Characterizing Finite Groups Using the Sum of the Orders of the Elements

For any nonempty subset X of a finite group G, we let ψ(X) denote the sum of the orders of all elements of X (counting multiplicities). Amiri et al. [1] gave a characterization of cyclic groups using ψ. They showed that if C is a cyclic group and G is a noncyclic group of the same order, then ψ(G) < ψ(C). In 2011, H. Amiri and S. M. J. Amiri [2] turned their attention to the minimum value of ψ(G), and they showed that, for nilpotent groups of a fixed order, theminimum value is attained when each Sylow subgroup has prime exponent. In addition, they showed that if there exists a nonnilpotent group of fixed order n, then the minimum value of ψ(G), as G ranges over all groups of order n, will be attained for a nonnilpotent group. Moreover, this minimum value of ψ on such a nonnilpotent group is strictly smaller than the value of ψ on any nilpotent group of order n. In the same article, H. Amiri and S. M. J. Amiri conjectured that ifH is a simple group and G is a nonsimple group of order |H|, then ψ(H) < ψ(G). However, this conjecture has recently been shown to be false [3]. In 2012, H. Amiri and S. M. J. Amiri [4] showed that


Introduction
For any nonempty subset  of a finite group , we let () denote the sum of the orders of all elements of  (counting multiplicities).Amiri et al. [1] gave a characterization of cyclic groups using .They showed that if  is a cyclic group and  is a noncyclic group of the same order, then () < ().In 2011, H. Amiri and S. M. J. Amiri [2] turned their attention to the minimum value of (), and they showed that, for nilpotent groups of a fixed order, the minimum value is attained when each Sylow subgroup has prime exponent.In addition, they showed that if there exists a nonnilpotent group of fixed order , then the minimum value of (), as  ranges over all groups of order , will be attained for a nonnilpotent group.Moreover, this minimum value of  on such a nonnilpotent group is strictly smaller than the value of  on any nilpotent group of order .In the same article, H. Amiri and S. M. J. Amiri conjectured that if  is a simple group and  is a nonsimple group of order ||, then () < ().However, this conjecture has recently been shown to be false [3].In 2012, H. Amiri and S. M. J. Amiri [4] showed that () < (  ) for all proper subgroups  ̸ =   of   , where   is the symmetric group and   is the alternating group.
In this paper, we give characterizations of various infinite sets of finite groups , given certain restrictions on () or () (or both), where  is a proper subgroup of .More precisely, we prove the following.Theorem 1.Let  be a finite abelian group.Then () divides () for every subgroup  of  if and only if  is cyclic of square-free order.Theorem 2. Let  be a finite abelian group.Then () < || for every proper subgroup  of  if and only if for any prime  and any positive integer .
Theorem 3.For any fixed  > 0, there exist infinitely many finite groups , both abelian and nonabelian that satisfy the conditions in each of the following situations: Theorem 4. Let  be a positive integer.Then Moreover, the number log(3)/ log(2) is the smallest possible constant  such that (  ) ≤ !  for all  ≥ 1.

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International Journal of Combinatorics

Preliminaries and Notation
We assume that all groups in this paper are finite.For an element  in a group , we let () denote the order of  in .For a group , a prime  that divides ||, and a positive integer , we define We let   and   denote, respectively, the symmetric group and the alternating group on the set {1, 2, . . ., }.We let  denote Euler's totient function.
We now state, without proof, two useful results from the literature, the first of which was mentioned previously in Section 1.
Theorem 5 (see [1]).If  is a cyclic group and  is a noncyclic group of the same order, then () < ().
The following lemma, which we also state without proof, is a special case of Theorem A in [2].

Lemma 7.
Let  be a prime and let  be a positive integer.If  is an abelian group with || =   , then Lemma 7 shows that the minimum value of (), as  ranges over all abelian -groups of fixed order   is achieved on the elementary abelian group of that order.Lemma 7 will be useful in proving Theorem 2.
While not difficult to prove, the next proposition does not currently appear in the literature and will be useful throughout this paper.This result determines formulas for () in specials cases.Proposition 8. Let  and  be positive integers, and let  be a prime.
Proof.Let  ≤  be a positive integer.Observe that ∑  =0 (  ) =   .Hence, the number of elements of order   in  is and therefore Thus,

The Proof of Theorem 1
For the sake of convenience, we say that  is a -divisible group if () divides () for all subgroups  of .We can deduce the following result concerning -divisible groups from Lemma 6.

