A question of Zhou, Shi and Duan on nonpower subgroups of finite groups

Abstract A subgroup H of a group G is called a power subgroup of G if there exists a non-negative integer m such that H = ⟨gm : g ∈ G⟩. Any subgroup of G which is not a power subgroup is called a nonpower subgroup of G. Zhou, Shi and Duan, in a 2006 paper, asked whether for every integer k (k ≥ 3), there exist groups possessing exactly k nonpower subgroups. We answer this question in the affirmative by giving an explicit construction that leads to at least one group with exactly k nonpower subgroups, for all k ≥ 3, and in_nitely many such groups when k is composite and greater than 4. Moreover, we describe the number of nonpower subgroups for the cases of elementary abelian groups, dihedral groups, and 2-groups of maximal class.

normal subgroups. A subgroup of G which is not a power subgroup is called a nonpower subgroup of G.
Let k be the number of nonpower subgroups of a group G. The authors (Zhou, Shi and Duan) of [4] proved the following: (a) k ∈ (0, ∞) if and only if G is a finite noncyclic group; (b) k = 0 if and only if G is a cyclic group; (c) k = ∞ if and only if G is an infinite noncyclic group.
They also remarked that neither k = 1 nor k = 2 is possible in any group. With respect to the case k ≥ 3, they asked (see [4,Problem]): In this paper, we show that the answer to this question is yes. In fact, we prove that there is at least one group possessing exactly k nonpower subgroups for each k ≥ 3 (see Theorem 5). Our method of proof also shows that there are infinitely many such groups for each k > 4 and k not prime. The constructions we used are given in Section 2; part of it involves the direct product of a dihedral group with a carefully chosen cyclic group.
There are further questions one could ask. For example, given a positive integer n, what is the maximum number of nonpower subgroups in a group of order n? To supply further examples of the possible numbers of nonpower subgroups in a group of a given order, we also explore in Section 3 some special cases: elementary abelian p-groups, dihedral groups, and 2-groups of maximal class. For example, we observe (see Corollary 10) that the elementary abelian p-group C p × C p (p prime) contains exactly p + 1 nonpower subgroups, and the generalised quaternion group Q 2 n (where n ≥ 3) contains exactly 2 n−1 − 1 nonpower subgroups (see Theorem 16). All the groups studied here are finite.
We end this introductory section by briefly establishing the notation we will use. For a positive integer n, we write C n for the cyclic group of order n, with D 2n being the dihedral group of order 2n.
Notation. Let G be a group. We write s(G) for the total number of subgroups in G. Also, we write ps(G) for the number of power subgroups, and nps(G) for the number of non-power subgroups. For example, in C 2 × C 2 we have s(G) = 5, ps(G) = 2 and nps(G) = 3.

Groups with exactly k nonpower subgroups.
In this section, we give constructions that supply, for each k ≥ 3, at least one finite group containing exactly k nonpower subgroups. Moreover, for k ̸ = 4 and k not prime, our constructions give infinitely many finite groups containing exactly k nonpower subgroups.
Remark 2. Let G be a finite group. If n is coprime to |G|, then G n = G as the map g → g n , while not a homomorphism, is certainly a bijection from G to itself in this case. More generally, G mn = G m for any positive integer m.
The fact that the subgroups of G in this case are the direct products of subgroups of A and B is well-known, but we include the proof for completeness. Suppose H ≤ G and let (a, b) ∈ H. Since |A| and |B| are coprime, the orders r and s of a and b respectively are also coprime. Therefore, there exist integers q and t such that rq + st  Let n be a positive integer. Zhou et al. showed that nps(C n ) = 0. We also note that s(C n ) = ps(C n ) = τ (n), where τ (n) is the number of divisors of n.
Corollary 4. Suppose G = A×C n , where n is a positive integer and A is a finite group whose order is coprime to n. Then nps(G) = τ (n) × nps(A).
Proof. We have that s(C n ) = ps(C n ) = τ (n). Therefore in Equation (2), we have Before the next result we note that if p is an odd prime, then nps(D 2p ) = p. This is because D 2p has exactly p + 3 subgroups; the p cyclic subgroups of order 2 are the nonpower subgroups. The remaining groups (the trivial subgroup, the cyclic subgroup of index 2, and the whole group) are the power subgroups D 2p 2p , D 2 2p and D 1 2p , respectively. For a full description of nonpower subgroups in arbitrary dihedral groups, see Section 3.
Theorem 5. Let k be a positive integer, with k ≥ 3. Then there exists a finite group G with exactly k nonpower subgroups. If k is composite and k > 4, then there are infinitely many such groups.

