GROUPS WITH FEW NONPOWER SUBGROUPS

Abstract

For a group G and $m\ge 1$, let $G^m$ denote the subgroup generated by the elements $g^m$, where g runs through G. The subgroups not of the form $G^m$ are the nonpower subgroups of G. We classify the groups with at most nine nonpower subgroups.

1 Introduction

For a group G and $m\ge 1$ , the power subgroup $G^m$ is the subgroup generated by the elements $g^m$ , where g runs through G. A subgroup that is not a power subgroup is a nonpower subgroup. Let $\operatorname {\mathrm {ps}}(G)$ and $\operatorname {\mathrm {nps}}(G)$ denote the number of power and nonpower subgroups of G. It is immediate that every power subgroup is a characteristic subgroup of G. But the converse is false, as illustrated by $M_{n,p}$ defined in Section 3: it has a unique maximal noncyclic subgroup, which is characteristic but not a power subgroup.

The study of nonpower subgroups was initiated by Szász [7] who proved that G is cyclic if and only if $\operatorname {\mathrm {nps}}(G) = 0$ . The terminology ‘nonpower subgroup’ was introduced by Zhou et al. [9]. They proved that a noncyclic group G is finite if and only if $\operatorname {\mathrm {nps}}(G)$ is finite. Furthermore, if G is a finite noncyclic group, it was proved by Zhou and Ping that $\operatorname {\mathrm {nps}}(G)\ge 3$ . Therefore, from now on, we assume that all groups under consideration are finite.

For the most part, our notation follows Gorenstein [6]. In particular, $\Phi (G)$ denotes the Frattini subgroup of G, and for subgroups H and K, $[H,K]$ is generated by the commutators $[x,y] = x^{-1}y^{-1}xy$ with $x\in H$ and $y\in K$ . For a finite p-group G, let $\Omega _{i}(G)$ be the subgroup $\langle x\in G\mid x^{p^{i}}=1 \rangle $ for $i \ge 1$ . We use [6, Ch. 5] as a reference for standard results about p-groups.

Anabanti et al. [1, 2] classified the groups G with $\operatorname {\mathrm {nps}}(G)\in \{3,4\}$ and showed that, for all $k> 4$ , there are infinitely many groups G with $\operatorname {\mathrm {nps}}(G) = k$ .

The following theorem extends the classification to the groups G with $\operatorname {\mathrm {nps}}(G)\le 9$ . For completeness, we include the results for $\operatorname {\mathrm {nps}}(G)\le 4$ .

Let $C_n$ denote the cyclic group of order n and let $\operatorname {\mathrm {Alt}}(n)$ and $\operatorname {\mathrm {Sym}}(n)$ denote the alternating and symmetric groups of a set of size n. See Definition 2.5 for descriptions of the other groups referred to in the following theorem.

Theorem 1.1. For $0\le k\le 9$ , a group has exactly k nonpower subgroups if and only if, up to isomorphism, it is one of the following:

1. k = 0 a cyclic group;

2. k = 1 no examples;

3. k = 2 no examples;

4. k = 3 $C_2\times C_2$ , $Q_8$ or $G_{n,3}$ for $n\ge 1$ ;

5. k = 4 $C_3\times C_3$ ;

6. k = 5 $C_2\times C_4$ or $G_{n,5}$ for $n\ge 1$ ;

7. k = 6 $C_5\times C_5$ , $C_2\times C_2\times C_p$ , $Q_8\times C_p$ , where $p> 2$ is a prime, or $G_{n,3}\times C_q$ for ${n\ge 1}$ , where $q> 3$ is a prime;

8. k = 7 $D_8$ , $\operatorname {\mathrm {Alt}}(4)$ , $C_2\times C_8$ , $Q_{16}$ , $M_{4,2}$ , $C_3\times C_9$ , $M_{3,3}$ , $G_{n,7}$ or $F_{n,7}$ for $n\ge 1$ ;

9. k = 8 $C_7\times C_7$ or $C_3\times C_3\times C_p$ , where $p\ne 3$ is a prime;

10. k = 9 $C_2\times C_{16}$ , $M_{5,2}$ , $C_2\times C_2\times C_{p^2}$ , $Q_8\times C_{p^2}$ , where $p> 2$ is a prime, or $G_{n,3}\times C_{q^2}$ , where $q> 3$ is a prime.

2 Preliminaries

Recall that the exponent $\exp (G)$ of a finite group G is the least positive integer e such that $g^e = 1$ for all $g\in G$ . The number of positive divisors of an integer n is denoted by $\tau (n)$ .

Lemma 2.1. The power subgroups of a finite group G are the subgroups $G^d$ , where d is a divisor of the exponent of G. Thus, $\operatorname {\mathrm {ps}}(G)\le \tau (\exp (G))$ .

Proof. Given $m\ge 1$ , we prove that $G^m = G^n$ , where $n = \gcd (m,e)$ and e is the exponent of G. To this end, we may write $n = am+be$ and $m = dn$ for some integers a, b and d. Then, for all g in G, $g^m = g^{nd}\in G^n$ and $g^n = g^{am+be} = g^{ma}\in G^m$ , from which we get $G^m = G^n$ .

Lemma 2.2 [1, Lemma 3].

If A and B are finite groups such that $|A|$ and $|B|$ are coprime, then $\operatorname {\mathrm {ps}}(A\times B) = \operatorname {\mathrm {ps}}(A)\operatorname {\mathrm {ps}}(B)$ and $\operatorname {\mathrm {nps}}(A\times B) = \operatorname {\mathrm {nps}}(A)s(B) + \operatorname {\mathrm {ps}}(A)\operatorname {\mathrm {nps}}(B)$ , where $s(B)$ is the number of subgroups of B.

Corollary 2.3. For any finite abelian group G, we have $\operatorname {\mathrm {ps}}(G) = \tau (\exp (G))$ .