Lemma
which implies that  is a -divisible group.
In light of Lemma 9, it is enough to focus our attention on -groups, where  is a prime.Recall for a given positive integer  and a prime  that divides || that    is the subset of  containing all elements of order   in .Note that (   ) ≡ 0 (mod ) when  ≥ 1.Two obvious, but useful, consequences of this observation are given below.
Lemma 10.Let  be a prime and let  be a -group.Then () ≡ 1 (mod ).
Proof.Let   be the exponent of .Then Corollary 11.Let  be a prime and let  be a -group.If  is a -divisible group and  is any proper subgroup of , then Lemma 12. Let  and  be positive integers, and let  be a prime.Suppose that  ≃ (Z   )  .Then  is a -divisible group if and only if  =  = 1.
Proof.Clearly,  is a -divisible group if  =  = 1, so assume that  ≥ 2. First suppose that  = 1, so that  ≥ 2.Then, using Proposition 8 with  = 1, we have Let  be a subgroup of  such that  ≃ (Z  ) −1 .Then () =   −  + 1 by Proposition 8, and thus since Hence,  is not a -divisible group when  = 1.Now assume that  ≥ 2, and let  be a subgroup of  such that  ≃ (Z  −1 )  .Again, using Proposition 8, we can write where Since  ≥ 2, it follows that  < , and hence ()/() ∉ Z.Thus,  is not a -divisible group and the proof is complete.
Lemma 13.Let  be a prime and suppose that  ≃ Z  × , where ) Proof.Let (, ) ∈ Z  × , and define Then Z  ×  =  ∪  ∪  and Thus since where Proof.Note that   ≥ 2 and  ≥ 2 since   are not all equal.Suppose that and define and observe that Thus, we may assume without loss of generality that  = .Let  be a subgroup of  such that Since () = ∑   −1 =0 (   ), we have that from which the lemma follows.
We are now in a position to establish Theorem 1.For the convenience of the reader, we restate Theorem 1 below as Theorem 15 using the -divisible terminology.
Theorem 15.Let  be an abelian group.Then  is a -divisible group if and only if  is cyclic of square-free order.

International Journal of Combinatorics
Proof.If  is cyclic of square-free order, then the fact that  is a -divisible group follows immediately from Lemma 6.So, now assume conversely that  is a -divisible group.To establish the theorem, it is enough, by Lemmas 6, 12, and 13, to show for a prime  that where  ≥ 2 with 2 ≤  1 ≤  2 ≤ ⋅ ⋅ ⋅ ≤   not all equal, is not a -divisible group.Let  be as in (29), and assume by way of contradiction that  is a -divisible group.Let  be a subgroup of  with Note that (    ) = (    ).Using this observation, Lemma 14, and Corollary 11, it follows that Using the fact that  > 0 to solve the inequality we get that 1 <  < 2, which is impossible, and the proof is complete.
The following is immediate from Theorem 15.
Corollary 16.If  is an abelian -divisible group, then every proper subgroup of  is a -divisible group.
Remark 17.The converse of Corollary 16 is false.For example, let  = Z  × Z  , where  is a prime.Then every proper subgroup of  is a -divisible group, but  is not a -divisible group by Lemma 12.

A Remark on Nonabelian 𝜓-Divisible
Groups.The nonabelian situation is much less clear than the abelian case.
In fact, we have been unable to find a single nonabelian -divisible group, although we are somewhat hesitant to conjecture their nonexistence.On the other hand, using Lemma 9, we are able to construct infinitely many nonabelian groups that are not -divisible.Since ( 3 ) = 13, we see that  3 is not -divisible.Then, by Lemma 9, it follows that  ≃ Z  ×  3 is not -divisible for any prime .

The Proof of Theorem 2
We say a group  is a B  -group if () < || for all proper subgroups  of .For the convenience of the reader, we use this new terminology to restate Theorem 2 below as Theorem 18.
Theorem 18.Let  be a finite abelian group.Then  is a B group if and only if for any prime  and any positive integer .
Proof.Using Proposition 8, it is easy to show that if for any prime  and any positive integer , then  is a B group.So, assume conversely that  is an abelian B  -group, and write where and  1 <  2 < ⋅ ⋅ ⋅ <   are primes.Assume first that  ≥ 2, and let  be a subgroup of  such that where . We claim that () > ||.To see this, we have which shows that  is not a B  -group if  ≥ 2. Hence, assume that  = 1 so that  is a -group for some prime .Suppose that || =   and write where Then Thus  = 1 and  ≃ (Z  )  , where  ≥ 2. Now suppose that  = 1 so that  ≃ Z   .Let  be the subgroup of  with  ≃ Z  −1 .Then, using Proposition 8, we have which implies that  + 1 >  −1 .Hence,  ≤ 2 and  ≃ Z  or  ≃ Z  2 , which completes the proof of the theorem.

