a has only three conjugates; therefore, $y^3 = 1$ . The group $\langle y \rangle P$ is generated by the conjugates of P and so $\langle y \rangle P \trianglelefteq G$ . Moreover, for some n, we have $\langle y \rangle P\simeq G_{n,3}$ .

We now have $G = \langle y \rangle P\times C_Q(P)$ . If $C_Q(P)$ is not cyclic, it contains an elementary abelian subgroup E of order 9. Then, $\langle y \rangle \times H$ is elementary abelian of order 27; it contains 13 subgroups of order 3, at most one of which can be a nonpower subgroup (Theorem 3.1). It follows that $C_Q(P)$ is cyclic. If the exponent of $C_Q(P)$ is $3^e$ , then $Q^3$ is a power subgroup of order $3^{e-1}$ and from Corollary 3.2, Q contains at least $2e + 2$ nonpower subgroups of G. If $e\ge 2$ , then P, $P\times C_Q(P)$ and $P\times Q^3$ each have three conjugates and thus, $\operatorname {\mathrm {nps}}(G)\ge 15$ , which contradicts our assumption. Therefore, $G\simeq G_{n,3}\times C_3$ and from Lemma 4.5, we have $n = 1$ ; that is, $G\simeq \operatorname {\mathrm {Sym}}(3)\times C_3$ .

Claim. If $p = 3$ , then $\operatorname {\mathrm {nps}}(G) = 11$ and $G\simeq \operatorname {\mathrm {SL}}(2,3)$ .

In this case, $|G : N_G(P)| = 4$ and the permutation action on $\operatorname {\mathrm {Syl}}_3(G)$ defines a homomorphism $G\to \operatorname {\mathrm {Sym}}(4)$ whose kernel K is a proper subgroup of $N_G(P)$ . As ${Q \trianglelefteq G}$ , we must have $G/K\simeq \operatorname {\mathrm {Alt}}(4)$ and $QK/K\simeq C_2\times C_2$ . The assumption ${10\le \operatorname {\mathrm {nps}}(G)}$ implies $K \ne 1$ . Let $x_1, x_2, x_3\in Q$ be elements such that $x_iK$ is an involution in $QK/K$ . From Lemma 3.4, $\langle x_i \rangle $ and $\langle x_i \rangle K$ are nonpower subgroups. The four conjugates of P and the four conjugates of $N_G(P)$ are nonpower subgroups, and since there are at most 13 nonpower subgroups in G, it follows that $\langle x_i \rangle = \langle x_i \rangle K$ . That is, $K\subseteq \langle x_i \rangle $ and so $K = \langle x_i^2 \rangle $ for $1\le i\le 3$ . We now see that $|P| = 3$ , $K\subseteq C_G(P)$ and P permutes the $x_i$ ; hence, $x_1^2 = x_2^2 = x_3^2$ . It follows that $Q\simeq Q_8$ and thus, $G\simeq \operatorname {\mathrm {SL}}(2,3)$ .

Proof. The Sylow subgroup P has $m = |G:N_G(P)|$ conjugates. Since $m\equiv 1\pmod p$ and $N_G(P)\ne G$ , we have $m \ge p+1$ and the conjugates of P are nonpower subgroups. The assumption $\operatorname {\mathrm {nps}}(G)\le 13$ implies $p\in \{2,3,5,7,11\}$ .

If $p = 2$ , then $m\in \{3,5,7,9,11,13\}$ ; if $p = 3$ , then $m\in \{4,7,10,13\}$ ; if $p = 5$ , then $m\in \{6,11\}$ ; if $p = 7$ , then $m = 8$ ; if $p = 11$ , then $m = 12$ .

Suppose that $p = 3$ and $m = 10$ or $p = 5$ and $m = 6$ . Then, G is a group of twice odd order and, therefore, has a subgroup H of index 2, which contains P. Then, ${|H:P| = m/2}$ , which is impossible. Suppose that $p = 11$ and $m = 12$ . The permutation action on $\operatorname {\mathrm {Syl}}_{11}(G)$ defines a homomorphism $G\to \operatorname {\mathrm {Sym}}(12)$ whose kernel $K = \bigcap _{g\in G}P^g$ is properly contained in P. Let $\overline G = G/K$ . Then, $N_{\overline G}(P/K) = C_{\overline G}(P/K)$ and, therefore, by Burnside’s normal p-complement theorem (see [Reference Gorenstein6, Theorem 7.4.3]), $P/K$ has a normal 11-complement in $\overline G$ , which is a contradiction.