Proof.
Let k be a positive integer with k ≥ 3. Then either k is divisible by 4, or k is divisible by an odd prime p (or both). Suppose first that k is divisible by an odd prime p. Let q be any odd prime other than p, and let r = k p − 1. Then τ (q r ) = k p . We observe that nps(D 2p ) = p. Therefore, by Corollary 4, we get nps(D 2p × C q r ) = k. On the other hand, if k is divisible by 4, then let r = k 4 − 1, and let q be any prime greater than 3. A quick calculation shows that We note that, in each case, if k > 4 and k is composite, then the exponent r is strictly positive. Therefore, since there are infinitely many choices for q, there are infinitely many finite groups G with exactly k nonpower subgroups.

Special cases.
Notation. For a prime p and a positive integer n, we write C n p for the elementary abelian p-group of finite rank n, and denote the number of subgroups of rank r in C n p by N p (n, r).
Remark. (a) The group G = C n p can be realised as an n-dimensional vector space (say V ) over GF (p). Now, the number of subgroups of rank r in C n p is equal to the number of subspaces of dimension r in V . In the light of Theorem 6 therefore, given any prime p and positive integers n and r, with n > r ≥ 2, we have that ) .
(b) N p (n, 0) = 1 = N p (n, n) for any prime p and natural number n, and for n > 1, Proposition 7. For prime p and positive integers n and r (with n > r ≥ 2), we have: Proof. Setting n = n − 1 and r = r − 1 in Equation (3), we have that .

A question on nonpower subgroups of finite groups
Setting n = n − 1 in Equation (3), we have that which settles the (a) part. For the (b) part, we multiply Equation (5) by p r , add the result to Equation (4) and regroup the terms to get the desired result. 2 The recurrence relations given in Proposition 7 would be a good source for OEIS https://oeis.org/. We now turn to the first main result of this study; see Theorem 8.
Theorem 8. For prime, p and a natural number n > 1,

Proof.
Let p be a prime and n > 1 be an integer. We write G = C n p . For m ∈ N ∪ {0}, This tells us that the only power subgroups of G are the unique subgroups of ranks 0 and n (viz; the two trivial subgroups). That is, nps(G) = s(G) − 2. In particular, the nonpower subgroups of G are the subgroups of ranks 1, 2, . . . , n − 1. Thus, the number of nonpower subgroups of G is ∑ n−1 r=1 N p (n, r). 2 The following result is an immediate consequence of Theorem 8.
Corollary 9. Let n > 1 and p be prime. Then the elementary abelian p-group C n p contains exactly ∑ n−1 r=1 N p (n, r) nonpower subgroups.
In particular, when n = 2, we have the following.
Corollary 10. Let p be prime. The elementary abelian p-group C 2 p contains exactly p + 1 nonpower subgroups.
Definition. A 2-group of maximal class is a group of order 2 n and nilpotency class n − 1 for n ≥ 3.
On the other hand, (i) Let n be odd. Then ⟨x 2m ⟩ is of the form ⟨x v ⟩, where v is a positive divisor of n. Therefore the set of all power subgroups of G is given as Thus ps(G) = τ + 1, and we conclude that nps(G) = (τ + u) − (τ + 1) = u − 1.
(ii) Let n be even. Then ⟨x 2m ⟩ is of the form ⟨x µ ⟩, where µ is an even proper divisor of n. Therefore the set of all power subgroups of G is given as {{1}, G} ∪ {⟨x µ ⟩ | µ is an even proper divisor of n}.
(iii) Let n = 2 ℓ ≥ 4. In the light of (6), the set of power subgroups of G is (iv) Let n = 2p for an odd prime p. In the light of (6), the set of power subgroups of G is where ⟨x 2 ⟩ ∼ = C p . Hence, ps(G) = 3, and we conclude that Corollary 13. Given an integer n ≥ 3, s(D 2 n ) = 2 n + n − 1 and nps(D 2 n ) = 2 n − 1.
Proof. The results follow from a direct application of Theorem 11 and Proposition 12(iii) since the number of positive divisors of 2 n−1 , which is the same as the number of subgroups of D 2 n in ⟨x⟩, is n, and the sum of positive divisors of 2 n−1 , which is the same as the number of subgroups of D 2 n not contained in ⟨x⟩, Definition. For n ≥ 3, we write for the generalised quaternion group of order 2 n .
Definition. For n ≥ 4, we write for the semidihedral group of order 2 n . Remark.