Investigating the Range of 𝜓
It is easy to get bounds on () as the next proposition shows.
Proposition 19.Let  be a finite group.Then with equality holding in both inequalities in (43) if and only if  is trivial.
Proof.For any  ∈ , we have that 1 ≤ () ≤ ||.Thus, the proposition follows from the definition of .
A natural question to ask is whether there exist nontrivial groups  with () arbitrarily close to the bounds given in (43).Theorem 3 provides an affirmative answer to this question.

The Proof of Theorem 3
Proof of Theorem 3. To prove part (1) of Theorem 3, first let  ≃  3 × (Z  )  , where  ≥ 13 is a prime and  is a positive integer.Then, using Lemma 6, and Proposition 8 with  = 1, we have that Solving the equation for  gives  = log (13 ( +1 −  + 1)) Since it follows that  > 1.Then, since  ≥ 13, we have A similar computation gives the same conclusion when  ≃ (Z  )  , and the proof of part (1) of Theorem 3 is complete.
To prove part (2) of Theorem 3, first let  ≃  3 × Z   , where  ≥ 13 is prime and  is a positive integer.Then, using Lemma 6, and Proposition 8 with  = 1, we have that Solving the equation for  gives  = log (13 (( 2+1 + 1) / ( + 1))) Since it follows that  < 2. Also, since  2 >  + 1, we see that Thus, International Journal of Combinatorics Therefore, A similar computation yields the same conclusion when  ≃ Z   , which completes the proof of Theorem 3.

The Proof of Theorem 4.
It is clear that estimating (  ) is, up to scaling, equivalent to estimating the average order   := (  )/! of an element in   .This problem was first raised by Erdös and Turán [5] in 1968 when they conjectured that log Conjecture (56) was proven by Schmutz [6] where The proof of (57) is nontrivial and requires some very technical results concerning partitions.In 1991, Goh and Schmutz [7] sharpened (57) by proving that where The proof of (60) also requires some partition theory.
In the spirit of Theorem 3, it is our goal here to determine, if possible, the smallest possible constant  such that in a nontrivial manner.In other words, does there exist a smallest real number  < 2 for which (62) is true, and can we determine it?Theorem 4 provides an affirmative answer to these questions.There are two advantages to the bound given for (  ) in Theorem 4 over the results in [6,7].First of all, Theorem 4 is proven for all  ≥ 1, not just for sufficiently large values of .The second advantage is that the proof of Theorem 4 that we present here does not rely on any results from partition theory.We use the following result, due to Massias [8] in 1984.
Theorem 20.Let   denote the largest order of an element in   .Then In 1903, Landau [9] proved that log(  ) ∼ √ log(), but Theorem 20 represents the first explicit upper bound for log(  ).Although somewhat technical, Massias' proof of Theorem 20 uses elementary analytic techniques requiring little more than calculus.
We also need the following lemmas.
Proof.Observe that if  fixes both  and , then So, suppose that  fixes , but not .Then we may write for some  with 2 ≤  ≤  − 1, where  1 and  2 are disjoint, and   ̸ =  for all .Thus, which completes the proof.
Lemma 23.If  ≥ 20 is an integer, then Proof.The proof is straightforward using calculus.
Remark 26.The sequence {(  )} ∞ =1 is listed in the On-Line Encyclopedia of Integer Sequences and can be found at http://oeis.org/A060014.We end with the following conjecture, which is supported by numerical evidence.Moreover, the number log(7)/ log(3) is the smallest possible constant  such that (  ) ≤ (!/2)  for all  ≥ 1.
Remark 28.The sequence {(  )} ∞ =1 is listed in the On-Line Encyclopedia of Integer Sequences and can be found at http://oeis.org/A060015.

Conjecture 27 .
Let  be a positive integer.Then  (  ) Suppose now that both  and  are -divisible, and let  be a subgroup of .Since gcd(||, ||) = 1, it follows that  ≃  × , where  is a subgroup of , and  is a subgroup of .Then, by Lemma 6, we have that