Thus, in all cases, m is a power of a prime q and for $Q\in \operatorname {\mathrm {Syl}}_q(G)$ , we have $G = PQ$ and $|Q| = m$ . We shall show that $Q \trianglelefteq G$ .

If $N_G(Q)\ne G$ , it follows from Lemma 5.1 that $Q = N_G(Q)$ , otherwise, $P \trianglelefteq G$ . If the order of Q is q or $q^2$ , then Q is abelian and it follows from Burnside’s normal p-complement theorem that $P \trianglelefteq G$ . Thus, the only possibility is $p = 7$ , $m = 8$ and ${q = 2}$ . However, then $|G : Q|$ must be a power of 7 and so $\operatorname {\mathrm {nps}}(G)\ge 15$ , in contrast to our assumption. This proves that $Q \trianglelefteq G$ .

The permutation action on $\operatorname {\mathrm {Syl}}_p(G)$ defines a homomorphism $G\to \operatorname {\mathrm {Sym}}(m)$ whose kernel $K = \bigcap _{g\in G}P^g$ is properly contained in P. Since $Q \trianglelefteq G$ , we have $[K,Q]\subseteq K\cap Q = 1$ . Moreover, Q acts transitively on the m conjugates of P and, therefore, ${K = C_P(Q)}$ .

Case 1: $p = 2$ and $m\in \{3,5,7,9,11,13\}$ .

We have $G = Q\rtimes P$ , where P is a 2-group and Q is an abelian group of order m. We treat each value of m separately.

Case 1a: $p = 2$ and $|Q| = 3$ .

In this case, $G/K\simeq \operatorname {\mathrm {Sym}}(3)$ and $|P/K| = 2$ . Therefore, $\Phi (P)\subseteq K = C_P(Q)$ and so $\Phi (P) \trianglelefteq G$ . Thus,

$$ \begin{align*} G/\Phi(P)\simeq\operatorname{\mathrm{Sym}}(3)\times\underbrace{C_2\times\cdots\times C_2}_n. \end{align*} $$

From Lemma 4.8, $n \le 1$ . If $n = 0$ , then P is cyclic, $G\simeq G_{n,3}$ and $\operatorname {\mathrm {nps}}(G) = 3$ , which contradicts the assumption $10\le \operatorname {\mathrm {nps}}(G)$ . Thus, $n = 1$ , $G/\Phi (P)\simeq \operatorname {\mathrm {Sym}}(3)\times C_2\simeq D_{12}$ and, from Lemma 4.8, $\operatorname {\mathrm {nps}}(G/\Phi (P)) = 13$ . The only nontrivial proper power subgroup of $\operatorname {\mathrm {Sym}}(3)\times C_2$ is the subgroup of order 3. Therefore, for all $H\subseteq P$ such that $|H\Phi (P)/\Phi (P)| = 2$ , it follows from Lemma 3.4 that both H and $H\Phi (P)$ are nonpower subgroups. Therefore, $H = H\Phi (P)$ and hence, $\Phi (P)\subseteq H$ .

If $P\setminus \Phi (P)$ contains an element of order 2, then $\Phi (P) {\kern-1pt}={\kern-1pt} 1$ and thus, ${G{\kern-1pt}\simeq{\kern-1pt} \operatorname {\mathrm {Sym}}(3){\kern-1pt}\times{\kern-1pt} {C}_2}$ . Otherwise, $\Phi (P) = \langle x^2 \rangle $ for all $x\in P\setminus \Phi (P)$ . Thus, P is nonabelian and it follows from Lemma 3.4 that $\operatorname {\mathrm {nps}}(G)\ge 2\operatorname {\mathrm {nps}}(G/Q)$ . Then, Theorems 1.1 and 1.2 show that the only possibility for $P\simeq G/Q$ is the quaternion group $Q_8$ . Thus, G is the semidirect product

$$ \begin{align*} C_3\rtimes Q_8 = \langle x,y,b\mid x^4 = y^4 = b^3 = [y,b] = 1,\, x^2 = y^2,\,[x,y] = x^2,\,b^x = b^{-1} \rangle. \end{align*} $$

Case 1b: $p = 2$ and $|Q| = 5$ .

In this case, $QK/K$ is a normal subgroup of $G/K\subseteq \operatorname {\mathrm {Sym}}(5)$ . Therefore, $|G/K|$ is either 10 or 20.

