Proof of Theorem 1.1.

Let p be an odd prime. Our goal is to show that for any positive integer n and any odd prime p, the group $G_{n,p} = \langle x, y: x^{2^n}=1=y^p, yx=xy^{-1} \rangle $ contains exactly p nonpower subgroups. We have $|G_{n,p}|=2^np$ . We first obtain a count of the number of subgroups in $G_{n,p}$ . Since the Sylow 2-subgroup $\langle x \rangle $ is not a normal subgroup, the number of Sylow $2$ -subgroups of $G_{n,p}$ must be p. On the other hand, since $y^x = y^{-1}$ , there is a unique normal Sylow p-subgroup, namely the cyclic subgroup $\langle y\rangle $ of order p. Since $x^2$ is central in $G_{n,p}$ and each Sylow $2$ -subgroup of $G_{n,p}$ is cyclic, there is a unique subgroup of order $2^k$ (for each $k\in \{0,\ldots ,n-1\})$ and a unique subgroup of order $2^kp$ (for each $k\in \{1,\ldots ,n\})$ . Along with the p subgroups of order $2^n$ , we see that $s(G_{n,p})=2n+p+1$ . As the subgroups of order $2^n$ are not normal, we know immediately that they are nonpower subgroups. Hence ${\textit {nps}}(G_{n,p})\geq p$ . We now show that any subgroup of $G_{n,p}$ that is not a Sylow $2$ -subgroup of $G_{n,p}$ is a power subgroup of $G_{n,p}$ . First, the unique subgroup of order p is $G_{n,p}^{2^n}$ . Secondly, for each $k\in \{0,\ldots , n-1\}$ , the subgroup of order $2^k$ is $G_{n,p}^{2^{n-k}p}$ . Finally, for each $k\in \{1,\ldots ,n\}$ , the subgroup of order $2^kp$ is $G_{n,p}^{2^{n-k}}$ . Therefore, ${\textit {ps}}(G_{n,p})=2n+1$ , whence ${\textit {nps}}(G_{n,p})=p$ .

Proof. Let G be noncyclic of order $p^n$ . It was shown in [Reference Zhou, Shi and Duan3] that if $N \unlhd G$ and $A/N$ is a nonpower subgroup of $G/N$ , then A is a nonpower subgroup of G. Suppose G has exactly k nonpower subgroups, where $k \in \{3,4\}$ . Then $G/\Phi (G) \cong C_p \times \cdots \times C_p$ (d times) and $d\geq 2$ as G is not cyclic. The ${(p^d-1)}/{(p-1)}$ cyclic subgroups of order p in $C_p^d$ are nonpower subgroups. It follows that $G/\Phi (G)$ , and hence G, has at least $1+p+\cdots +p^{d-1}$ nonpower subgroups. Hence, $d=2$ , either $p=2$ or $p=3$ , and G has $p+1$ maximal subgroups that are nonpower subgroups.

The power subgroups of G are $G^1=G$ , $G^p$ , $G^{p^2}, \ldots , G^{p^m}$ , where $p^m$ is the exponent of G. There are thus at most $m+1$ distinct power subgroups. Since G is not cyclic, this means $m<n$ ; so ${\textit {ps}}(G)\leq n$ .

What about $s(G)$ ? There is at least one subgroup of order $p^i$ for $0\leq i\leq n$ (just take any composition series). This gives at least $n+1$ subgroups. But there are $p+1$ maximal subgroups (of order $p^{n-1})$ arising from the $p+1$ nontrivial proper subgroups of $G/\Phi (G)$ . Thus $s(G) \geq n + p + 1$ .

Suppose $p=2$ . If G is not generalised quaternion (and by assumption G is not cyclic), then G has at least three involutions and hence at least three subgroups of order $2$ . So, if $n>2$ , then $s(G)\geq n+5$ , meaning that ${\textit {nps}}(G)\geq 5$ , a contradiction. Thus, either G is generalised quaternion or $n=2$ , which means $G \cong C_2 \times C_2$ , and in this case ${\textit {nps}}(G)=3$ . If G is generalised quaternion, then G has $2^{n-1}+2$ elements of order $4$ , resulting in $2^{n-2}+1$ subgroups of order $4$ . On the other hand, if $n>3$ , then $s(G)\geq n+1+2^{n-2}\geq n+5$ . Again, this means that ${\textit {nps}}(G)\geq 5$ . Thus, $n=3$ and then $G\cong Q_8$ . Again, ${\textit {nps}}(Q_8)=3$ .

The remaining case is $p=3$ . By Lemma 1.6, there are at least four subgroups of order 3 in G. If $n>2$ , then these are distinct from the four maximal subgroups and so $s(G) \geq n+7$ . This forces ${\textit {nps}}(G) \geq 7$ , a contradiction. The only possibility is that $n=2$ . A quick check shows that ${\textit {nps}}(C_3\times C_3) = 4$ .

Thus, ${\textit {nps}}(G)=3$ if and only if G is $C_2 \times C_2$ or $Q_8$ , and ${\textit {nps}}(G) = 4$ if and only if G is $C_3\times C_3$ .