Proof. From Lemma 2.2, it is no loss to assume that G is an abelian p-group. Then, by Lemma 2.1, it suffices to prove that, for different divisors m and n of $\exp (G)$ , we have $G^n\neq G^m$ . Let $\exp (G) = p^e$ and $e\ge i> j\ge 0$ . Then $G^{p^i} = (G^{p^{j}})^{p^{i-j}} < G^{p^j}$ .

Lemma 2.4 [9, Lemma 2].

Suppose that N and H are subgroups of G such that $N\trianglelefteq G$ and $N\subseteq H$ . If $H/N$ is a nonpower subgroup of $G/N$ , then H is a nonpower subgroup of G. Therefore, $\operatorname {\mathrm {nps}}(G)\ge \operatorname {\mathrm {nps}}(G/N)$ .

Definition 2.5.

1. (i) For $n\ge 3$ , $\langle a,b\mid a^n = b^2 = 1,\ b^{-1}ab = a^{-1} \rangle $ is a presentation for the dihedral group $D_{2n}$ of order $2n$ .

2. (ii) For $n\ge 3$ , $\langle a,b\mid a^{2^{n-1}} = b^2 = z,\ z^2 = 1,\ b^{-1}ab = a^{-1} \rangle $ is a presentation for the generalised quaternion group $Q_{2^n}$ of order $2^n$ .

3. (iii) For $n\ge 4$ , $\langle a,b \mid a^{2^{n-1}} = b^2 = 1,\ b^{-1}ab = a^{-1+2^{n-2}} \rangle $ is a presentation for the semidihedral group $S_{2^n}$ of order $2^n$ .

4. (iv) For $n\ge 4$ when $p = 2$ and $n\ge 3$ when p is an odd prime, a presentation for the quasidihedral group $M_{n,p}$ of order $p^n$ is $\langle a,b \mid a^{p^{n-1}} = b^p = 1,\ b^{-1}ab = a^{1+p^{n-2}} \rangle $ . The group $M_{3,p}$ is the extraspecial group of order $p^3$ and exponent $p^2$ .

5. (v) For an odd prime p, $\langle x,y,z \mid x^p = y^p = z^p = 1,\ [x,z] = [y,z] = 1,\ [x,y] = z \rangle $ is a presentation for the extraspecial group $M(p)$ of order $p^3$ and exponent p.

6. (vi) For $k\ge 1$ and $n\ge 2$ , $\langle a,b\mid a^{2^n}= b^k = 1,\ a^{-1}ba = b^{-1} \rangle $ is a presentation for the group $G_{n,k}$ of order $2^n k$ . Note that $G_{1,k} = D_{2k}$ and $G_{n,2} = C_2\times C_{2^n}$ .

7. (vii) For $n\ge 1$ and a prime $p\equiv 1\pmod 3$ , choose $i\not \equiv ~1\pmod p$ such that ${i^3\equiv 1\pmod p}$ . Then $\langle a,b\mid a^{3^n} = b^p = 1,\ a^{-1}ba = b^i \rangle $ is a presentation for the group $F_{n,p}$ of order $3^np$ .

8. (viii) For $n\ge 1$ , $\langle a,b,c \mid a^{3^n} = b^2 = 1,\ bc = cb,\ b^a = c,\ c^a = bc \rangle $ is a presentation for the group $A_n = (C_2\times C_2)\rtimes C_{3^n}$ of order $2^2 3^n$ . When $n = 1$ , $A_1 = \operatorname {\mathrm {Alt}}(4)$ .

9. (ix) For $n\ge 1$ and a prime p,

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

   is a presentation for the group $B^1_{n,p}$ of order $p^{n+2}$ . Except for $B^1_{1,2} = D_8$ , it is nonmetacyclic (see [4, Lemma 2.5]). The quotient mod $\langle c \rangle $ is $C_p\times C_{p^n}$ and, for p odd, $B^1_{1,p}= M(p)$ .

10. (x) For $n\ge 1$ and a prime p,

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

    is a presentation for the group $B^2_{n,p}$ of order $p^{n+2}$ . It is metacyclic: $\langle a \rangle $ is a normal cyclic subgroup with cyclic quotient. The quotient mod $\langle c \rangle $ is $C_p\times C_{p^n}$ , $B^2_{1,2} = D_8$ and, for p odd, $B^2_{1,p}= M_{3,p}$ .

There are some basic facts about a $p'$ -group acting on a p-group in [6]. For the reader’s convenience, we give these theorems as lemmas.

Lemma 2.6 [6, Theorem 5.2.3].

If A is a $p'$ -group of automorphisms of the abelian p-group P, then $ P=C_{P}(A)\times [P,A]$ .

Lemma 2.7 [6, Theorem 5.3.5].

If A is a $p'$ -group of automorphisms of the p-group P, then $P=CH$ , where $C=C_{P}(A)$ and $H=[P,A]$ . In particular, if $H\subseteq \Phi (P)$ , then $A=1$ .

The p-groups with a cyclic maximal subgroup are well known. It is clear that $C_{p^n}$ and $C_p\times C_{p^n}$ are the only abelian groups of this type. For the nonabelian case, we have the following lemma.

Lemma 2.8 [6, Theorem 5.4.4].

Let P be a nonabelian p-group of order $p^n$ that contains a cyclic subgroup of index p. Then one of the following holds.

1. (i) p is odd and P is isomorphic to $M_{n,p}$ , for $n \ge 3$ .

2. (ii) $p = 2$ , $n = 3$ and P is isomorphic to $D_8$ or $Q_8$ .

3. (iii) $p = 2$ , $n> 3$ and P is isomorphic to $M_{n,2}$ , $D_{2^n}$ , $Q_{2^n}$ or $S_{2^n}$ .

The p-groups in Lemma 2.8 are well studied. We collect some basic facts in the following lemmas.

Lemma 2.9 [6, Theorem 5.4.3].

For $P=M_{n,p}$ :

1. (i) $P'=\langle a^{p^{n-1}} \rangle $ ;

2. (ii) $\Phi (P)=Z(P)=\langle a^{p} \rangle $ ; and

3. (iii) $\Omega _{i}(P)= \langle a^{p^{n-i-1}},b \rangle $ is abelian of type ( $p^{i},p$ ), $1\leq i \leq n-2$ .