Suppose $|G/K| = 10$ . If P is not cyclic, then $\Phi (P)\subseteq K$ and, therefore, $\Phi (P) \trianglelefteq G$ . We have $G/\Phi (P)\simeq D_{10}\times C_2\times \cdots \times C_2$ , and Lemmas 3.4 and 4.8 imply $\operatorname {\mathrm {nps}}(G)\ge 19$ , whereas we assume that $\operatorname {\mathrm {nps}}(G)\le 13$ ; hence, P is cyclic. If a generates P and b generates Q, then $b^a = b^{-1}$ and $G\simeq G_{n,5}$ . However, then $\operatorname {\mathrm {nps}}(G) = 5$ and so G does not satisfy the assumption $10\le \operatorname {\mathrm {nps}}(G)$ .

Thus, $|G/K| = 20$ , $G/K\simeq G^{(2)}_{2,n;\,5,1}$ and $\operatorname {\mathrm {nps}}(G/K) = 10$ . Choose $a\in P$ such that $aK$ generates $P/K$ . The five conjugates of $\langle a \rangle $ and the five conjugates of $\langle a^2 \rangle $ are nonpower subgroups of G. Similarly, the five conjugates of $\langle a \rangle K$ and the five conjugates of $\langle a^2 \rangle K$ are nonpower subgroups of G. Since $\operatorname {\mathrm {nps}}(G)\le 13$ , we must have $K\subseteq \langle a \rangle $ and, therefore, P is cyclic. Therefore, $G\simeq G^{(2)}_{2,n;\,5,1}$ .

Case 1c: $p = 2$ and $|Q| = 7$ .

In this case, $G/K\simeq D_{14}$ and P has seven conjugates. Suppose that $a\in P\setminus K$ . Then, P, $\langle a \rangle $ and their conjugates are nonpower subgroups. However, $\operatorname {\mathrm {nps}}(G)\le 13$ and so $P = \langle a \rangle $ . Therefore, $G\simeq G_{n,7}$ whence $\operatorname {\mathrm {nps}}(G) = 7$ and there are no examples with ${10\le \operatorname {\mathrm {nps}}(G)}$ .

Case 1d: $p = 2$ and $|Q| = 9$ .

Suppose that there is a subgroup R of order 3 in Q such that $R \trianglelefteq G$ . Then, $RP$ is not normal in G, otherwise, the Frattini argument [Reference Gorenstein6, Theorem 1.3.7] implies $G = RN_G(P) = RP$ , which is a contradiction. The nine conjugates of P and the three conjugates of $RP$ are nonpower subgroups. Therefore, all other subgroups are normal. Thus, R is the unique subgroup of Q of order 3 and hence, Q is cyclic. If $a\in P\setminus K$ and b generates Q, then $b^a = b^{-1}$ . If $P\ne \langle a \rangle $ , then $\langle a \rangle \trianglelefteq G$ and so $ab = ba$ , which is a contradiction. Therefore, P is cyclic, $\operatorname {\mathrm {nps}}(G) = 12$ and $G\simeq G_{n,9}$ .

This leaves us with the possibility that no subgroup of order 3 is normal in G. This implies Q is not cyclic, otherwise the subgroup of order 3 in Q would be a normal subgroup of G. Consequently, $Q\simeq C_3\times C_3$ and its four subgroups of order 3 are not normal in G. Since $\operatorname {\mathrm {nps}}(G)\le 13$ and the nine conjugates of P are nonpower subgroups, it follows that the proper subgroups of P are normal in G. As above, this implies that P is cyclic. Furthermore, if $P = \langle a \rangle $ , then $\langle a^2 \rangle \trianglelefteq G$ and so $a^2\in K$ . Thus, if x is an element of order 3 in Q and $y = x^a\ne x$ , then $(xy)^a = xy$ and $\langle xy \rangle \trianglelefteq G$ , which is a contradiction.

Case 1e: $p = 2$ and $|Q| = 11$ .

In this case, $G/K\simeq D_{22}$ and P has 11 conjugates. Thus, P is cyclic and $G\simeq G_{n,11}$ .

Case 1f: $p = 2$ and $|Q| = 13$ .

In this case, $G/K\simeq D_{26}$ and P has 13 conjugates. Thus, P is cyclic and $G\simeq G_{n,13}$ .

Case 2: $p = 3$ and $m\in \{4,7,13\}$ .

In this case, $G = Q\rtimes P$ , where P is a 3-group and Q is a q-group with $q\in \{2,7,13\}$ .

Case 2a: $p = 3$ and $|Q| = 4$ .