Proof. Recall that a finite group is nilpotent if and only if it is the direct product of its Sylow subgroups, each of which is normal. Since G is noncyclic, at least one of these Sylow subgroups is noncyclic. Let $p_1,\ldots ,p_r$ be the primes dividing $|G|$ and let $P_1, \ldots , P_r$ be the respective Sylow subgroups. Assume, without loss of generality, that $P_1$ is noncyclic. Write $Q=P_2 \times \cdots \times P_r$ ; so $G\cong P_1 \times Q$ . Since G is not a p-group, we have $Q\neq \{1\}$ . Therefore, by Lemma 1.4,

$$\begin{align*}{\textit{nps}}(G)={\textit{nps}}(P_1)s(Q) + {\textit{ps}}(P_1){\textit{nps}}(Q)\geq {\textit{nps}}(P_1)s(Q).\end{align*}$$

As $Q\neq \{1\}$ , we have $s(Q)\geq 2$ . As $P_1$ is not cyclic, ${\textit {nps}}(P_1)\geq 3$ . Hence ${\textit {nps}}(G)\geq 6$ .

Proof. Suppose G is finite, nonnilpotent and ${\textit {nps}}(G) = k\in \{3,4\}$ . If G had a unique Sylow p-subgroup for each p dividing $|G|$ , then G would be nilpotent. So there is at least one such p for which G has more than one Sylow p-subgroup. For any such p, the number, $n_p$ , of Sylow p-subgroups is congruent to $1$ mod p. So $n_p\geq p+1$ . These groups are not normal, so are not power subgroups. Therefore, as ${\textit {nps}}(G)\in \{3,4\}$ , either $p=2$ and $n_2=3$ , or $p=3$ and $n_3 = 4$ . For all other primes q dividing $|G|$ , there must be a unique Sylow q-subgroup. If any subgroup of G, other than the Sylow p-subgroups, were nonnormal, then it and its conjugates could not be power subgroups. Thus there would be at least two further nonpower subgroups, forcing ${\textit {nps}}(G) \geq 5$ , a contradiction. Therefore, every subgroup of G, other than the Sylow p-subgroups, is normal.

Let P be one of the Sylow p-subgroups. Let $q_1,\ldots ,q_r$ be the primes other than p dividing $|G|$ . Let $Q_1,\ldots ,Q_r$ be the corresponding normal Sylow subgroups. Each $Q_i$ is normal and the $Q_i$ intersect trivially. Therefore, defining $H=Q_1Q_2\cdots Q_r$ , we have that $H\cong Q_1 \times Q_2 \times \cdots \times Q_r$ is a normal subgroup of G, with $G=PH$ . Now $P\unlhd N_G(P)$ and, setting $K=H \cap N_G(P)$ , we have $K\unlhd G$ (because certainly K is not a Sylow p-subgroup). But P is normal in $N_G(P)=PK$ ; so $N_G(P)\cong P \times K$ . Let $h\in H-N_G(P)$ . Then $(PK)^h=P^hK\neq PK$ . This means that $PK$ is not normal in G, a contradiction unless $K=\{1\}$ . Therefore, $K=\{1\}$ and $P=N_G(P)$ . In particular, $n_p = |G:P| = |H|$ .

Suppose first that $p=3$ . Then $|H| = 4$ . If $H \cong C_2\times C_2$ , then each of its cyclic subgroups would be normal, and hence the involutions they contain would be central. But that would imply that P is normal in G, a contradiction. Therefore $H \cong C_4$ . Let z be a generator of H. We have $H \leq C_G(z) \leq G$ . Thus, $|z^G| = 3^i$ for some i with $0\leq i \leq n$ . But $z^G \subseteq \{z, z^{-1}\}$ . The only possibility is that $z^G = \{z\}$ , and z is central in G. Again, this implies that P is normal in G, a contradiction. Therefore, $p\neq 3$ .

The remaining case is when $p=2$ . In this case, $H\cong C_3$ . Let $A_1$ , $A_2$ and $A_3$ be the three Sylow 2-subgroups. Every proper subgroup of P is not one of $A_1$ , $A_2$ and $A_3$ , so is normal in G and hence contained in all of $A_1$ , $A_2$ and $A_3$ . If P were not cyclic, then each of its generators would generate a proper cyclic subgroup, and would hence be contained in $A_1$ , $A_2$ and $A_3$ . This implies $P\leq A_1 \cap A_2 \cap A_3$ , a contradiction. Therefore, P is cyclic of order $2^n$ . Write $P=\langle x\rangle $ and $H=\langle y \rangle $ . Certainly, $y^x\neq y$ ; so the only possibility is that $y^x=y^{-1}$ . Therefore,

$$\begin{align*}G=\langle x, y: x^{2^n}=1, y^3=1, yx=xy^{-1} \rangle\end{align*}$$

for some integer $n\geq 1$ . That is, $G \cong G_{n,3}$ . By Theorem 1.1, we have ${\textit {nps}}(G) = 3$ .