Lemma 2.10 [6, Theorem 5.4.5].

Let P be a nonabelian $2$ -group of order $2^{n}$ in which $|P/P'|=4$ . Then P is isomorphic to $D_{2^n}$ , $Q_{2^n}$ or $S_{2^n}$ .

Lemma 2.11 [6, Theorem 5.5.1].

A nonabelian p-group P of order $p^3$ is extraspecial and is isomorphic to one of the groups $M_{3,p}$ , $M(p)$ , $D_{8}$ or $Q_{8}$ .

Theorem 2.12. There is no finite p-group G such that $G/N \simeq M_{n,p}$ , where N is a central subgroup of G of order p contained in $G'$ .

Proof. From the presentation of $M_{n,p}$ in Definition 2.5, we may suppose that G has a presentation of the form

$$ \begin{align*} \langle a,b,c \mid a^{p^{n-1}} = c^i, b^p = c^j,\ c^p = 1,\ b^{-1}ab = a^{1+p^{n-2}}c^k, ac = ca,\ bc = cb \rangle, \end{align*} $$

where $N = \langle c \rangle $ , $0\le i,j,k < p$ and not all i, j, k are zero. Since $c\in Z(G)$ , it is clear that $a^p$ commutes with b and hence $z = a^{p^{n-2}}\in Z(G)$ . Therefore, $[a,b] = zc^k \in Z(G)$ and it follows from elementary properties of commutators that $G' = \langle zc^k \rangle $ , which is a contradiction.

From Lemmas 2.2, 2.5 and Theorem 2.3 in [4], we deduce the following lemma.

Lemma 2.13. For a nonabelian p-group G generated by two elements, let $R=\Phi (G')G_{3}$ , where $G_{3}=[[G,G],G]$ . Then:

1. (i) R is the only maximal subgroup of $G'$ that is normal in G;

2. (ii) G is metacyclic if and only if $G/R$ is metacyclic; and

3. (iii) if the type of $G/G'$ is $(p,p^{n})$ and $G/R$ has no cyclic maximal subgroup, then $G/R$ is isomorphic to $B_{n,p}^{1}$ or $B_{n,p}^{2}$ .

Proof. (i) and (ii) are the statements of Lemma 2.2 and Theorem 2.3 in [4].

(iii) Let $H=G/R$ . Since $H'=(G/R)'=G'R/R=G'/R$ , we have $H/H'\simeq G/G'$ and $H'\subseteq Z(H)$ . Thus, we may assume that $H/H'=\langle aH',bH' \rangle $ . Then $H=\langle a,b \rangle $ and ${H'=\langle c \rangle }$ , where $c=[a,b]$ and $|c|=p$ . Thus, $c\in Z(H)$ . Since the type of $H/H'$ is $(p,p^{n})$ , we have $a^p=c^i$ , $b^{p^{n}}=c^j$ for suitable integers i, j. If $p\nmid j$ , then $\langle b \rangle $ is a cyclic maximal subgroup of $G/R$ , contrary to our assumption. Thus, $b^{p^{n}}=1$ . If $p\nmid i$ , replacing c with $c^{i}$ , we have $H\simeq B_{n,p}^{2}$ . If $p\mid i$ , then $H\simeq B_{n,p}^{1}$ .

3 A catalogue of nonpower values

The value of $\operatorname {\mathrm {nps}}(G)$ for the groups that occur in the proof of Theorem 1.1 can be computed from their presentation or from the Small Groups Database using the computer algebra system Magma [5]. For ease of reference, we include some general formulas here.

Proposition 3.1. For an integer n and a prime p we have:

1. (i) for $n\ge 3$ , $\operatorname {\mathrm {nps}}(D_{2^n}) = 2^n - 1$ ;

2. (ii) for $n\ge 3$ , $\operatorname {\mathrm {nps}}(Q_{2^n}) = 2^{n-1} - 1$ ;

3. (iii) for $n\ge 4$ , $\operatorname {\mathrm {nps}}(S_{2^n}) = 3\cdot 2^{n-2} - 1$ ;

4. (iv) for $n\ge 3$ , $\operatorname {\mathrm {nps}}(M_{n,p}) = p(n-1) + 1$ (when $p = 2$ , assume that $n\ge 4$ );

5. (v) $\operatorname {\mathrm {nps}}(M(p)) = p^2 + 2p + 2$ ;

6. (vi) if $p> 2$ , then $\operatorname {\mathrm {nps}}(G_{n,p^k}) = p(p^k-1)/(p-1)$ ;

7. (vii) if $p\equiv 1\pmod 3$ , then $\operatorname {\mathrm {nps}}(F_{n,p}) = p$ ;

8. (viii) for $n\ge 1$ , $\operatorname {\mathrm {nps}}(A_n) = 3n + 4$ ;

9. (ix) for $n\ge 1$ , $\operatorname {\mathrm {nps}}(B^1_{n,p})\ge 17$ except that $\operatorname {\mathrm {nps}}(B^1_{1,2}) = 7$ ;

10. (x) for $n\ge 1$ , $\operatorname {\mathrm {nps}}(B^2_{n,p})\ge 11$ except that $\operatorname {\mathrm {nps}}(B^2_{1,2}) = \operatorname {\mathrm {nps}}(B^2_{1,3}) = 7$ .

Proof. (i), (ii) and (iii) follow from Proposition 11 and Theorems 16 and 17 of [1].

(iv) Suppose $G = M_{n,p}$ . From Lemma 2.9, $Z(G) = \Phi (G) = G^p = \langle a^p \rangle $ and $\Omega _i(G) = \langle a^{p^{n-i-1}},b \rangle $ . The commutator $c = [a,b] = a^{p^{n-2}}$ has order p, $[a^i,b] = c^i$ , $G' = \langle c \rangle $ and $(ba^j)^p = c^{jp(p-1)/2} a^{jp}$ .