Considering the homomorphism $G\to \operatorname {\mathrm {Sym}}(4)$ (with kernel $K\subset P$ ), we have ${G/K\simeq \operatorname {\mathrm {Alt}}(4)}$ . Thus, $Q\simeq C_{2}\times C_{2}$ . From $\operatorname {\mathrm {nps}}(\operatorname {\mathrm {Alt}}(4)) = 7$ and $10\le \operatorname {\mathrm {nps}}(G)$ , it follows that $K\ne 1$ . Then, G has seven nonpower subgroups that properly contain K and three nonpower subgroups that are nontrivial proper subgroups of Q. By $\operatorname {\mathrm {nps}}(G)\leq 13$ , we have $\operatorname {\mathrm {nps}}(G/Q)\leq 3$ . This implies $P\simeq C_{3^n}$ and hence, $G\simeq A_{n}$ . From Lemma 4.1, we have $G\simeq A_2$ or $A_3$ .

Case 2b: $p = 3$ and $|Q| = 7$ .

From Theorems 1.1 and 1.2, Lemma 3.4, and the assumption $\operatorname {\mathrm {nps}}(G)\le 13$ , P is either cyclic or isomorphic to $C_3\times C_3$ . If $P\simeq C_{3} \times C_{3}$ and H is a noncentral subgroup of order 3, then H has seven conjugates. This contradicts $\operatorname {\mathrm {nps}}(G)\le 13$ because P has three subgroups of order 3 not in the centre of G. Thus, P is cyclic. Let a be a generator of P. Then, there exists a generator b of Q such that $a^{-1}ba = b^2$ . Therefore, for some $n\ge 1$ , $G\simeq F_{n,7}$ . However, then $\operatorname {\mathrm {nps}}(G) = 7$ , whereas we assume that $10\le \operatorname {\mathrm {nps}}(G)$ .

Case 2c: $p = 3$ and $|Q| = 13$ .

As in previous cases, P is cyclic and thus, $G\simeq F_{n,13}$ .

Case 3: $p = 5$ and $|Q| = 11$ .

As in previous cases, P is cyclic and thus, $G\simeq G_{5,n;\,11,1}^{(3)}$ .

Case 4: $p = 7$ and $|Q| = 8$ .

The subgroup P acts transitively on the seven subgroups of order 2 of Q and on the seven subgroups of order 4, otherwise $Q\subseteq C_G(P)$ . Thus, $\operatorname {\mathrm {nps}}(G)\ge 14$ and so there are no examples with $p = 7$ .

Proof of Theorem 1.3.

From the above discussion, we may suppose that G is nilpotent so that $G = P\times H$ , where P is a noncyclic p-subgroup and $H\ne 1$ is a group whose order is not divisible by p. If H is also noncyclic, then H has at least three subgroups and from Lemma 3.3, we have $\operatorname {\mathrm {nps}}(G)\ge 15$ . Thus, H is cyclic and $\operatorname {\mathrm {nps}}(P\times H) = \operatorname {\mathrm {nps}}(P)s(H)$ .

If $H\ne 1$ , then $s(H)\ge 2$ and, therefore, the assumption $\operatorname {\mathrm {nps}}(G)\le 13$ implies $\operatorname {\mathrm {nps}}(P)\le 6$ . Referring to Theorems 1.1 and 1.2, P is isomorphic to $C_2\times C_2$ , $C_3\times C_3$ , $C_2\times C_4$ or $C_5\times C_5$ . If $\operatorname {\mathrm {nps}}(G) = 10$ , then $G\simeq C_2\times C_4\times C_p$ , where $p> 3$ is a prime; if $\operatorname {\mathrm {nps}}(G) = 12$ , then $G\simeq Q_8\times C_{qr}$ , $C_2\times C_2\times C_{qr}$ , $C_3\times C_3\times C_{s^2}$ or $C_5\times C_5\times C_p$ , where $p,q,r$ and s are primes such that $p\ne 5$ , $3 < q < r$ and $s\ne 3$ .

From now on, we may suppose that G is a noncyclic p-group. Then, $G/\Phi (G)$ is an elementary abelian group of order $p^d$ and every proper nontrivial subgroup of $G/\Phi (G)$ is a nonpower subgroup. Since ${\operatorname {\mathrm {nps}}(C_p\times C_p\times C_p)\ge \operatorname {\mathrm {nps}}(C_2\times C_2\times C_2) = 14}$ , it follows that $d = 2$ and G is generated by two elements. Thus, from Lemma 4.1(ix), we have $G/G'\simeq C_2\times C_{2^n}$ for $1\le n\le 6$ , $C_3\times C_{3^n}$ for $1\le n\le 4$ , $C_5\times C_{5^n}$ for ${n\in \{1,2\}}$ , $C_7\times C_7$ , $C_{11}\times C_{11}$ or $C_4\times C_4$ . If G is abelian, the proof is complete. From now on, we assume that $G'\ne 1$ .