Proof.
Let G be any of the three 2-groups of maximal class and of order 2 n , and let A be a noncyclic proper normal subgroup of G. Clearly, A ̸ ⊂ ⟨x⟩. Let a ∈ A be such that a ∈ {y, xy, . . . , x 2 n−1 −1 y}. Now, suppose G is either dihedral or generalised quaternion. We have that a = x i y for some i ∈ {0, 1, . . . , 2 n−1 − 1}. Using the relation xy = yx −1 , we obtain that xax −1 = x 2 (x i y) = x 2 a. As A is normal in G and a ∈ A, we deduce that (xax −1 )a −1 = x 2 ∈ A. So ⟨x 2 ⟩ ⊆ A. Let G be a semidihedral group. If a = x 2i+1 y for some i ∈ {0, 1, . . . , 2 n−2 − 1}, then using the relation xy = yx 2 n−2 −1 , we obtain that {0, 1, . . . , 2 n−2 − 1}, then using the relation xy = yx 2 n−2 −1 , we obtain that But the order of x 2 n−2 −2 is the same as the order of x 2 ; whence ⟨x 2 n−2 −2 ⟩ = ⟨x 2 ⟩ ⊆ A. In all the cases, we have these three in common: [G : Proposition 15. Let G be any of the three 2-groups of maximal class, and of order 2 n for some n ≥ 4. Given k ∈ {1, 2, . . . , n − 2}, the number of subgroups of order 2 n−k is 2 k + 1.
Proof. Let G = G 2 n be any of the three 2-groups of maximal class, and of order 2 n for some n ≥ 4, and let k ∈ {1, 2, . . . , n − 2} be arbitrary. We show that there are 2 k + 1 subgroups of size 2 n−k . The first case (k = 1) follows from the well-known fact that there are 3 subgroups of index 2 in G; the subgroups of index 2 in G are ⟨x⟩, ⟨x 2 , y⟩ and ⟨x 2 , xy⟩, where ⟨x⟩ ∼ = C 2 n−1 and ⟨x 2 , y⟩ ∼ = G 2 n−1 ∼ = ⟨x 2 , xy⟩.
Let H be a non-trivial subgroup of G. Recall that every non-trivial subgroup of a 2-group is contained in an index 2-subgroup of the group. Let k ∈ {1, 2,. . . , n−2}, and suppose H is a subgroup of size 2 n−k in G. In the light of Lemma 14, H is contained in either ⟨x⟩ or one of the noncyclic subgroups of index 2 in any (noncyclic) subgroup of G which is isomorphic to G 2 n−k+1 . But there are 2 k noncyclic subgroups of index 2 k in G 2 n for any k ∈ {1, 2,. . . , n − 2}, where n ≥ 4. Thus, the subgroups of size 2 n−k (i.e., subgroups of index 2 k ) in G 2 n are the unique cyclic subgroup of size 2 n−k and the 2 k non-cyclic subgroups of index 2 k . Therefore there are 1 + 2 k subgroups of size 2 n−k in G 2 n . 2 Theorem 16. Given an integer n ≥ 3, s(Q 2 n ) = 2 n−1 + n − 1 and nps(Q 2 n ) = 2 n−1 − 1.
Proof. In the light of Proposition 15, the number of subgroups of size 2 k in Q 2 n and D 2 n are equal for each k ∈ {2, 3, . . . , n−1}. As the the number of subgroups of index 2 in both D 8 and Q 8 is 3, one sees immediately that the assertion is also true for both D 8 and Q 8 . The distinction between the number of subgroups of various sizes in Q 2 n and D 2 n (where n ≥ 3) is in the subgroups of size 2. In particular, we have only one subgroup of size 2 in Q 2 n as opposed in D 2 n , where there are 2 n−1 + 1 subgroups of size 2. Thus, For the second part, let m ∈ N ∪ {0} be arbitrary, and G = Q 2 n for n ≥ 3. Firstly, G 4m+1 = ⟨1, x 4m+1 , . . . , x −(4m+1) , y, xy, . . . , x 2 n−1 −1 y⟩. But {1, y, xy, . . . , Therefore, ps(Q 2 n ) = n; whence nps(Q 2 n ) = 2 n−1 + (n − 1) − n = 2 n−1 − 1. Proof. In the light of Proposition 15, the number of subgroups of size 2 k in SD 2 n and D 2 n are equal for each k ∈ {2, 3, . . . , n − 1}. The distinction between the number of subgroups of various sizes in SD 2 n and D 2 n is in the subgroups of size 2. In particular, we have only 2 n−2 + 1 subgroups of size 2 in SD 2 n whilst there are 2 n−1 + 1 subgroups of size 2 in D 2 n . Thus, s(SD 2 n ) =s(D 2 n ) − (2 n−1 + 1) + (2 n−2 + 1) =3(2 n−2 ) + n − 1 (by Corollary 13).
For the second part, let m ∈ N ∪ {0} be arbitrary, and G = SD 2 n for n ≥ 4. Then follows from similar arguments as in the proof of Theorem 16. On the other hand, the results for G 4m and G 4m+2 are also the same with the results for the generalised quaternion cases. Thus, ps(SD 2 n ) = n; whence nps(SD 2 n ) = 3(2 n−2 ) + (n − 1) − n = 3(2 n−2 ) − 1.