Therefore, for $1\le i\le n-1$ , we have $G^{p^i} = \langle a^{p^i} \rangle $ and so $\operatorname {\mathrm {ps}}(G) = n$ . The maximal subgroups of G are the cyclic subgroups $\langle a \rangle $ and $\langle a^ib \rangle $ , $1\le i< p$ , and the noncyclic subgroup $\langle a^p, b \rangle $ . Therefore, the proper noncyclic subgroups of G are the abelian groups $\Omega _i(G)$ of type $(p^i,p)$ , $1\le i\le n-2$ . Thus, G has $n-1$ noncyclic subgroups each of which, except $\Omega _1(G)$ , has p maximal cyclic subgroups. There are $p+1$ cyclic subgroups in $\Omega _1(G)$ . Therefore, $s(G) = (n-1) + p(n-2) + (p+1) + 1 = p(n-1)+n+1$ and $\operatorname {\mathrm {nps}}(G) = p(n-1)+1$ .

(v) The exponent of $M(p)$ is p; therefore, it has $(p^3-1)/(p-1)$ subgroups of order p. Every subgroup of order $p^2$ is normal and hence contains the centre (of order p). Therefore, there are $(p^3-p)/(p^2-p)$ subgroups of order $p^2$ . In total there are ${p^2+2p+2}$ proper subgroups all of which are nonpower subgroups.

(vi), (vii) We only prove that $\operatorname {\mathrm {nps}}(G_{n,p^k}) = p(p^k-1)/(p-1)$ . Then $\operatorname {\mathrm {nps}}(F_{n,p})=p$ is obtained similarly.

Recall that $G_{n,p^k}=\langle a,b\mid a^{2^{n}}= b^{p^{k}} = 1,\ a^{-1}ba = b^{-1} \rangle $ . Thus, $\langle a^2 \rangle \subseteq Z(G_{n,p^k})$ . Then we have $G_{n,p^k}/\langle a^2 \rangle \simeq D_{2p^k}$ . Since the number of Sylow $2$ -subgroups of $D_{2p^k}$ is $p^k$ , the number of Sylow $2$ -subgroups of $G_{n,p^k}$ is also $p^k$ . Thus, the Sylow $2$ -subgroups of $G_{n,p^k}$ are self-normalising. For $0\leq j\leq k-1$ , $\langle b^{p^{k-j}} \rangle $ is the unique subgroup of order $p^j$ of $G_{n,p^k}$ . Let $H_{j}$ be a subgroup of $G_{n,p^k}$ and $|H_{j}|=2^{n}p^j$ . Then the Sylow $2$ -subgroups of $H_{j}$ are self-normalising. Thus, every subgroup of order $2^{n}p^j$ in $G_{n,p^k}$ contains $p^j$ Sylow $2$ -subgroups of $G_{n,p^k}$ . Thus, there are exactly $p^{k-j}$ subgroups of order $2^{n}p^j$ in $G_{n,p^k}$ and they are conjugate to each other in $G_{n,p^k}$ . The number of those subgroups is ${p(p^k-1)/(p-1)}$ . Now we prove that the other subgroups of $G_{n,p^k}$ are power subgroups. Since $\langle a^2 \rangle \subseteq Z(G_{n,p^k})$ , for $0\leq i\leq n-1$ and $0\leq s \leq k$ , $\langle a^{2^{n-i}}, b^{p^{k-s}} \rangle $ is the unique subgroup of order $2^{i}p^{s}$ in $G_{n,p^k}$ and $\langle a^{2^{n-i}}, b^{p^{k-s}} \rangle = G_{n,p^k}^{2^{n-i}p^{k-s}}$ . This completes the proof.

(viii) The power subgroups of $A_n$ are distinct except that $A_n^2 = A_n^1$ . The $2n - 1$ subgroups Q, $\langle a^{3^i} \rangle $ and $Q\langle a^{3^i} \rangle $ , where $1\le i < n$ and $Q = \langle b,c \rangle $ , are the proper nontrivial normal subgroups of $A_n$ . The other subgroups are $\langle a \rangle $ , $\langle a^{3^i}b \rangle $ , $\langle a^{3^i}c \rangle $ , $\langle a^{3^i}bc \rangle $ for $0\le i\le n$ . Thus, $A_n$ has $5(n+1)$ subgroups; therefore, $\operatorname {\mathrm {nps}}(A_n) = 3n+4$ .

(ix) There are n proper power subgroups of $B^1_{n,p}$ ; their orders are $p^i$ for $0\le i < n$ . We claim that $\operatorname {\mathrm {nps}}(B_{n,p}^1)$ is an increasing function of n and p by counting the subgroups. Since $B_{1,p}^1$ is well known, we only consider the case $n\ge 2$ . We count subgroups by considering their exponent. First, notice that $\Omega _{n-1}(B_{n,p}^1)=\langle a \rangle \times \langle b^{p} \rangle \times \langle c \rangle \simeq C_p \times C_{p^{n-1}} \times C_p$ , which implies that all the subgroups of exponent $\le p^{n-1}$ are in $\Omega _{n-1}(B_{n,p}^1)$ . Thus, the number of subgroups with exponent $\le p^{n-1}$ is $s(C_p\times C_{p^{n-1}} \times C_p)$ , which is an increasing function of n and p. Next, we consider the subgroups of exponent $p^n$ . Considering the number of elements of order $p^n$ , we see that there are $p^2$ cyclic subgroups of order $p^n$ . Let H be a subgroup of order $p^{n+1}$ with $\exp (H)=p^n$ . Then $c \in H$ and $\exp (H/\langle c \rangle )=p^n$ . Since $B_{n,p}^1/\langle c \rangle \simeq C_p\times C_{p^n}$ , we see that the number of such subgroups H is p. Therefore, the number of subgroups of exponent $p^n$ is $p^2+p+1$ . Thus, $\operatorname {\mathrm {nps}}(B_{n,p}^1)$ is an increasing function of n and p, as claimed. By direct calculation or from Magma [5], we find that $\operatorname {\mathrm {nps}}(B^1_{2,2}) = 20$ and $\operatorname {\mathrm {nps}}(B^1_{1,3}) = \operatorname {\mathrm {nps}}(M(3)) = 17$ .