If $G/G'= C_2\times C_2$ , it follows from [Reference Gorenstein6, Theorem 5.4.5] that G is isomorphic to $D_{2^m}$ , $S_{2^m}$ or $Q_{2^m}$ , and then from Lemma 4.1, the only possibility is $S_{16}$ .

Since $G'\ne 1$ , there exists $R \trianglelefteq G$ such that $|G'/R| = p$ . We shall determine the structure of $G/R$ for each choice of $G/G'$ .

If $G/G' = C_p\times C_p$ and p is odd, it follows from [Reference Gorenstein6, Theorem 5.5.1] that $G/R$ is an extraspecial group of order $p^3$ and then from Lemma 4.1, the only possibility for $G/R$ is $M_{3,5}$ . We argue as in [Reference Zheng, Zhou and Taylor15]. The group $M_{3,5}$ has a cyclic subgroup of order 25 and, therefore, it is metacyclic. It follows from Lemma 3.7 that G is metacyclic and so G has a cyclic normal subgroup that properly contains $G'$ ; that is, G has a cyclic subgroup of index 5 and, therefore, by [Reference Gorenstein6, Theorem 5.4.4], we have $G\simeq M_{n,5}$ . From Lemma 4.1, it follows that $n = 3$ and $G\simeq M_{3,5}$ .

If $G/G'\simeq C_2\times C_{2^6}$ , $C_3\times C_{3^4}$ , $C_{11}\times C_{11}$ or $C_4\times C_4$ , then $\operatorname {\mathrm {nps}}(G/G')\in \{12,13\}$ and all subgroups of G that do not contain $G'$ are normal. However, then every subgroup of G is normal because all subgroups that contain $G'$ are normal. Thus, G is a Hamiltonian group whence $p = 2$ and $G\simeq Q_8\times C_2\times \cdots \times C_2$ (see [Reference Huppert9, III, 7.12]). It follows from Lemmas 4.1 and 4.6 that no groups of this type satisfy ${10\le \operatorname {\mathrm {nps}}(G)\le 13}$ . Similarly, because $\operatorname {\mathrm {nps}}(C_5\times C_{5^2}) = 11$ , the same argument shows that there are no examples with $G/G' \simeq C_5\times C_{5^2}$ .

Finally, we consider the cases $G/G'\simeq C_2\times C_{2^n}$ for $2\le n\le 5$ and $C_3\times C_{3^{n}}$ for ${n = 2,3}$ . If $G/R$ has a cyclic maximal subgroup, it follows from [Reference Gorenstein6, Theorem 5.4.4] that $G\simeq M_{n+2,2}$ or $M_{n+2,3}$ . From Lemmas 3.6 and 4.1, G is isomorphic to $M_{6,2}$ , $M_{7,2}$ , $M_{4,3}$ or $M_{5,3}$ . If the exponent of $G/R$ is $2^n$ or $3^n$ , then from Lemmas 3.7 and 5.2, $G/R\simeq B^2_{2,2}$ and $\operatorname {\mathrm {nps}}(G/R) = 12$ .

We may choose generators a and b for G such that their images $\overline a$ and $\overline b$ in $G/R$ satisfy the relations $\overline a^4 = 1$ , $\overline b^4 = 1$ and ${\overline a}^{-1}\overline b\overline a = {\overline b}^{-1}$ . The nonpower subgroup $\langle \overline a \rangle $ is the image in $G/R$ of both $\langle a \rangle $ and $\langle a \rangle R$ . Since $\langle a \rangle $ is not normal in G, we must have $R\subseteq \langle a \rangle $ and, therefore, $R = \langle a^4 \rangle $ is cyclic. The group G has at most one power subgroup of order 2 and since G is neither cyclic nor a generalised quaternion group, it has at least three subgroups of order 2. Therefore, there are elements $x,y\in G\setminus R$ of order 2 such that $\langle x \rangle $ and $\langle y \rangle $ are nonpower subgroups. However, then $\langle x \rangle R$ and $\langle y \rangle R$ are also nonpower subgroups and this is a contradiction. Therefore, $R = 1$ , $G\simeq B^2_{2,2}$ and this completes the proof.