(x) There are n proper power subgroups of $B_{n,p}^2$ ; their orders are $p^i$ for $0\le i\le n-2$ and $p^n$ . Similarly, $\operatorname {\mathrm {nps}}(B^2_{n,p})$ is an increasing function of n and p. By direct calculation or from Magma [5], we find that $\operatorname {\mathrm {nps}}(B^2_{2,2}) = 12$ , $\operatorname {\mathrm {nps}}(B^2_{2,3}) = 20$ and $\operatorname {\mathrm {nps}}(B^2_{1,5}) =\operatorname {\mathrm {nps}}(M_{3,5})= 11$ .

Remark 3.2. It can be shown that:

1. (i) for $n\ge 2$ , $\operatorname {\mathrm {nps}}(B^1_{n,p}) = p^2(2n-1) + p(n+1) + 2$ ; and

2. (ii) for $n\ge 2$ , $\operatorname {\mathrm {nps}}(B^2_{n,p}) = (p+1)(2 + p(n-1))$ .

Lemma 3.3. Let $G = D_{2p}\times \underbrace {C_2\times \cdots \times C_2}_n$ . For $n \ge 1$ and a prime $p> 2$ , we have $\operatorname {\mathrm {nps}}(G)\ge 3p+4$ and equality holds when $n = 1$ .

Proof. For all $n\ge 1$ , the exponent of G is $2p$ and the only proper nontrivial power subgroup is $G^2$ of order p. The group $D_{2p}\times C_2$ has $3p+7$ subgroups; therefore, G has at least $3p+4$ nonpower subgroups.

Lemma 3.4. Let $X_{n,p} = D_{2p}\times \underbrace {C_3\times \cdots \times C_3}_n$ . For $n \ge 1$ and a prime $p> 3$ , we have $\operatorname {\mathrm {nps}}(X_{n,p})=(p+3)s(C_{3}^{n})-6\geq 10$ , where $\underbrace {C_3\times \cdots \times C_3}_n$ is denoted by $C_3^n$ for short. For $n \ge 1$ , $\operatorname {\mathrm {nps}}(X_{n,3}) \geq \operatorname {\mathrm {nps}}(X_{1,3})=10$ .

Proof. By simple calculation, $s(D_{2p})=p+3$ and $\operatorname {\mathrm {ps}}(D_{2p}) = 3$ . Let $p>3$ . Then, from Lemma 2.2, $\operatorname {\mathrm {nps}}(X_{n,p})=\operatorname {\mathrm {nps}}(D_{2p})s(C_{3}^{n})+ps(D_{2p})\operatorname {\mathrm {nps}}(C_{3}^{n})= \operatorname {\mathrm {nps}}(D_{2p})s(C_{3}^{n})+ps(D_{2p})(s(C_{3}^{n})-2)=(p+3)s(C_{3}^{n})-6$ . In particular, $\operatorname {\mathrm {nps}}(X_{1,5})=10$ . For any $n\geq 1$ , $\operatorname {\mathrm {ps}}(X_{n,3})=3$ . Thus, $\operatorname {\mathrm {nps}}(X_{n,3})$ is an increasing function for n. Then a straightforward calculation shows that $\operatorname {\mathrm {nps}}(X_{1,3}) = 10$ , and this completes the proof.

Lemma 3.5 [8, Theorem 3.3].

For $n_{2}\ge n_{1}\ge 1$ and a prime p, the total number of subgroups of $C_{p^{n_{1}}}\times C_{p^{n_{2}}}$ is

$$ \begin{align*} \dfrac{(n_{2}-n_{1}+1)p^{n_{1}+2}-(n_{2}-n_{1}-1)p^{n_{1}+1}-(n_{2}+n_{1}+3)p+(n_{2}+n_{1}+1)}{(p-1)^{2}}. \end{align*} $$

Lemma 3.6. For $n_{2}\ge n_{1}\ge 1$ and a prime p, the value of $\operatorname {\mathrm {nps}}(C_{p^{n_{1}}}\times C_{p^{n_{2}}})$ is

$$ \begin{align*} \frac{(n_{2}-n_{1}+1)p^{n_{1}+2}-(n_{2}-n_{1}-1)p^{n_{1}+1}-(n_{2}+n_{1}+3)p+(n_{2}+n_{1}+1)}{(p-1)^{2}}-(n_{2}+1). \end{align*} $$

Proof. This follows from Lemma 3.5 and Corollary 2.3.

Example 3.7. We have $\operatorname {\mathrm {nps}}(C_p\times C_{p^n}) = pn+1$ and $\operatorname {\mathrm {nps}}(C_{p^2}\times C_{p^2}) = (p+1)(p+2)$ .

4 The groups with at most nine nonpower subgroups

In proving Theorem 1.1, we use Theorem 1.3 of [2] and the theorems of [7] and [9], which we summarise in the following lemma.

Lemma 4.1. For a finite group G:

1. (1) G is cyclic if and only if $\operatorname {\mathrm {nps}}(G) = 0$ ;

2. (2) if G is noncyclic, then $\operatorname {\mathrm {nps}}(G) \ge 3$ ;

3. (3) if $\operatorname {\mathrm {nps}}(G) = 3$ , then G is $C_2\times C_2$ , $Q_8$ or $G_{n,3}$ for $n\ge 1$ ; and

4. (4) if $\operatorname {\mathrm {nps}}(G) = 4$ , then G is $C_3\times C_3$ .

Let $\operatorname {\mathrm {Syl}}_p(G)$ denote the set of Sylow subgroups of G. Recall that $P\in \operatorname {\mathrm {Syl}}_p(G)$ has $|G:N_G(P)|$ conjugates and we have $|G:N_G(P)|\equiv 1\pmod p$ . Moreover, $N_G(P)$ is self-normalising; therefore, it also has $|G:N_G(P)|$ conjugates.

Lemma 4.2. Let G be a finite group and let P be a Sylow p-subgroup of G such that $P\ne N_G(P)\ne G$ . Then $\operatorname {\mathrm {nps}}(G)\le 9$ if and only if, for some $n\ge 1$ and a prime $q> 3$ :

1. (1) $\operatorname {\mathrm {nps}}(G) = 6$ , $p = 2$ and $G\simeq G_{n,3}\times C_q$ ; or

2. (2) $\operatorname {\mathrm {nps}}(G) = 9$ , $p = 2$ and $G\simeq G_{n,3}\times C_{q^2}$ .

Proof. Since $P\ne N_G(P)\ne G$ , both P and $N_G(P)$ have at least $p+1$ conjugates and so $2(p+1)\le \operatorname {\mathrm {nps}}(G)\le 9$ , from which we get p is 2 or 3. If $Q\in \operatorname {\mathrm {Syl}}_q(G)$ for some prime $q\ne p$ and $N_G(Q)\ne G$ , then $2(p+1)+q+1\le \operatorname {\mathrm {nps}}(G)\le 9$ , which is impossible. Therefore, for $q\ne p$ , all Sylow q-subgroups are normal. Thus, $G = NP$ , where N is a nilpotent normal subgroup such that $N\cap P = 1$ . Then $[N_N(P),P]\subseteq N\cap P = 1$ and, consequently, $N_G(P) = PC_N(P)$ . If $|G : N_G(P)|> p+1$ , then $2(2p+1) \le 9$ , which is a contradiction. Thus, $|G:N_G(P)| = p + 1$ .

If $p = 3$ , then $\operatorname {\mathrm {nps}}(G)$ is 8 or 9. Thus, for $Q\in \operatorname {\mathrm {Syl}}_2(G)$ , we have $Q\trianglelefteq G$ and ${G = N_G(P)Q}$ . If $[P,Q]\subseteq \Phi (Q)$ , it follows from Lemma 2.7 that $[P,Q] = 1$ . But then ${Q\subseteq C_G(P)\subseteq N_G(P)}$ , which is a contradiction. Thus, Q has at least three subgroups that are not normal in G. Hence, G has at least 11 nonpower subgroups, contrary to $\operatorname {\mathrm {nps}}(G)\le 9$ . Therefore, for the remainder of the proof, we take $p = 2$ .

If P is not cyclic, it follows from Lemma 4.1 that $\operatorname {\mathrm {nps}}(P)\ge 3$ . Then, in addition to the three conjugates of P and the three conjugates of $N_G(P)$ , there would be at least three nonpower subgroups H such that $H/N$ is a nonpower subgroup of $G/N\simeq P$ . In this case, $\operatorname {\mathrm {nps}}(P) = 3$ , $\operatorname {\mathrm {nps}}(G) = 9$ and P is either $C_2\times C_2$ or $Q_8$ . But $\operatorname {\mathrm {nps}}(G) = 9$ implies that the proper subgroups of P are normal in G, which is a contradiction since P is generated by its proper subgroups. Therefore, P is cyclic.

Let $R\in \operatorname {\mathrm {Syl}}_r(N)$ . Then R acts by conjugation on the three conjugates of P. If $r> 3$ , then $R\subseteq N_G(P)$ , from which we get $R\subseteq N_G(P)\cap N\subseteq C_G(P)$ . For $Q\in \operatorname {\mathrm {Syl}}_3(G)$ , we have $[R,Q]\subseteq R\cap Q = 1$ and so $R\subseteq C_G(Q)$ . Consequently, $G = PQ\times A$ , where A is a nilpotent group whose order is not divisible by two or three. From Lemma 2.2,

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

It follows from $\operatorname {\mathrm {nps}}(G)\le 9$ that $\operatorname {\mathrm {nps}}(A) = 0$ and so A is cyclic. Furthermore, $s(A)\le 3$ ; therefore, A is either trivial or a cyclic group of order r or $r^2$ for some prime $r> 3$ .

The permutation action of G on the conjugates of P defines a homomorphism $G\to \operatorname {\mathrm {Sym}}(3)$ with kernel $K = \bigcap _{g\in G}N_G(P)^g$ . Since $|Q : C_Q(P)| = 3$ , we have $\Phi (Q)\subseteq C_Q(P)\subseteq K$ . Therefore, $M = (P\cap K)\times \Phi (Q)\times A$ is a normal subgroup of G. The group P acts on the elementary abelian group $\overline Q = Q/\Phi (Q)$ and it follows from Lemma 2.6 that $\overline Q = C_{\overline Q}(P)\times [P,\overline Q]$ . Thus, $[P,\overline Q] \simeq C_3$ and

$$ \begin{align*} G/M\simeq \operatorname{\mathrm{Sym}}(3)\times\underbrace{C_3\times\cdots\times C_3}_n. \end{align*} $$

If $n \ge 1$ , it follows from Lemmas 3.4 and 2.4 that $\operatorname {\mathrm {nps}}(G)\ge 10$ , contrary to assumption. Thus, $Q/\Phi (Q) \simeq C_3$ ; therefore, Q is cyclic. But now $Q = [P,Q]\times C_Q(P)$ ; therefore, $C_Q(P) = 1$ and $|Q| = 3$ .

Let a be a generator of P and let b a generator of Q. Then $a^{-1}ba = b^{-1}$ and so $QP\simeq G_{n,3}$ for some n. The assumption $N_G(P)\ne P$ implies that $A\ne 1$ . Thus, G is either $G_{n,3}\times C_r$ or $G_{n,3}\times C_{r^2}$ for some prime $r> 3$ .

Remark 4.3. The group $G = \operatorname {\mathrm {Sym}}(3)\times C_3$ satisfies the hypothesis of the lemma (with $p=2$ ) except that $\operatorname {\mathrm {nps}}(G) = 10$ .

Lemma 4.4. Let G be a finite group and let P be a Sylow p-subgroup of G such that $P = N_G(P)\ne G$ . Then $\operatorname {\mathrm {nps}}(G)\le 9$ if and only if, for some $n\ge 1$ , one of the following holds.

1. (1) $\operatorname {\mathrm {nps}}(G) = 3$ , $p = 2$ and $G\simeq G_{n,3}$ .

2. (2) $\operatorname {\mathrm {nps}}(G) = 5$ , $p = 2$ and $G\simeq G_{n,5}$ .

3. (3) $\operatorname {\mathrm {nps}}(G) = 7$ , $p = 2$ and $G\simeq G_{n,7}$ .

4. (4) $\operatorname {\mathrm {nps}}(G) = 7$ , $p = 3$ and $G\simeq F_{n,7}$ .

5. (5) $\operatorname {\mathrm {nps}}(G) = 7$ , $p = 3$ and $G\simeq \operatorname {\mathrm {Alt}}(4)$ .

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 $\operatorname {\mathrm {nps}}(G)\le 9$ implies that $p\in \{2,3,5,7\}$ .

If $p = 2$ , then $m\in \{3,5,7,9\}$ ; if $p = 3$ , then $m\in \{4,7\}$ ; if $p = 7$ , then $m = 8$ . However, if $p = 5$ , then $m = 6$ and $G$ is a group of twice odd order. It is an elementary fact that a group of twice odd order has a subgroup H of index two, which, in this case, contains P. Then $|H:P| = 3$ , which is impossible. 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 $|Q| = m$ .

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 a proper subgroup of P.

If $N_G(Q)\ne G$ , then Q has at least $q+1$ conjugates; therefore, $m+q+1\le 9$ . In this case, either $p = 2$ and $|Q| = 3$ or $p = 3$ and $|Q| = 4$ . Furthermore, we must have $Q = N_G(Q)$ . Otherwise, both Q and $N_G(Q)$ would have at least $q+1$ conjugates. From the structure of $\operatorname {\mathrm {Sym}}(3)$ and $\operatorname {\mathrm {Sym}}(4)$ , we have $KQ\trianglelefteq G$ and, by the Frattini argument [6, Theorem 1.3.7], $G = KN_G(Q)$ and $Q\ne N_G(Q)$ , which is a contradiction. Therefore, $Q\trianglelefteq G$ and $[K,Q]\subseteq K\cap Q = 1$ and hence $K\subseteq C_G(Q)$ .

The order of Q is either q or $q^2$ ; therefore, Q is abelian.

Case 1: $p = 2$ and $m\in \{3,5,7,9\}$ . 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, $|Q| = 3$ and $G/K\simeq \operatorname {\mathrm {Sym}}(3)$ . Since ${|P/K| = 2}$ , we have $\Phi (P)\subseteq K$ and so $\Phi (P)\trianglelefteq G$ . If P is not cyclic, then $G/\Phi (P)\simeq \operatorname {\mathrm {Sym}}(3)\times C_2\times \cdots \times C_2$ and it follows from Lemmas 2.4 and 3.3 that $\operatorname {\mathrm {nps}}(G)\ge 13$ , contrary to assumption. Thus, P is cyclic. If a generates P and b generates Q, then $b^a = b^{-1}$ and $G\simeq G_{n,3}$ .

Case 1b: $p = 2$ and $|Q| = 5$ . In this case, $|Q| = 5$ and $QK/K$ is a normal subgroup of $G/K$ . Therefore, $|G/K|$ is either 10 or 20. If $|G/K| = 20$ , then $G/K = \langle x, b\mid x^4 = b^5 =1, x^{-1}bx = b^2 \rangle $ . This group has 14 subgroups, 4 of which are power subgroups; therefore, there are 10 nonpower subgroups. From Lemma 2.4, $\operatorname {\mathrm {nps}}(G)\ge 10$ , contrary to our assumption. Thus, $|G/K| = 10$ . If P is not cyclic, then $G/\Phi (P)\simeq D_{10}\times C_2\times \cdots \times C_2$ and, using Lemma 3.3, we arrive at a contradiction, as in Case 1a. Thus, P is cyclic. If a generates P and b generates Q, then $b^a = b^{-1}$ and so $G\simeq G_{n,5}$ .

Case 1c: $p = 2$ and $|Q| = 7$ . In this case, $|Q| = 7$ , $G/K\simeq D_{14}$ and P has seven conjugates. If P is not cyclic, then $\operatorname {\mathrm {nps}}(P)\ge 3$ and it follows from Lemma 2.4 that $\operatorname {\mathrm {nps}}(G)\ge 10$ , contrary to assumption. Thus, P is cyclic and $G\simeq G_{n,7}$ .

Case 1d: $p = 2$ and $|Q| = 9$ . Since $\operatorname {\mathrm {nps}}(G) \le 9$ , all subgroups are normal except the Sylow 2-subgroups. In particular, if R is a subgroup of Q of order three, then  ${R\trianglelefteq G}$ ; therefore, $RP\trianglelefteq G$ . But then P is not maximal. Thus, all maximal subgroups are normal. This implies that G is nilpotent and this contradiction shows that there are no examples in this case.

Case 2: $p = 3$ . In this case, $G = Q\rtimes P$ , where P is a 3-group and $|Q| \in \{4,7\}$ .

Case 2a: $p = 3$ and $|Q| = 4$ . We must have $Q\simeq C_2\times C_2$ , otherwise $Q\simeq C_4$ and then $Q\subseteq C_G(P)$ , which contradicts the assumption that $P = N_G(P)$ .

It follows that Q has three subgroups $R_1$ , $R_2$ and $R_3$ of order two and P acts transitively on them. Then $R_1$ , $R_2$ , $R_3$ and the four conjugates of P are not normal in G, from which we get $\operatorname {\mathrm {nps}}(G)\ge 7$ . Let K be the kernel of the action of P on $\{R_1, R_2, R_3\}$ . Then $K = C_P(Q)$ , $|P : K| = 3$ and so $K\trianglelefteq G$ . Then $R_1K$ , $R_2K$ and $R_3K$ are permuted by P and, if $K\ne 1$ , we would have $\operatorname {\mathrm {nps}}(G)\ge 10$ . Therefore, $K = 1$ , $|P| = 3$ and so $G\simeq \operatorname {\mathrm {Alt}}(4)$ .

Case 2b: $p = 3$ and $|Q| = 7$ . For $Q\in \operatorname {\mathrm {Syl}}_7(G)$ , we have $|Q| = 7$ and P is cyclic; otherwise, $\operatorname {\mathrm {nps}}(P) \ge 3$ and we obtain a contradiction by applying Lemma 2.4 to $G/Q$ . The image of the homomorphism $G\to \operatorname {\mathrm {Sym}}(7)$ is the group $F_{1,7}$ of order $21$ . We may write $P = \langle a \rangle $ and $Q = \langle b \rangle $ , where $a^{3^n} = 1$ , $b^7 = 1$ and $b^a = b^2$ . Thus, $G\simeq F_{n,7}$ .

Case 3: $p = 7$ and $|Q| = 8$ . All subgroups are normal except the Sylow 7-subgroups. As in Case 1d, we see that all maximal subgroups of G are normal and so G is nilpotent, which is a contradiction.

Remark 4.5. For $n \ge 2$ , the groups $\langle x, b\mid x^{2^n} = b^5 =1, x^{-1}bx = b^2 \rangle $ have exactly 10 nonpower subgroups and their Sylow 2-subgroups are self-normalising.

Proof of Theorem 1.1.

From Lemmas 4.2 and 4.4, we may suppose that G is nilpotent. If P is a noncyclic p-group, then $\operatorname {\mathrm {nps}}(P)\ge 3$ . Suppose that $H\ne 1$ is a group whose order is not divisible by p. If $\operatorname {\mathrm {nps}}(P\times H)\le 9$ , then, from Lemma 2.2, there are two possibilities: (i) $\operatorname {\mathrm {nps}}(P) = 3$ and H is a cyclic group whose order is a prime or the square of a prime; (ii) $\operatorname {\mathrm {nps}}(P) = 4$ and H is a cyclic group of prime order.

From Lemma 4.1, if $\operatorname {\mathrm {nps}}(P) = 3$ , then G is $C_2\times C_2\times C_r$ , $C_2\times C_2\times C_{r^2}$ , $Q_8\times C_r$ or $Q_8\times C_{r^2}$ , where $r> 2$ is a prime. If $\operatorname {\mathrm {nps}}(P) = 4$ , then G is $C_3\times C_3\times C_r$ , where $r\ne 3$ is a prime.

Thus, from now on, we may suppose that G is a noncyclic p-group. Then $G/\Phi (G)$ is an elementary abelian group of order $p^d$ . The proper subgroups of $G/\Phi (G)$ are nonpower subgroups; therefore, $\operatorname {\mathrm {nps}}(G)\le 9$ implies that $d = 2$ and $p\in \{2,3,5,7\}$ .

The group G can be generated by two elements; therefore, $G/G' = C_{p^m}\times C_{p^n}$ for some $m\le n$ . It follows from Example 3.7 and Lemma 2.4 that $m = 1$ . Thus, $G/G'$ is one of $C_p\times C_p$ for $p\in \{2,3,5,7\}$ , $C_2\times C_{2^n}$ for $n\in \{2,3,4\}$ or $C_3\times C_9$ . If G is abelian, this completes the proof. From now on, we assume that $G'\ne 1$ .

Suppose that $G/G'\simeq C_2\times C_2$ . It follows from Lemma 2.10 that G is isomorphic to $D_{2^n}$ , $S_{2^n}$ or $Q_{2^n}$ . From Proposition 3.1, the only possibilities are $D_8$ , $Q_8$ and $Q_{16}$ .

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 $G/G'$ .

Suppose that p is odd and $G/G'\simeq C_p\times C_p$ . From Lemma 2.11, $G/R$ is an extraspecial group of order $p^3$ : that is, $M_{3,p}$ or $M(p)$ . From Proposition 3.1(iv) and (v), $\operatorname {\mathrm {nps}}(M_{3,p}) = 2p+1$ and $\operatorname {\mathrm {nps}}(M(p)) = p^2 + 2p + 2$ . Thus, $p = 3$ and $G/R\simeq M_{3,3}$ . The group $M_{3,3}$ has a cyclic subgroup of order nine; therefore, it is metacyclic. It follows from Lemma 2.13 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 three. Therefore, by Lemma 2.8, $G\simeq M_{n,3}$ . (This argument is based on the MathSciNet review of [3] by Marty Isaacs.) But, from Lemma 2.9, if $M = M_{n,p}$ , then $M'$ is its unique normal subgroup of order p and $M/M' \simeq C_p\times C_{p^{n-2}}$ . Thus, $R = 1$ and $G\simeq M_{3,3}$ .

Suppose that $G/G'\simeq C_2\times C_{2^n}$ ( $n = 2,3,4$ ) or $C_3\times C_9$ . If $G/R$ has a cyclic subgroup of prime index, it follows from Lemma 2.8 that $G/R$ is isomorphic to $M_{n+2,2}$ or $M_{4,3}$ . The assumption that $\operatorname {\mathrm {nps}}(G)\le 9$ excludes $M_{6,2}$ and $M_{4,3}$ . Then, from Theorem 2.12, $R = 1$ and hence G is isomorphic to $M_{4,2}$ or $M_{5,2}$ .

We may suppose that the exponent of $G/R$ is $2^n$ or 9. Lemma 2.13 shows that $G/R$ is either $B^1_{n,p}$ or $B^2_{n,p}$ for $p\in \{2,3\}$ and $n\ge 2$ . Proposition 3.1(ix) and (x) shows that none of these groups satisfy our assumptions. This completes the proof.